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82 Principles and first examples of sieve methods
3.2 The Selberg lambda-squared method
Let /Lambda1 n be a real-valued arithmetic function such that /Lambda1 1 = 1. Then
(∑
d |n
/Lambda1 d
)2
≥
{ 1i f n = 1,
0i f n > 1.
This simple observation can be used to obtain an upper bound for S(x ,y; P );
namely
S(x ,y; P ) ≤
∑
x <n≤x +y
⎛
⎜
⎝
∑
d |n
d | P
/Lambda1 d
⎞
⎟
⎠
2
=
∑
d | P
e| P
/Lambda1 d /Lambda1 e
∑
x <n≤x +y
d |n,e|n
1
=
∑
d | P
e| P
/Lambda1 d /Lambda1 e
([x + y
[d ,e]
]
−
[ x
[d ,e]
])
= y
∑
d | P
e| P
/Lambda1 d /Lambda1 e
[d ,e] + O
⎛
⎝
(∑
d | P
|/Lambda1 d |
)2 ⎞
⎠. (3.10)
In the general framework of the preceding section this amounts to taking
λ+
n=
∑
d ,e
[d ,e]=n
/Lambda1 d /Lambda1 e ,
since it then follows that
∑
d |n
λ+
d=
(∑
d |n
/Lambda1 d
)2
.
W e now suppose that /Lambda1 n = 0 for n > z where z is a parameter at our disposal, in
the hope that this will restrict the size of the error term. As for the main term, we
see that we wish to minimize a quadratic form subject to the constraint/Lambda1 1 = 1.
In fact we can diagonalize this quadratic form and determine the optimal /Lambda1 n
exactly; this permits us to prove
Theorem 3.2Let x , y, and z be real numbers such that y > 0 and z ≥ 1.F o r
any positive integer P we have
S(x ,y; P ) ≤ y
L P (z) + O (z2 L P (z)−2 ) | {
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3.2 The Selberg lambda-squared method 83
where
L P (z) =
∑
n≤z
n| P
µ(n)2
ϕ(n) .
Proof Clearly we may assume that P is square-free. Since [ d ,e](d ,e) = de
and ∑
d |n ϕ(d ) = n, we see that
1
[d ,e] = (d ,e)
de = 1
de
∑
f |d ,f |e
ϕ( f ).
Hence
∑
d | P,e| P
/Lambda1 d /Lambda1 e
[d ,e] =
∑
f | P
ϕ( f )
∑
d
f |d | P
/Lambda1 d
d
∑
e
f |e| P
/Lambda1 e
e
=
∑
f | P
ϕ( f ) y2
f
where
yf =
∑
d
f |d | P
/Lambda1 d
d . (3.11)
This linear change of variables, from /Lambda1 d to yf , is non-singular. That is, if the yf
are given then there exist unique /Lambda1 d such that the above holds. Indeed, by a form
of the M ¨ obius inversion formula (cf. Exercise 2.1.6) the above is equivalent to
the relation
/Lambda1 d = d
∑
f
d | f | P
yf µ( f /d ). (3.12)
Moreover, from these formulæ we see that /Lambda1 d = 0 for all d > z if and only if
yf = 0 for all f > z. Thus we have diagonalized the quadratic form in (3.10),
and by (3.12) we see that the constraint /Lambda1 1 = 1 is equivalent to the linear
condition
∑
f | P
yf µ( f ) = 1. (3.13)
W e determine the value of the constrained minimum by completing squares. If
theyf satisfy (3.13), then
∑
f | P
ϕ( f ) y2
f =
∑
f | P
f ≤z
ϕ( f )
(
yf − µ( f )
ϕ( f )L P (z)
)2
+ 1
L P (z) . (3.14) | {
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84 Principles and first examples of sieve methods
Here the right-hand side is minimized by taking
yf = µ( f )
ϕ( f )L P (z) (3.15)
for f ≤ z, and we note that these yf satisfy (3.13). Hence the minimum of the
quadratic form in (3.10), subject to /Lambda1 1 = 1, is precisely 1 /L P (z); this gives the
main term.
W e now treat the error term. Since P is square-free, from (3.12) and (3.15)
we see that
/Lambda1 d = d
L P (z)
∑
f
d | f | P
f ≤z
µ( f )µ( f /d )
ϕ( f ) = d µ(d )
L P (z)ϕ(d )
∑
m| P
(m,d )=1
m≤z/d
µ(m)2
ϕ(m) ; (3.16)
here we have put m = f /d . Thus
∑
d ≤z
|/Lambda1 d |≤ 1
L P (z)
∑
d ≤z
d
ϕ(d )
∑
m≤z/d
1
ϕ(m) = 1
L P (z)
∑
m≤z
1
ϕ(m)
∑
d ≤z/m
d
ϕ(d ) .
Since d /ϕ(d ) = ∑
r |d µ2 (r )/ϕ(r ), it follows by the method of Section 2.1 that
∑
d ≤y
d
ϕ(d ) =
∑
r ≤y
µ2 (r )
ϕ(r ) [ y/r ] ≤ y
∑
r
µ2 (r )
r ϕ(r ) ≪ y.
On inserting this in our former estimate, we find that
∑
d ≤z
|/Lambda1 d |≪ z
L P (z)
∑
m≤z
1
mϕ(m) ≪ z
L P (z) . (3.17)
This gives the stated error term, so the proof is complete. □
In order to apply Theorem 3.2, we require a lower bound for the sum L P (z).
T o this end we show that
∑
n≤z
µ(n)2
ϕ(n) > log z (3.18)
for all z ≥ 1. Let s(n) denote the largest square-free number dividing n (some-
times called the ‘square-free kernel of n’). Then for square-free n,
1
ϕ(n) = 1
n
∏
p|n
(
1 + 1
p + 1
p2 +···
)
=
∑
m
s(m)=n
1
m ,
so that the sum in (3.18) is
∑
m
s(m)≤z
1
m . | {
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3.2 The Selberg lambda-squared method 85
Since s(m) ≤ m, this latter sum is
≥
∑
m≤z
1
m > log z.
Here the last inequality is obtained by the integral test. With more work one can
derive an asymptotic formula for the the sum in (3.18) (recall Exercise 2.1.17).
By taking z = y1/2 in Theorem 3.2, and appealing to (3.18), we obtain
Theorem 3.3 Let P = ∏
p≤√
y p. Then for any x and any y ≥ 2,
S(x ,y; P ) ≤ 2 y
log y
(
1 + O
(1
log y
))
.
By combining the above with (3.3) we obtain an immediate application to
the distribution of prime numbers.
Corollary 3.4F or any x ≥ 0 and any y ≥ 2,
π(x + y) − π(x ) ≤ 2 y
log y
(
1 + O
(1
log y
))
.
In Theorem 3.3 we consider only a very special sort of P , but the following
lemma enables us to obtain corresponding results for more general P .
Lemma 3.5 Put M ( y; P ) = maxx S(x ,y; P ).I f ( P,q ) = 1, then
M ( y; P ) ≤ q
ϕ(q ) M ( y; qP ).
Proof It suffices to show that
ϕ(q )S(x ,y; P ) =
q∑
m=1
S(x + Pm ,y; qP ), (3.19)
since the right-hand side is bounded above by qM ( y; qP ). Suppose that x +
Pm < n ≤ x + Pm + y and that ( n,qP ) = 1. Put r = n − Pm . Then x <
r ≤ x + y,( r,P ) = 1, and ( r + Pm ,q ) = 1. Thus the right-hand side above is
∑
m
∑
r
1 =
∑
x <r ≤x +y
(r,P )=1
∑
1≤m≤q
(r + Pm ,q )=1
1.
Since ( P,q ) = 1, the map m ↦→r + Pm permutes the residue classes (mod q ).
Hence the inner sum above is ϕ(q ), and we have (3.19). □
Theorem 3.6 F or any real x and any y ≥ 2,
S(x ,y; P ) ≤ eC0 y
⎛
⎜
⎜
⎝
∏
p| P
p≤√
y
(
1 − 1
p
)
⎞
⎟
⎟
⎠
(
1 + O
(1
log y
))
. | {
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86 Principles and first examples of sieve methods
Proof Let
P1 =
∏
p| P
p≤√y
p, q1 =
∏
p∤ P
p≤√y
p.
Theorem 3.3 provides an upper bound for M ( y; q1 P1 ), and hence by Lemma
3.5 we have an upper bound for M ( y; P1 ). T o complete the argument it suffices
to note that S(x ,y; P ) ≤ S(x ,y; P1 ) ≤ M ( y; P1 ), and to appeal to Mertens’
formula (Theorem 2.7(e)). □
W e note that Theorem 3.3 is a special case of Theorem 3.6. Although we have
taken great care to derive uniform estimates, for many purposes it is enough to
know that
S(x ,y; P ) ≪ y
∏
p| P
p≤y
(
1 − 1
p
)
. (3.20)
This follows from Theorem 3.6 since ∏
√y<p≤y (1 − 1/p)−1 ≪ 1 by Mertens’
formula. T o obtain an estimate in the opposite direction, write P = P1 q1 where
P1 is composed entirely of primes > y, and q1 is composed entirely of primes
≤ y. Since the integers in the interval (0 ,y] have no prime factor > y, we see
that M ( y; P1 ) ≥ [ y] . Hence by Lemma 3.5,
M ( y; P ) ≥ [ y]
∏
p| P
p≤y
(
1 − 1
p
)
. (3.21)
Thus the bound (3.20) is of the correct order of magnitude.
The advantage of Theorem 3.6 lies in its uniformity . On the other hand, the
use of Lemma 3.5 is wasteful if the P in Theorem 3.6 is much smaller than in
Theorem 3.3. For example, if P = ∏
p≤y1/4 p, then by Theorem 3.6 we find that
S(x ,y; P ) ≤ cy
log y
(
1 + O
(1
log y
))
with c = 4, whereas by Theorem 3.2 with z = y1/2 we obtain the above with
the better constant
c = 4
3 − 2 log 2 = 2.4787668 ....
T o see this, we note that
L P (z) =
∑
n≤z
µ(n)2
ϕ(n) −
∑
z1/2 <p≤z
1
p − 1
∑
n≤z/p
µ(n)2
ϕ(n) . (3.22) | {
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3.2 The Selberg lambda-squared method 87
Then by Exercise 2.1.17 and Mertens’ estimates (Theorem 2.7) it follows that
this is1
4 (3 − 2 log 2) log y + O (1).
3.2.1 Exercises
1. Let /Lambda1 d be defined as in the proof of Theorem 3.2.
(a) Show that
/Lambda1 d ≪ d
L P (z)ϕ(d ) log 2z
d
for d ≤ z.
(b) Use the above to give a second proof of (3.17).
2. Show that for y ≥ 2 the number of prime powers pk in the interval
(x ,x + y]i s
≤ 2 y
log y
(
1 + O
(1
log y
))
.
3. (Chowla 1932) Let f (n) be an arithmetic function, put
g(n) =
∑
[d ,e]=n
f (d ) f (e),
and let σc denote the abscissa of convergence of the Dirichlet series∑ g(n)n−s .
(a) Show that if σ> max(1,σc ), then
ζ(s)
∑
d ,e
f (d ) f (e)
[d ,e]s =
∞∑
n=1
⏐
⏐
⏐
⏐
∑
d |n
f (d )
⏐
⏐
⏐
⏐
2
n−s .
(b) Show that
∑
d ,e
µ(d )µ(e)
[d ,e]2 = 6
π2 .
(c) Show that
∑
d ,e
[d ,e]=n
µ(d )µ(e) = µ(n)
for all positive integers n.
4. Let f (n) be an arithmetic function such that f (1) = 1. Show that f is
multiplicative if and only if f (m) f (n) = f ((m,n)) f ([m,n]) for all pairs
of positive integers m, n. | {
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88 Principles and first examples of sieve methods
5. (Hensley 1978)
(a) Let P = ∏
p≤√y p. Show that the number of n, x < n ≤ x + y, such
that /Omega1 (n) = 2, is
≤ S(x ,y; P ) +
∑
p≤√y
(
π
(x + y
p
)
− π
(x
p
))
.
(b) By using Theorem 3.3 and Corollary 3.4, show that for y ≥ 2,
∑
x <n≤x +y
/Omega1 (n)=2
1 ≤ 2 y log log y
log y
(
1 + O
( 1
log log y
))
.
6. (H.-E. Richert, unpublished)
(a) Show that
∑
x <n≤x +y
(∑
d 2 |n
/Lambda1 d
)2
= y
∑
d ,e
/Lambda1 d /Lambda1 e
[d ,e]2 + O
⎛
⎝
(∑
d
|/Lambda1 d |
)2 ⎞
⎠.
(b) Let f (n) = n2 ∏
p|n (1 − p−2 ). Show that ∑
d |n f (d ) = n2 .
(c) For 1 ≤ d ≤ z let /Lambda1 d be real numbers such that /Lambda1 1 = 1. Show that the
minimum of ∑
d ,e /Lambda1 d /Lambda1 e /[d ,e]2 is 1 /L where L = ∑
n≤z µ(n)2 /f (n).
Show also that /Lambda1 d ≪ 1 for the extremal /Lambda1 d .
(d) Show that ζ(2) − 1/z ≤ L ≤ ζ(2).
(e) Let Q(x ) denote the number of square-free numbers not exceeding x .
Show that for x ≥ 0, y ≥ 1,
Q(x + y) − Q(x ) ≤ y
ζ(2) + O
(
y2/3 )
.
7. Let m( y; P ) = minx S(x ,y; P ). Show that if ( q ,P ) = 1, then
m( y; P ) ≥ q
ϕ(q ) m( y; qP ).
8. (N. G. de Bruijn, unpublished; cf. van Lint & Richert 1964) Let M be an
arbitrary set of natural numbers, and let s(n) denote the largest square-free
divisor of n. Show that
0 ≤
∑
n≤x
n∈M
µ(n)2
ϕ(n) −
∑
n≤x
s(n)∈M
1
n ≤
∑
n≤x
µ(n)2
ϕ(n) −
∑
n≤x
1
n ≪ 1.
9. (van Lint & Richert 1965)
(a) Show that
∑
n≤z
µ(n)2
ϕ(n) ≤
(∑
d |q
µ(d )2
ϕ(d )
)⎛
⎜
⎝
∑
m≤z
(m,q )=1
µ(m)2
ϕ(m)
⎞
⎟
⎠. | {
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3.3 Sifting an arithmetic progression 89
(b) Deduce that
∑
n≤z
(n,q )=1
µ(n)2
ϕ(n) ≥ ϕ(q )
q
∑
n≤z
µ(n)2
ϕ(n) .
10. (Hooley 1972; Montgomery & V aughan 1979)
(a) Let λ+
d be an upper bound sifting function such that λ+
d= 0 for all
d > z. Show that for any q ,
0 ≤ ϕ(q )
q
∑
d
(d ,q )=1
λ+
d
d ≤
∑
d
λ+
d
d .
(Hint: Multiply both sides by P /ϕ( P ) = ∑ 1/m where m runs over
all integers composed of the primes dividing P , and P = ∏
p≤z p.)
(b) Let /Lambda1 d be real with /Lambda1 d = 0 for d > z. Show that for any q ,
0 ≤ ϕ(q )
q
∑
d ,e
(de ,q )=1
/Lambda1 d /Lambda1 e
[d ,e] ≤
∑
d ,e
/Lambda1 d /Lambda1 e
[d ,e] .
(c) Let λ−
dbe a lower bound sifting function such that λ−
d= 0 for d > z.
Show that for any q ,
ϕ(q )
q
∑
d
(d ,q )=1
λ−
d
d ≥
∑
d
λ−
d
d .
3.3 Sifting an arithmetic progression
Thus far we have sifted only the zero residue class from a set of consecutive
integers. W e now widen the situation slightly .
Lemma 3.7Let P be a positive integer , and for each prime p dividing P
suppose that one particular residue class a p has been chosen. Let S ′(x ,y; P )
denote the number of integers m, x < m ≤ x + y, such that for each p | P,
m ̸≡a p (mod p). Then
max
x
S′(x ,y; P ) = max
x
S(x ,y; P ).
Since S′(x ,y; P ) reduces to S(x ,y; P ) when we take a p = 0 for all p| P ,
we see that there is no loss of generality in sifting only the zero residue class,
when the initial set of numbers consists of consecutive integers. Also, we note
that the value of the maximum taken above is independent of the choice of the
a p . | {
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90 Principles and first examples of sieve methods
Proof By the Chinese remainder theorem there is a number c such that c ≡ a p
(mod p) for every p| P . Put n = m − c. Thus the inequality x < m ≤ x + y is
equivalent to x − c < n ≤ x − c + y, and the condition that p| P implies m ̸≡
a p (mod p) is equivalent to ( n,P ) = 1. Hence S′(x ,y; P ) = S(x − c,y; P ),
so that
max
x
S′(x ,y; P ) = max
x
S(x − c,y; P ) = max
x
S(x ,y; P ),
and the proof is complete. □
Theorem 3.8 Suppose that (a,q ) = 1, that ( P,q ) = 1, and that x and y are
real numbers with y ≥ 2q . The number of n, x < n ≤ x + y, such that n ≡ a
(mod q ) and (n,P ) = 1 is
≤ eC0 y
q
⎛
⎜
⎜
⎝
∏
p| P
p≤√
y/q
(
1 − 1
p
)
⎞
⎟
⎟
⎠
(
1 + O
( 1
log y/q
))
.
Proof Write n = mq + a, so that x ′ < m ≤ x ′ + y′ where x ′ = (x − a)/q
and y′ = y/q . For each p| P let a p be the unique residue class (mod p) such
that a p q + a ≡ 0 (mod p). Thus p|n if and only if m ≡ a p (mod p). Hence
the number of n in question is S′(x ′,y′; P ), in the language of Lemma 3.7. The
stated bound now follows from this lemma and Theorem 3.6. □
Using the estimate above, we generalize Corollary 3.4 to arithmetic progres-
sions. W e let π(x ; q ,a) denote the number of prime numbers p ≤ x such that
p ≡ a(mod q ).
Theorem 3.9 (Brun–Titchmarsh) Let a and q be integers with (a,q ) = 1, and
let x and y be real numbers with x ≥ 0 and y ≥ 2q . Then
π(x + y; q ,a) − π(x ; q ,a) ≤ 2 y
ϕ(q ) log y/q
(
1 + O
( 1
log y/q
))
. (3.23)
Proof Ta k e P to be the product of those primes p ≤ √y/q such that p∤q .
Then
∏
p| P
(
1 − 1
p
)
=
∏
p|q
p≤√y/q
(
1 − 1
p
)−1 ∏
p≤√y/q
(
1 − 1
p
)
≤
∏
p|q
(
1 − 1
p
)−1 ∏
p≤√y/q
(
1 − 1
p
)
.
By Mertens’ estimate this is
= q
ϕ(q ) · 2e−C0
log y/q
(
1 + O
( 1
log y/q
))
. | {
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3.4 T win primes 91
Thus by Theorem 3.8, the number of primes p, x < p ≤ x + y, such that p ≡ a
(mod q ) and ( p,P ) = 1 satisfies the bound (3.23). T o complete the proof it
remains to note that the number of primes p, x < p ≤ x + y, such that p ≡ a
(mod q ) and p| P is at most ω( P ) ≤ √y/q , which can be absorbed in the error
term in (3.23). □
3.4 T win primes
Thus far we have removed at most one residue class per prime. More generally ,
we might wish to delete from an interval (x ,x + y] those numbers n that lie
in a certain set B( p) of ‘bad’ residue classes modulo p. Let b( p) = card B( p)
denote the number of residue classes to be removed, for p| P where P i sag i v e n
square-free number, and set
a(n) =
∏
p| P
n∈B( p) (mod p)
p .
Thus the n that remain after sifting are precisely the n for which ( a(n),P ) = 1.
By the sieve we obtain upper and lower bounds for the number of remaining n
of the form
∑
x <n≤x +y
∑
m|(a(n),P )
λm =
∑
m| P
λm
∑
x <n≤x +y
m|a(n)
1 . (3.24)
Now p|a(n) if and only if n ∈ B( p) (mod p). By the Chinese remainder theo-
rem, this will be the case for all p|m when n lies in one of precisely ∏
p|m b( p)
residue classes modulo m. The b( p) are defined only for primes, but it is con-
venient now to extend the definition to all positive integers by putting
b(m) =
∏
pα∥ m
b( p)α .
Thus b(m) is the totally multiplicative function generated by the b( p). For
square-free m, b(m) represents the number of deleted residue classes modulo
m. W e are now in a position to estimate the inner sum above. W e partition the
interval ( x ,x + y] into [ y/m] intervals of length m, and one interval of length
{y/m}m. In each interval of length m there are precisely b(m) values of n for
which m|a(n). In the final shorter interval, the number of such n lies between
0 and b(m). Thus the inner sum on the right above is = yb (m)/m + O (b(m)),
and hence the expression (3.24) is
= y
∑
m| P
b(m)λm
m + O
(∑
m| P
b(m)|λm |
)
. (3.25) | {
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92 Principles and first examples of sieve methods
T o continue from this point, one should specify the choice of λm , and then
estimate the main term and error term. In the context of Selberg’s /Lambda1 2 method,
we have real /Lambda1 d with /Lambda1 1 and /Lambda1 d = 0 for d > z. The number of n ∈ (x ,x + y]
that survive sifting is
≤
∑
x <n≤x +y
(∑
d |(a(n),P )
/Lambda1 d
)2
=
∑
d | P
∑
e| P
/Lambda1 d /Lambda1 e
∑
x <n≤x +y
[d ,e]|a(n)
1
= y
∑
d | P
∑
e| P
b([d ,e])
[d ,e] /Lambda1 d /Lambda1 e + O
(∑
d | P
∑
e| P
g([d ,e])|/Lambda1 d /Lambda1 e |
)
. (3.26)
This is (3.25) with λm = ∑
[d ,e]=m /Lambda1 d /Lambda1 e .
W e consider first the main term above. Clearly [ d ,e] = de /(d ,e) and
b([d ,e]) = b(d )b(e)/b((d ,e)). For square-free m put
g(m) =
∏
p|m
b( p)
p − b( p) . (3.27)
Here we have 0 in the denominator if there is a prime p for which b( p) = p.
However, in that case all residues modulo p are removed, and no integer survives
sifting. Thus we may confine our attention to b( p) such that b( p) < p for all
p.I f m is square-free, then
∑
d |m
1
g(d ) =
∏
p|m
(
1 + p − b( p)
b( p)
)
= m
b(m) .
By applying this with m = (d ,e) we see that the first sum in (3.26) is
∑
d | P
e| P
b(d )/Lambda1 d
d · b(e)/Lambda1 e
e · (d ,e)
b((d ,e)) =
∑
d | P
e| P
b(d )/Lambda1 d
d · b(e)/Lambda1 e
e
∑
f |d
f |e
1
g( f )
=
∑
f | P
1
g( f )
∑
d
f |d | P
b(d )
d /Lambda1 d
∑
e
f |e| P
b(e)
e /Lambda1 e
=
∑
f | P
1
g( f ) y2
f (3.28)
where
yf =
∑
d
f |d | P
b(d )
d /Lambda1 d . (3.29) | {
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3.4 T win primes 93
The linear change of variables from /Lambda1 d to yf is invertible:
/Lambda1 d = d
b(d )
∑
f
d | f | P
yf µ( f /d ) . (3.30)
By the above formulæ we see that the condition that /Lambda1 d = 0 for d > z is
equivalent to the condition that yf = 0 for f > z. Also, the condition that
/Lambda1 1 = 1 is equivalent to
∑
f | P
yf µ( f ) = 1. (3.31)
For such yf we see that
∑
f | P
1
g( f ) y2
f =
∑
f | P
f ≤z
1
g( f )
(
yf − µ( f )g( f )/L
)2
+ 1
L (3.32)
where
L =
∑
f ≤z
f | P
µ( f )2 g( f ) . (3.33)
Thus our main term is minimized by taking
y f =
{µ( f )g( f )/L ( f ≤ z),
0 (otherwise) , (3.34)
and we note that these yf satisfy (3.31). The size of L depends on P , z, and the
b( p). In the case of twin primes we obtain the following estimate.
Theorem 3.10 Let P = ∏
p≤√y p where y ≥ 4. The number of integers n ∈
(x ,x + y], such that (n,P ) = (n + 2,P ) = 1 does not exceed
8cy
(log y)2
(
1 + O
(log log y
log y
))
where
c = 2
∏
p>2
(
1 − 1
( p − 1)2
)
.
The number of primes p ∈ (x ,x + y] for which p| P is ≤ π(√y). Likewise,
the number of primes p ∈ (x ,x + y] for which p + 2 is prime and ( p + 2)| P
is ≤ π(√y). Otherwise, if p ∈ (x ,x + y] and p + 2 is prime, then ( p,P ) =
( p + 2,P ) = 1; the number of such p is bounded by the above. Since π(√y)
is negligible by comparison, the above bound applies also to the number of
primesp ∈ (x ,x + y] for which p + 2 is prime. | {
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94 Principles and first examples of sieve methods
Proof W e first estimate L as given in (3.33). W e have b(2) = 1 and b( p) = 2
for p > 2. Since µ(m)2 g(m) is a multiplicative function that takes the value
2/( p − 2) when m = p > 2, and since d (n)/n is a multiplicative function that
takes the value 2 /p when n = p, we expect that d (n)/n and µ(m)2 g(m) are
‘close’ in the sense that we can obtain the latter function by convolving d (n)/n
with a fairly tame function c(k). On comparing the Euler products of the re-
spective Dirichlet series generating functions, we see that if the c(k) are defined
so that
∞∑
k=1
c(k)k−s = (1 + 2−s )(1 − 2−s−1 )2 ∏
p>2
(
1 + 2
( p − 2) ps
)(
1 − 1
ps+1
)2
,
(3.35)
then
µ(m)2 g(m) =
∑
k,n
kn =m
c(k)d (n)/n.
Hence
L =
∑
m≤z
µ(m)2 g(m) =
∑
k≤z
c(k)
∑
n≤z/k
d (n)/n.
By Theorem 2.3 and (Riemann–Stieltjes) integration by parts we see that
N∑
n=1
d (n)
n = 1
2 (log N )2 + O (log N ).
Hence
L =
∑
k≤z
c(k)((log z/k)2 /2 + O (log z))
= 1
2 (log z)2 ∑
k≤z
c(k) + O
(
(log z)
∑
k
|c(k)| log 2 k
)
+ O
(∑
k
|c(k)|(log k)2
)
.
The Euler product in (3.35) is absolutely convergent for σ> −1/2. Hence∑ |c(k)|k−σ < ∞ for σ> −1/2. Thus the two sums in the error terms above
are convergent. Also,
∑
k>z
|c(k)|≤ 1
log z
∞∑
k=1
|c(k)| log k ≪ 1
log z .
Thus by taking s = 0 in (3.35) we find that
L = 1
2c (log z)2 + O (log z). (3.36) | {
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3.4 T win primes 95
It remains to bound the error term in (3.26). Since 0 ≤ b([d ,e]) ≤ b(d )b(e),
the error term is
≪
(∑
d ≤z
b(d )|/Lambda1 d |
)2
.
From (3.30) and (3.34) we see that
/Lambda1 d = d
b(d )L
∑
f ≤z
d | f
µ( f )g( f )µ( f /d ) = µ(d )dg (d )
b(d )L
∑
m≤z/d
(m,d )=1
µ(m)2 g(m) .
Hence
∑
d ≤z
b(d )|/Lambda1 d |≪ 1
L
∑
d ≤z
µ(d )2 dg (d )
∑
m≤z/d
µ(m)2 g(m)
= 1
L
∑
m≤z
µ(m)2 g(m)
∑
d ≤z/m
µ(d )2 dg (d ) .
By Corollary 2.15 we see that
∑
d ≤D
µ(d )2 dg (d ) ≪ D
log D
∏
p≤D
(1 + g( p))
≪ D
log D
∏
p≤D
(
1 − 1
p
)−2
≪ D log D .
Since L ≍ (log z)2 , it follows that
∑
d ≤z
b(d )|/Lambda1 d |≪ z
log z
∑
m≤z
µ(m)2 g(m)/m ≪ z
log z .
On combining our estimates, we see that the number of n, x < n ≤ x + y, such
that ( a(n),P ) = 1i s
≤ 2cy
(log z)2 + O
( y
(log z)3
)
+ O
( z2
(log z)2
)
.
In order that the last error term is majorized by the one before it, we take
z = ( y/log y)1/2 . Then
log z = 1
2 log y + O (log log y),
so we obtain the stated result. □
Corollary 3.11 (Brun) Let ∑ ∗
p denote a sum over those primes p for which
p + 2 is prime. Then ∑ ∗
p 1/p converges. | {
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96 Principles and first examples of sieve methods
Proof The number of twin primes for which 2 k−1 < p ≤ 2k is ≪ 2k /k2 .
Hence the contribution of such primes to the sum in question is ≪ 1/k2 . But∑ 1/k2 < ∞, so we obtain the stated result. □
Let r be an even non-zero integer. T o bound the number of primes p for
which p + r is also prime, it suffices to establish the following monotonicity
principle, which is a natural generalization of Lemma 3.5.
Lemma 3.12F or each prime p let B( p) be the union of b ( p) arithmetic
progressions with common difference p. Put B = ⋃
p| P B( p), and set
M (x ,y; b) = max
B
∑
x <n≤x +y
n /∈B
1
where the maximum is over all choices of the B( p) with b ( p) fixed. If 0 ≤
b1 ( p) ≤ b2 ( p) < p for all p, then
M (x ,y; b1 )
∏
p| P
(
1 − b1 ( p)
p
)−1
≤ M (x ,y; b2 )
∏
p| P
(
1 − b2 ( p)
p
)−1
.
Proof W e induct on ∑
p| P (b2 ( p) − b1 ( p)). If b1 ( p) = b2 ( p) for all p| P , then
we have equality in the above. Let p′| P be a prime for which b1 ( p′) < b2 ( p′).
Suppose that the B1 ( p) are chosen so that card B1 ( p) = b1 ( p) and
∑
x <n≤x +y
n /∈B1
1 = M (x ,y; b1 ) .
W e note that
p′
∑
b=1
b /∈B1 ( p′ )
∑
x <n≤x +y
n /∈B1
n̸≡b ( p′ )
1 =
∑
x <n≤x +y
n /∈B1
p′
∑
b=1
b /∈B1 ( p′ )
b̸≡n ( p′ )
1 . (3.37)
Consider the inner sum on the right. Since n /∈ B1 ( p′), the variable b is restricted
to lie in one of p′ − b1 ( p′) − 1 residue classes. Hence the right-hand side above
is
= ( p′ − b1 ( p′) − 1) M (x ,y; b1 ).
Since there are p′ − b1 ( p′) values of b in the outer sum on the left-hand side of
(3.37), it follows that there is a choice of b such that b /∈ B1 ( p′) and
∑
x <n≤x +y
n /∈B1
n̸≡b ( p′ )
1 ≥ p′ − b1 ( p′) − 1
p′ − b1 ( p′) M (x ,y; b1 ) . | {
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3.4 T win primes 97
Let b′
1 ( p) = b1 ( p) for p ̸=p′, b′
1 ( p′) = b1 ( p′) + 1. The left-hand side above
is ≤ M (x ,y; b′
1), which by the inductive hypothesis is
≤ M (x ,y; b2 ) p − b1 ( p′) − 1
p − b2 ( p′)
∏
p| P
p̸=p′
(p − b1 ( p)
p − b2 ( p)
)
.
Thus
M (x ,y; b1 ) ≤ M (x ,y; b2 )
∏
p| P
(p − b1 ( p)
p − b2 ( p)
)
,
and the induction is complete. □
By combining Theorem 3.10 and Lemma 3.12, we obtain
Theorem 3.13 Suppose that y ≥ 4. Let B( p) be the union of b ( p) arithmetic
progressions with common difference p, and put B = ⋃
p| P B( p).I f b (2) ≤ 1
and b ( p) ≤ 2 for p > 2, then the number of n ∈ (x ,x + y] such that n /∈ B is
≤ 8 y
(log y)2
(∏
p| P
(
1 − b( p)
p
)(
1 − 1
p
)−2 )(
1 + O
(log log y
log y
))
.
Corollary 3.14 Let r be an even non-zero integer , and suppose that y ≥ 4.
The number of primes p ∈ (x ,x + y] such that p + r is also prime is
≤ 8c(r ) y
(log y)2
(
1 + O
(log log y
log y
))
uniformly in r where
c(r ) =
(∏
p|r
(
1 − 1
p
)−1 )⎛
⎝∏
p∤r
(
1 − 2
p
)(
1 − 1
p
)−2
⎞
⎠=
⎛
⎜
⎝
∏
p|r
p>2
p − 1
p − 2
⎞
⎟
⎠c
and c is the constant in Theorem 3.10.
Suppose that r is a fixed even non-zero integer. It is conjectured that the
number of primes p ≤ y such that p + r is also prime is asymptotic to
c(r ) y
(log y)2
as y tends to infinity . Thus the bound we have derived is larger than this by a
factor of 8. W e conclude with an application of the above.
Theorem 3.15(Romanoff) Let N (x ) denote the number of integers n ≤ x
that can be expressed as a sum of a prime and a power of 2. Then N (x ) ≫ x
for x ≥ 4. | {
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98 Principles and first examples of sieve methods
Proof Let r (n) denote the number of solutions of n = p + 2k . By Cauchy’s
inequality ,
(∑
n≤x
r (n)
)2
≤ N (x )
∑
n≤x
r (n)2 .
Thus to complete the proof it suffices to show that
∑
n≤x
r (n) ≫ x (x ≥ 4), (3.38)
and that
∑
n≤x
r (n)2 ≪ x . (3.39)
The first of these estimates is easy: Put y = [(log x )/log 2]. If 0 ≤ k ≤ y − 1,
then 2 k ≤ x /2, and if also p ≤ x /2, then p + 2k ≤ x . Thus the sum in (3.38)
is
≥ π(x /2) y ≫ x
log x log x ≫ x
for x ≥ 4.
T o prove (3.39), we first observe that the sum on the left-hand side is
=
∑
p1 ,p2 ,j,k
p1 +2 j ≤x
p2 +2k ≤x
p1 +2 j = p2 +2k
1 .
This sum includes ‘diagonal’ terms, in which p1 = p2 and j = k; there are
≪ x /log x choices for p1 and ≪ log x choices for j , so there are ≪ x such
terms. The remaining terms above contribute an amount that is
≪
∑
0≤ j <k≤y
π2 (x ,2k − 2 j ) (3.40)
where π2 (x ,r ) denotes the number of primes p ≤ x for which p + r is also
prime. From Corollary 3.14 we know that if r ̸=0, then
π2 (x ,r ) ≪ x
(log x )2
∏
p|r
p>2
(
1 + 1
p
)
≪ x
(log x )2
∑
m|r
2∤m
1
m ,
uniformly in r . Thus the expression (3.40) is
≪ x
(log x )2
∑
0≤ j <k≤y
∑
m|(2k −2 j )
2∤m
1
m . | {
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3.4 T win primes 99
Put n = k − j . Thus 0 < n ≤ y. Let h2 (m) denote the order of 2 modulo m,
which is to say that h2 (m) is the least positive integer h such that 2 h ≡ 1
(mod m). W e note that m|(2n − 1) if and only if h2 (m)|n. The number of such
n,0 < n ≤ y,i s ≤ y/h2 (m). There are also ≤ y choices of j . Thus to complete
the proof of (3.39) it suffices to show that
∑
m
2∤m
1
mh 2 (m) < ∞ . (3.41)
T o this end, let
an =
∑
m
2∤m
h2 (m)=n
1
m ,
and set A(x ) = ∑
n≤x an . W e shall show that
A(x ) ≪ log x . (3.42)
By summation by parts it follows that ∑ an /n converges. (Alternatively , we
could appeal to Theorem 1.3, from which we see that ∑ an /ns converges for
σ> 0.) This suffices, since the sum in (3.41) is ∑ an /n.
It remains to establish (3.42). Set
P = P (x ) =
∏
n≤x
(2n − 1) .
If h2 (m) = n ≤ x , then m| P . Hence
A(x ) ≤
∑
m| P
1
m ≤
∏
p| P
(
1 + 1
p + 1
p2 +···
)
= P
ϕ( P ) ≪ log log P
by Theorem 2.9. But P ≤ 2x 2
, so we have (3.42), and the proof is complete. □
3.4.1 Exercises
1. For each prime p let B( p) be the union of b( p) ‘bad’ arithmetic progressions
with common difference p. Put B = ⋃
p| P B( p), and let
m(x ,y; b) = min
B
∑
x <n≤x +y
n /∈B
1
where the minimum is over all choices of the B( p) with b( p) fixed. Show
that if b1 ( p) ≤ b2 ( p) for all p, then
m(x ,y; b1 )
∏
p
(
1 − b1 ( p)
p
)−1
≥ m(x ,y; b2 )
∏
p
(
1 − b2 ( p)
p
)−1
. | {
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100 Principles and first examples of sieve methods
2. Show that the number of primes p ≤ 2n such that 2 n − p is prime is
≤ 8c
⎛
⎜
⎝
∏
p|n
p>2
p − 1
p − 2
⎞
⎟
⎠2n
(log 2 n)2
(
1 + O
(log log 4 n
log 2 n
))
where c is the constant in Theorem 3.10.
3. (Erd ˝ os 1940, Ricci 1954)
(a) Show that
∑
r ≤x
c(r ) = x + O (log x )
where c(r ) is defined as in Corollary 3.14.
(b) Let p′ denote the least prime > p, and put d ( p) = p′ − p. Show that
if a and b are fixed real numbers with a < b, then
∑
p≤x
a log p≤d ( p)≤b log p
log p ≲ 8(b − a)x .
(c) Suppose that f is a non-negative, properly Riemann-integrable function
on a finite interval [ a,b]. Show that
∑
p≤x
f
(d ( p)
log p
)
log p ≤ (8 + o(1))x
∫ b
a
f (u) du .
(d) Show that if a and b are fixed real numbers with a < b, then
∑
p≤x
a log p≤d ( p)≤b log p
(b log p − d ( p)) ≲ 4(b − a)2 x .
(e) Explain why
∑
p≤x
d ( p)>b log p
(d ( p) − b log p) ≥ 0 .
(f) Deduce that
∑
p≤x
d ( p)≥a log p
(b log p − d ( p)) ≲ 4(b − a)2 x .
(g) Show that
∑
p≤x
d ( p) ∼ x .
(h) Show that
∑
p≤x
(b log p − d ( p)) = (b − 1 + o(1))x . | {
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3.5 Notes 101
(i) T ake b = a + 1/8, and suppose that d ( p) ≥ a log p for all p > p0 .
Show that the estimates of (f) and (h) are inconsistent if a > 15/16.
Thus conclude that
lim inf
p→∞
d ( p)
log p ≤ 15
16 .
4. Let r (n) be defined as in the proof of Theorem 3.15. Show that
∑
n≤x
r (n) ∼ x
log 2 .
5. Let r (n) be defined as in the proof of Theorem 3.15. Show that
∑
n≤x
2|n
r (n) ≪ x
log x .
6. (Erd ˝ os 1950)
(a) Show that if n ≡ 1 (mod 3) and k ≡ 0 (mod 2), then 3 |(n − 2k ).
(b) Show that if n ≡ 1 (mod 7) and k ≡ 0 (mod 3), then 7 |(n − 2k ).
(c) Show that if n ≡ 2 (mod 5) and k ≡ 1 (mod 4), then 5 |(n − 2k ).
(d) Show that if n ≡ 8 (mod 17) and k ≡ 3 (mod 8), then 17 |(n − 2k ).
(e) Show that if n ≡ 11 (mod 13) and k ≡ 7 (mod 12), then 13 |(n − 2k ).
(f) Show that if n ≡ 121 (mod 241) and k ≡ 23 (mod 24), then 241 |
(n − 2k ).
(g) Show that every integer k satisfies at least one of the congruences
k ≡ 0 (mod 2), k ≡ 0 (mod 3), k ≡ 1 (mod 4), k ≡ 3 (mod 8), k ≡
7 (mod 12), k ≡ 23 (mod 24).
(h) Show that if n satisfies all the congruences n ≡ 1 (mod 3), n ≡ 1
(mod 7), n ≡ 2 (mod 5), n ≡ 8 (mod 17), n ≡ 11 (mod 13), n ≡
121 (mod 241), then n − 2k is divisible by at least one of the primes
3, 7, 5, 17, 13, 241.
(i) Show that these congruential conditions are equivalent to the single
condition n ≡ 172677 (mod 3728270).
(j) An integer n satisfying the above might still be representable in the
form p + 2k , but if it is, then the prime in question must be one of the
six primes listed. Show that if in addition, n ≡ 9 or 11 or 15 (mod 16),
then n cannot be expressed as a sum of a prime and a power of 2.
3.5 Notes
Sections 3.1, 3.2. The modern era of sieve methods began with the work
of Brun (1915, 1919). Hardy & Littlewood (1922) used Brun’s method to
establish the estimate (3.9). The sharp form of this in Corollary 3.4 is due | {
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102 Principles and first examples of sieve methods
to Selberg (1952a,b). The /Lambda12 method of Selberg (1947) provides only upper
bounds, but lower bounds can also be derived from it by using ideas of Buchstab
(1938).
In contrast to the elegance of the Selberg /Lambda12 method, the further study of
sieves leads us to construct asymptotic estimates for complicated sums over
integers whose prime factors are distributed in certain ways. In this connection,
the argument (3.22) is a simple foretaste of more complicated things to come.
Hence further discussion of sieves is possible only after the appropriate technical
tools are in place.
In this chapter we have applied the sieve only to arithmetic progressions,
but it can be shown that the sieve is applicable to much more general sets. This
makes sieves very versatile, but it also means that they are subject to certain
unfortunate limitations. In order to estimate the number of elements of a setS
that remain after sifting, it suffices to have a reasonably precise estimate of the
numberX d of multiples of d in the set, say of the form X d = f (d ) X /d + O ( Rd )
where X is an estimate for the cardinality of S, and f is a multiplicative function.
Thus Theorem 3.3 can be generalized to much more general sets, and in that
more general setting it is known that the constant 2 is best-possible. It may be
true that the constant 2 can be improved in the special case that one is sifting
an interval, but this has not been achieved thus far.
When sifting an interval, the error terms can be avoided by using Fourier
analysis as in Selberg (1991, Sections 19–22), or by using the large sieve as
in Montgomery & V aughan (1973). In particular, the number of integers in
[M + 1,M + N ] remaining after sifting is at most N /L where
L =
∑
q ≤Q
µ(q )2
1 + 3
2 qQ /N
∏
p|q
b( p)
p − b( p) . (3.43)
Here b( p) is the number of residue classes modulo p that are deleted. This is
both a generalization and a sharpening of Theorem 3.2.
Section 3.3. Titchmarsh (1930) used Brun’s method to obtain Theorem 3.9,
but with a larger constant instead of 2. Montgomery & V aughan (1973) have
shown that Corollary 3.4 and Theorem 3.9 are still valid when the error terms are
omitted. See also Selberg (1991, Section 22). The first significant improvement
of Theorem 3.9 was obtained by Motohashi (1973). Other improvements of
various kinds have been derived by Motohashi (1974), Hooley (1972, 1975),
Goldfeld (1975), Iwaniec (1982), and Friedlander & Iwaniec (1997).
In Lemmas 3.5 and 3.12, and in Exercises 3.2.7, 3.2.9, 3.2.10, 3.4.1 we see
evidence of a monotonicity principle that permeates sieve theory; cf. Selberg
(1991, pp. 72–73). | {
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3.5 Notes 103
Hooley (1994) has shown that quite sharp sieve bounds can be derived using
the interrupted inclusion–exclusion idea that Brun started with. This approach
has been developed further by Ford & Halberstam (2000). An exposition of
sieves based on these ideas is given by Bateman & Diamond (2004, Chapters 12,
13). Still more extensive accounts of sieve methods have been given by Greaves
(2001), Halberstam & Richert (1974), Iwaniec & Kowalski (2004, Chapter
6), Motohashi (1983), and Selberg (1971, 1991). In addition, a collection of
applications of sieves to arithmetic problems has been given by Hooley (1976),
and additional sieve ideas are found in Bombieri (1977), Bombieri, Friedlander
& Iwaniec (1986, 1987, 1989), Fouvry & Iwaniec (1997), Friedlander & Iwaniec
(1998a, b), and Iwaniec (1978, 1980a, b, 1981).
Section 3.4. The twin prime conjecture is a special case of the prime k-tuple
conjecture. Suppose that d1 ,..., dk are distinct integers, and let b( p) denote
the number of distinct residue classes modulo p found among the di . The prime
k-tuple conjecture asserts that if b( p) < p for every prime number p, then there
exist infinitely many positive integers n such that the k numbers n + di are all
prime. Hardy & Littlewood (1922) put this in a quantitative form: If b( p) < p
for all p, then the number of n ≤ N for which the k numbers n + di are all
prime is conjectured to be
∼ S(d ) N
(log N )k (3.44)
as N →∞ where
S(d ) =
∏
p
(
1 − b( p)
p
)(
1 − 1
p
)−k
. (3.45)
This product is absolutely convergent, since b( p) = k for all sufficiently large
primes p. Although this remains unproved, by sifting we can obtain an upper
bound of the expected order of magnitude. In particular, from (3.43) it can be
shown that the number ofn, M + 1 ≤ n ≤ M + N , for which the numbers
n + di are all prime is
≲ 2k k!S(d ) N
(log N )k . (3.46)
Corollarys 3.4 and 3.14 are special cases of this.
Theorem 3.15 is due to Romanoff (1934). Once the bound for the number
of twin primes is in place, the hardest part of the proof is to establish the
estimate (3.41). Romanoff’s original proof of this was rather difficult. Erd ˝ os
& Tur´an (1935) gave a simpler proof, but the clever proof we have given is
due to Erd ˝ os (1951). Let r (n) be defined as in the proof of Theorem 3.15.
Erd ˝ os (1950) showed that r (n) = /Omega1 (log log n), and that ∑
n≤x r (n)k ≪k x for | {
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104 Principles and first examples of sieve methods
any positive k. Presumably r (n) = o(log n), but for all we know there could be,
although it seems unlikely , infinitely many n such that n − 2k is prime whenever
0 < 2k < n. The number n = 105 has this property , and is probably the largest
such number. The best upper bound we have for the number of such n not
exceeding X is (V aughan 1973),
X exp
(
− c log X log log log X
log log X
)
.
For generalizations of Romanoff’s theorem, see Erd ˝ os (1950, 1951).
3.6 References
Ankeny , N. C. & Onishi, H. (1964/1965). The general sieve, Acta Arith. 10, 31–62.
Bateman, P . T . & Diamond, H. (2004). Analytic Number Theory , Hackensack: W orld
Scientific.
Behrend, F . A. (1948). Generalization of an inequality of Heilbronn and Rohrbach, Bull.
Amer . Math. Soc. 54, 681–684.
Bombieri, E. (1977). The asymptotic sieve, Rend. Accad. Naz . XL (5) 1/2 (1975/76),
243–269.
Bombieri, E., Friedlander, J. B., & Iwaniec, H. (1986). Primes in arithmetic progressions
to large moduli, Acta Math . 156, 203–251.
(1987). Primes in arithmetic progressions to large moduli, II, Math. Ann . 277, 361–
393.
(1989). Primes in arithmetic progressions to large moduli, III, J. Amer . Math. Soc . 2,
215–224.
Brun, V . (1915). ¨Uber das Goldbachsche Gesetz und die Anzahl der Primzahlpaare,
Archiv for Math. og Naturvid .B 34, no. 8, 19 pp.
(1919). La s´ erie 1 /5 + 1/7 + 1/11 + 1/13 + 1/17 + 1/19 + 1/29 + 1/31 +
1/41 + 1/43 + 1/59 + 1/61 +··· o`u les d´ enominateurs sont “nombres premiers
jumeaus” est convergente ou finie, Bull. Sci. Math . (2) 43, 100–104; 124–128.
(1967). Reflections on the sieve of Eratosthenes, Norske V id. Selsk . Skr . Trondheim,
no. 1, 9 pp.
Buchstab, A. A. (1938). New improvements in the method of the sieve of Eratosthenes,
Mat. Sb. (N. S.) 4 (46), 375–387.
Chowla, S. (1932). Contributions to the analytic theory of numbers, Math. Z. 35, 279–
299.
Chung, K.-L. (1941). A generalization of an inequality in the elementary theory of
numbers, J. Reine Angew . Math. 183, 193–196.
van der Corput, J. G. (1958). Inequalities involving least common multiple and other
arithmetical functions, Nederl. Akad. W etensch. Proc. Ser . A 61 (= Indag. Math.
20), 5–15.
Erd ˝ os, P . (1940). The difference of consecutive primes, Duke Math. J. 6, 438–441.
(1946). On the coefficients of the cyclotomic polynomial, Bull. Amer . Math. Soc . 52,
179–184. | {
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3.6 References 105
(1950). On integers of the form 2 k + p and some related problems, Summa Brasil.
Math. 2, 113–123.
(1951). On some problems of Bellman and a theorem of Romanoff, J. Chinese Math.
Soc. (N. S.) 1, 409–421.
Erd ˝ os, P . & Tur´ an, P . (1935). Ein zahlentheoretischer Satz, Mitt. F orsch. Inst. Math.
Mech. Univ . T omsk 1, 101–103.
Ford, K. & Halberstam, H. (2000). The Brun–Hooley sieve, J. Number Theory 81,
335–350.
Fouvry , E. & Iwaniec, H. (1997). Gaussian primes, Acta Arith. 79 (1997), 249–287.
Friedlander, J. B. & Iwaniec, H. (1997). The Brun–Titchmarsh theorem, Analytic Number
Theory (Kyoto, 1996). London Math. Soc. Lecture Note Ser. 247, Cambridge:
Cambridge University Press, pp. 85–93.
(1998a). The polynomial X 2 + Y 4 captures its primes, Ann. of Math. (2) 148, 945–
1040.
(1998b). Asymptotic sieve for primes, Ann. of Math. (2) 148, 1041–1065.
Goldfeld, D. M. (1975). A further improvement of the Brun–Titchmarsh theorem, J.
London Math. Soc. (2) 11, 434–444.
Greaves, G. (2001). Sieves in Number Theory . Berlin: Springer.
Halberstam, H. (1985). Lectures on the linear sieve , T opics in Analytic Number Theory
(Austin, 1982). Austin: University of T exas Press, pp. 165–220.
Halberstam, H. & Richert, H.-E. (1973). Brun’s method and the fundamental lemma,
Acta Arith. 24, 113–133.
(1974). Sieve Methods . London: Academic Press.
(1975). Brun’s method and the fundamental lemma. II, Acta Arith. 27, 51–59.
Hardy , G. H. & Littlewood, J. E. (1922). Some problems of ‘Partitio Numerorum’: III.
On the expression of a number as a sum of primes, Acta Math. 44, 1–70; Collected
P apers, V ol. I, London: Oxford University Press, 1966, pp. 561–630.
Heilbronn, H. (1937). On an inequality in the elementary theory of numbers, Proc.
Cambridge Philos. Soc. 33, 207–209.
Hensley , D. (1978). An almost-prime sieve, J. Number Theory 10, 250–262; Corrigen-
dum, 12, (1980), 437.
Hooley , C. (1972). On the Brun–Titchmarsh theorem, J. Reine Angew . Math. 255,
60–79.
(1975). On the Brun–Titchmarsh theorem, II, Proc. London Math. Soc. (3) 30, 114–
128.
(1976). Applications of Sieve Methods to the Theory of Prime Numbers, Cambridge
Tract 70. Cambridge: Cambridge University Press.
(1994). An almost pure sieve, Acta Arith. 66, 359–368.
Iwaniec, H. (1978). Almost-primes represented by quadratic polynomials, Invent. Math.
47, 171–188.
(1980a). Rosser’s sieve, Acta Arith. 36, 171–202.
(1980b). A new form of the error term in the linear sieve, Acta Arith. 37, 307–320.
(1981). Rosser’s sieve – bilinear forms of the remainder terms – some applications.
Recent Progress in Analytic Number Theory , V ol. 1. New Y ork: Academic Press,
pp. 203–230.
(1982). On the Brun–Titchmarsh theorem, J. Math. Soc. Japan 34, 95–123.
Iwaniec, H. & Kowalski, E. (2004). Analytic Number Theory , Colloquium Publications
53. Providence: Amer. Math. Soc. | {
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106 Principles and first examples of sieve methods
Jurkat, W . B. & Richert, H.-E. (1965). An improvement in Selberg’s sieve method, I,
Acta Arith. 11, 217–240.
Lehmer, D. H. (1955). The distribution of totatives, Canad. J. Math. 7, 347–357.
van Lint, J. H. & Richert, H.-E. (1964). ¨Uber die Summe ∑
n≦x
p(n)<y
µ2 (n)
ϕ(n) Nederl. Akad.
W etensch. Proc. Ser .A 67 (= Indag. Math. 26), 582–587.
(1965). On primes in artihmetic progressions, Acta Arith. 11, 209–216.
Montgomery , H. L. (1968). A note on the large sieve, J. London Math. Soc. 43,
93–98.
Montgomery , H. L. & V aughan, R. C. (1973). The large sieve, Mathematika 20, 119–134.
(1979). Mean values of character sums, Canad. J. Math. 31, 476–487.
Motohashi, Y . (1973). On some improvements of the Brun–Titchmarsh theorem, II,
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1973), Søurikaisekikenkyøusho K ´ okyøuroku, No. 193, 97–109.
(1974). On some improvements of the Brun–Titchmarsh theorem, J. Math. Soc. Japan
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(1975). On some improvements of the Brun–Titchmarsh theorem, III, J. Math. Soc.
Japan 27, 444–453.
(1983). Lectures on Sieve Methods and Prime Number theory . T ata Institute of Fun-
damental Research (Bombay). Berlin: Springer-V erlag.
Ricci, G. (1954). Sull’andamento della differenza di numeri primi consecutivi, Riv . Mat.
Univ . P arma 5, 3–54.
Riesel, H. & V aughan, R. C. (1983). On sums of primes, Ark. Mat. 21, 46–74.
Rohrbach, H. (1937). Beweis einer zahlentheoretischen Ungleichung, J. Reine Angew .
Math. 177, 193–196.
Romanoff, N. P . (1934). ¨Uber einige S¨ atze der additiven Zahlentheorie, Math. Ann. 109,
668–678.
Selberg, A. (1947). On an elementary method in the theory of primes, Norske V id. Selsk.
F orh., Trondhjem 19, no. 18, 64–67; Collected P apers , V ol. 1. Berlin: Springer-
V erlag, 1989, pp. 363–366.
(1952a). On elementary methods in primenumber-theory and their limitations, Den
11te Skandinaviske Matematikerkongress (Trondheim, 1949), Oslo: Johan Grundt
T anums Forlag, pp. 13–22; Collected P apers, V ol. 1. Berlin: Springer-V erlag, 1989,
pp. 388–397.
(1952b). The general sieve-method and its place in prime-number theory. Proceedings
of the International Congress of Mathematicians (Cambridge MA, 1950), V ol. 1,
Providence: Amer. Math. Soc., pp. 286–292;Collected P apers , V ol. 1. Berlin:
Springer-V erlag, 1989, pp. 411–417.
(1971). Sieve methods , Proceedings of Symposium on Pure Mathematics (SUNY
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(1972). Remarks on sieves , Proceedings of the Number Theory Conference (Boulder
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1989, pp. 609–615.
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and Discrete Groups (Oslo, 1987), K. E. Aubert, E. Bombieri, D. Goldfeld, eds. | {
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3.6 References 107
Boston: Academic Press, pp. 467–484; Collected P apers, V ol. 1. Berlin: Springer-
V erlag, 1989, pp. 675–69.
(1991). Lectures on Sieves, Collected P apers , V ol. 2. Berlin: Springer-V erlag,
pp. 65–247.
Titchmarsh, E. C. (1930). A divisor problem, Rend. Circ. Math . Palermo 54, 414–429.
Tsang, K. M. (1989). Remarks on the sieving limit of the Buchstab–Rosser sieve , Number
Theory , Trace Formulas and Discrete Groups (Oslo, 1987). Boston: Academic
Press, pp. 485–502.
V aughan, R. C. (1973). Some applications of Montgomery’s sieve, J. Number Theory 5,
64–79.
V ijayaraghavan, T . (1951). On a problem in elementary number theory , J. Indian Math.
Soc. (N.S.) 15, 51–56. | {
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4
Primes in arithmetic progressions: I
4.1 Additive characters
If f (z) = ∑ ∞
n=0 cn zn is a power series, we can restrict our attention to terms
for which n has prescribed parity by considering
1
2 f (z) + 1
2 f (−z) =
∞∑
n=0
n≡ 0 (2)
cn zn
or
1
2 f (z) − 1
2 f (−z) =
∞∑
n=0
n≡1 (2)
cn zn .
That is, we can express the characteristic function of an arithmetic progression
(mod 2) as a linear combination1
2 1n ± 1
2 (−1)n of 1 n and ( −1)n . Here 1 and
−1 are the square-roots of 1, and we can similarly express the characteristic
function of an arithmetic progression (mod q ) as a linear combination of the
sequences ζn where ζ runs over the q different q th roots of unity . W e write
e(θ) = e2πi θ, and then the q th roots of unity are the numbers ζ = e(a/q ) for
1 ≤ a ≤ q .I f( a,q ) = 1 then the least integer n such that ζn = 1i s q , and we
say that ζ is a primitive q th root of unity . From the formula
q −1∑
k=0
ζk = 1 − ζq
1 − ζ
for the sum of a geometric progression, we see that if ζ is a q th root of unity
then
q∑
k=1
ζk = 0
108 | {
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4.1 Additive characters 109
unless ζ = 1. Hence
1
q
q∑
k=1
e(−ka /q )e(kn /q ) =
{ 1i f n ≡ a (mod q ),
0 otherwise, (4.1)
and thus the characteristic function of an arithmetic progression (mod q ) can be
expressed as a linear combination of the sequences e(kn /q ). These functions
are called the additive characters (mod q ) because they are the homomorphisms
from the additive group ( Z/q Z)+ of integers (mod q ) to the multiplicative group
C× of non-zero complex numbers.
In the language of linear algebra we see that the arithmetic functions of
period q form a vector space of dimension q . For any k,1 ≤ k ≤ q , the se-
quence {e(kn /q )}∞
n=−∞ has period q , and these q sequences form a basis
for the space of q -periodic arithmetic functions. Indeed, the formula (4.1)
expresses the ath elementary vector as a linear combination of the vectors
[e(n/q ),e(2n/q ),..., e((q − 1)n/q ),1].
If f (n) is an arithmetic function with period q then we define the finite
F ourier transform of f to be the function
ˆf (k) = 1
q
q∑
n=1
f (n)e(−kn /q ). (4.2)
T o obtain a Fourier representation of f we multiply both sides of (4.1) by f (n)
and sum over n to see that
f (a) =
q∑
n=1
f (n)
q
q∑
k=1
e(−ka /q )e(kn /q )
=
q∑
k=1
e(−ka /q ) 1
q
q∑
n=1
f (n)e(kn /q )
=
q∑
k=1
e(−ka /q ) ˆf (−k).
Here the exact values that k runs through are immaterial, as long as the set of
these values forms a complete residue system modulo q . Hence we may replace
k by −k in the above, and so we see that
f (n) =
q∑
k=1
ˆf (k)e(kn /q ). (4.3)
This includes (4.1) as a special case, for if we take f to be the characteris-
tic function of the arithmetic progression a (mod q ) then by (4.2) we have
ˆf (k) = e(−ka /q )/q , and then (4.3) coincides with (4.1). The pair (4.2), (4.3)
of inversion formulæ are analogous to the formula for the Fourier coefficients | {
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110 Primes in arithmetic progressions: I
and Fourier expansion of a function f ∈ L 1 (T), but the situation here is simpler
because our sums have only finitely many terms.
Let v (h) be the vector v (h) = [e(h/q ),e(2h/q ),..., e((q − 1)h/q ),1].
From (4.1) we see that two such vectors v (h1 ) and v (h2 ) are orthogonal un-
less h1 ≡ h2 (mod q ). These vectors are not normalized, but they all have the
same length √q , so apart from some rescaling, the transformation from f to ˆf
is an isometry . More precisely , if f has period q and ˆf is given by (4.2), then
by (4.3),
q∑
n=1
| f (n)|2 =
q∑
n=1
⏐
⏐
⏐
⏐
q∑
k=1
ˆf (k)e(kn /q )
⏐
⏐
⏐
⏐
2
.
By expanding and taking the sum over n inside, we see that this is
=
q∑
j =1
q∑
k=1
ˆf ( j )
ˆf (k)
q∑
n=1
e( jn /q )e(−kn /q ).
By (4.1) the innermost sum is q if j = k and is 0 otherwise. Hence
q∑
n=1
| f (n)|2 = q
q∑
k=1
| ˆf (k)|2 . (4.4)
This is analogous to Parseval’s identity for functions f ∈ L 2 (T), or to
Plancherel’s identity for functions f ∈ L 2 (R).
Among the exponential sums that we shall have occasion to consider is
Ramanujan’s sum
cq (n) =
q∑
a=1
(a,q )=1
e(an /q ). (4.5)
W e now establish some of the interesting properties of this quantity .
Theorem 4.1As a function of n, c q (n) has period q . F or any given n, c q (n)
is a multiplicative function of q . Also,
∑
d |q
cd (n) =
{ qi f q |n,
0 otherwise. (4.6)
Finally,
cq (n) =
∑
d |(q ,n)
d µ(q /d ) = µ(q /(q ,n))
ϕ(q /(q ,n)) ϕ(q ). (4.7)
The case n = 1 of this last formula is especially memorable:
q∑
a=1
(a,q )=1
e(a/q ) = µ(q ). | {
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4.1 Additive characters 111
Proof The first assertion is evident, as each term in the sum (4.5) has period
q . As for the second, suppose that q = q1 q2 where ( q1 ,q2 ) = 1. By the Chinese
Remainder Theorem, for each a (mod q ) there is a unique pair a1 ,a2 with ai
determined (mod qi ), so that a ≡ a1 q2 + a2 q1 (mod q ). Moreover, under this
correspondence we see that ( a,q ) = 1 if and only if ( ai ,qi ) = 1 for i = 1,2.
Then
cq (n) =
q1∑
a1 =1
(a1 ,q1 )=1
q2∑
a2 =1
(a2 ,q2 )=1
e((a1 q2 + a2 q1 )n/(q1 q2 ))
=
⎛
⎜
⎝
q1∑
a1 =1
(a1 ,q1 )=1
e(a1 n/q1 )
⎞
⎟
⎠
⎛
⎜
⎝
q2∑
a2 =1
(a2 ,q2 )=1
e(a2 n/q2 )
⎞
⎟
⎠
= cq1 (n)cq2 (n).
T o establish (4.6), suppose that d |q , and consider those a,1 ≤ a ≤ q , such
that ( a,q ) = d . Put b = a/d . Then the numbers a are in one-to-one correspon-
dence with those b,1 ≤ b ≤ q /d , for which ( b,q /d ) = 1. Hence
q∑
a=1
e(na /q ) =
∑
d |q
q∑
a=1
(a,q )=d
e(na /q )
=
∑
d |q
q /d∑
b=1
(b,q /d )=1
e(nb /(q /d ))
=
∑
d |q
cq /d (n).
By (4.1), the left-hand side above is q when q |n, and is 0 otherwise. Thus we
have (4.6).
The first formula in (4.7) is merely the M ¨ obius inverse of (4.6). T o obtain
the second formula in (4.7), we begin by considering the special case in which
q is a prime power, q = pk .
c pk (n) =
pk
∑
a=1
p∤a
e(na /pk )
=
pk
∑
a=1
e(na /pk ) −
pk−1
∑
a=1
e(na /pk−1 ). | {
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112 Primes in arithmetic progressions: I
Here the first sum is pk if pk |n, and is 0 otherwise. Similarly , the second
sum is pk−1 if pk−1 |n, and is 0 otherwise. Hence the above is
=
⎧
⎨
⎩
0i f pk−1 ∤ n,
− pk−1 if pk−1 ∥ n,
pk − pk−1 if pk |n
= µ
(
pk /(n, pk )
)
ϕ
(
pk /(n, pk )
)ϕ( pk ).
The general case of (4.7) now follows because cq (n) is a multiplicative function
of q . □
4.1.1 Exercises
1. Let U = [ukn ] be the q × q matrix with elements ukn = e(kn /q )/√q . Show
that UU ∗ = U ∗U = I , i.e., that U is unitary .
2. (Friedman 1957; cf. Reznick 1995)
(a) Show that
∫ 1
0
(
ue (θ/2) + ve(−θ/2)
)2r
d θ =
(2r
r
)
ur vr
for any non-negative integer r and arbitrary complex numbers u,v.
(b) Show that if u = (x − iy )/2, v = (x + iy )/2, then
x cos πθ + y sin πθ = ue (θ/2) + ve(−θ/2)
for all θ.
(c) Show that∫ 1
0
(
x cos πθ + y sin πθ
)2r
d θ =
(2r
r
)
2−2r (x 2 + y2 )r
for any non-negative integer r and arbitrary real or complex numbers
x ,y.
(d) Show that
q∑
a=1
(
ue πia /q + ve−πia /q )2r
= q
(2r
r
)
ur vr
if r is an integer, 0 ≤ r < q .
(e) Show that
q∑
a=1
(x cos πa/q + y sin πa/q )2r = q
(2r
r
)
2−2r (x 2 + y2 )r
if r is an integer, 0 ≤ r < q . | {
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4.1 Additive characters 113
3. Show that |cq (n)|≤ (q ,n).
4. (Carmichael 1932)
(a) Show that if q > 1, then
q∑
n=1
cq (n) = 0.
(b) Show that if q1 ̸=q2 and [ q1 ,q2 ]|N , then
N∑
n=1
cq1 (n)cq2 (n) = 0.
(c) Show that if q |N , then
N∑
n=1
cq (n)2 = N ϕ(q ).
5. (Grytczuk 1981; cf. Redmond 1983) Show that
∑
d |q
|cd (n)|= 2ω(q /(q ,n)) (q ,n).
6. (Ramanujan 1918) Show that
ϕ(n)
n =
∞∑
d =1
µ(d )
d 2
∑
q |d
cq (n) =
∞∑
q =1
aq cq (n)
where
aq = 6µ(q )
π2 q 2
∏
p|q
(
1 − 1
p2
)−1
.
7. (Wintner 1943, Sections 33–35) The orthogonality relations of Exercise 4
give us hope that it might be possible to represent an arithmetic function
F (n) in the form
F (n) =
∞∑
q =1
aq cq (n) (4.8)
by taking
aq = 1
ϕ(q ) lim
x →∞
1
x
∑
n≤x
F (n)cq (n) . (4.9)
In the following, suppose that f (r ) is chosen so that F (n) = ∑
r |n f (r ) for
all n.
(a) Suppose that
∞∑
r =1
| f (r )|
r < ∞ . (4.10) | {
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114 Primes in arithmetic progressions: I
Let d be a fixed positive integer. Show that
∑
n≤x
d |n
F (n) = x
d
∞∑
r =1
f (r )
r (d ,r ) + o(x )
as x →∞ .
(b) Suppose that (4.10) holds. Show that
lim
x →∞
1
x
∑
n≤x
F (n)cq (n) = ϕ(q )
∞∑
r =1
q |r
f (r )
r .
(c) Put
aq =
∞∑
r =1
q |r
f (r )
r .
Show that if
∞∑
r =1
| f (r )|d (r )
r < ∞ (4.11)
then (4.8) and (4.9) hold, and moreover that ∑ ∞
q =1 |aq cq (n)| < ∞.
8. (Ramanujan 1918) Show that if q > 1, then ∑ ∞
n=1 cq (n)/n =− /Lambda1 (q ). (See
also Exercise 8.3.4.)
9. Let /Phi1 q (z) denote the q th cyclotomic polynomial, i.e., the monic polynomial
whose roots are precisely the primitive q th roots of unity , so that
/Phi1 q (z) =
q∏
n=1
(n,q )=1
(z − e(n/q )).
(a) Show that
/Phi1 q (z) =
∏
d |q
(zd − 1)µ(q /d )
and that ( zd − 1)µ(q /d ) has a power series expansion, valid when |z| < 1,
with integer coefficients. Deduce that /Phi1 q (z) ∈ Z[z].
(b) Suppose that z ∈ Z and p | /Phi1 q (z) and let e denote the order of z modulo
p. Show that e | q and that if p | (zd − 1) then e | d .
(c) Choose t so that pt ∥ (ze − 1). Show that for m ∈ N with p ∤ m one has
pt ∥ (zme − 1).
(d) Show that if p ∤ q , then pht ∥ /Phi1 q (z) where h = ∑
e|d |q
µ(q /d ). Deduce that
e = q and that q | ( p − 1).
(e) By taking z to be a suitable multiple of q , or otherwise, show that there
are infinitely many primes p with p ≡ 1 (mod q ). | {
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4.2 Dirichlet characters 115
4.2 Dirichlet characters
In the preceding section we expressed the characteristic function of an arithmetic
progression as a linear combination of additive characters. For purposes of
multiplicative number theory we shall similarly represent the characteristic
function of a reduced residue class (modq ) as a linear combination of totally
multiplicative functions χ(n) each one supported on the reduced residue classes
and having period q . These are the Dirichlet characters . Since χ(n) has period
q we may think of it as mapping from residue classes, and since χ(n) ̸=0 if and
only if ( n,q ) = 1, we may think of χ as mapping from the multiplicative group
of reduced residue classes to the multiplicative group C× of non-zero complex
numbers. As χ is totally multiplicative, χ(mn ) = χ(m)χ(n) for all m, n,w es e e
that the map χ :( Z/q Z)× −→ C× is a homomorphism. The method we use to
describe these characters applies when ( Z/q Z)× is replaced by an arbitrary finite
abelian group G , so we consider the slightly more general problem of finding
all homomorphisms χ : G → C× from such a group G to C×. W e call these
homomorphisms the characters of G , and let ˆG denote the set of all characters
of G . W e let χ0 denote the principal character , whose value is identically 1.
W e note that if χ ∈ ˆG , then χ(e) = 1 where e denotes the identity in G . Let n
denote the order of G .I f g ∈ G and χ ∈ ˆG , then gn = e, and hence χ(gn ) = 1.
Consequently χ(g)n = 1, and so we see that all values taken by characters are
nth roots of unity . In particular, this implies that ˆG is finite, since there can be at
most nn such maps. If χ1 and χ2 are two characters of G , then we can define
a product character χ1 χ2 by χ1 χ2 (g) = χ1 (g)χ2 (g). For χ ∈ ˆG , let
χ be the
character χ(g). Then χ · χ = χ0 , and we see that ˆG is a finite abelian group
with identity χ0 . The following lemmas prepare for a full description of ˆG in
Theorem 4.4.
Lemma 4.2Suppose that G is cyclic of order n, say G = (a). Then there are
exactly n characters of G , namely χk (am ) = e(km /n) for 1 ≤ k ≤ n. Moreover ,
∑
g∈G
χ(g) =
{ ni f χ = χ0 ,
0 otherwise, (4.12)
and
∑
χ∈ˆG
χ(g) =
{ ni f g = e,
0 otherwise. (4.13)
In this situation, ˆG is cyclic, ˆG = (χ1 ).
Proof Suppose that χ ∈ ˆG . As we have observed, χ(a)i sa n nth root of unity ,
say χ(a) = e(k/n) for some k,1 ≤ k ≤ n. Hence χ(am ) = χ(a)m = e(km /n). | {
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116 Primes in arithmetic progressions: I
Since the characters are now known explicitly , the remaining assertions are
easily verified.□
Next we describe the characters of the direct product of two groups in terms
of the characters of the factors.
Lemma 4.3Suppose that G 1 and G 2 are finite abelian groups, and that G =
G 1 ⊗ G 2 .I f χi is a character of G i ,i = 1,2, and g ∈ G is written g = (g1 ,g2 ),
gi ∈ Gi , then χ(g) = χ1 (g1 )χ2 (g2 ) is a character of G . Conversely, if χ ∈ ˆG,
then there exist unique χi ∈ Gi such that χ(g) = χ1 (g1 )χ2 (g2 ). The identities
(4.12) and (4.13) hold for G if they hold for both G 1 and G 2 .
W e see here that each χ ∈ ˆG corresponds to a pair ( χ1 ,χ2 ) ∈ ˆG 1 × ˆG 2 . Thus
G ∼
=ˆG 1 ⊗ ˆG 2 .
Proof The first assertion is clear. As for the second, put χ1 (g1 ) = χ((g1 ,e2 )),
χ2 (g2 ) = χ((e1 ,g2 )). Then χi ∈ ˆGi for i = 1,2, and χ1 (g1 )χ2 (g2 ) = χ(g). The
χi are unique, for if g = (g1 ,e2 ), then
χ(g) = χ((g1 ,e2 )) = χ1 (g1 )χ2 (e2 ) = χ1 (g1 ),
and similarly for χ2 .I f χ(g) = χ1 (g1 )χ2 (g2 ), then
∑
g∈G
χ(g) =
(∑
g1 ∈G 1
χ1 (g1 )
)(∑
g2 ∈G 2
χ2 (g2 )
)
,
so that (4.12) holds for G if it holds for G 1 and for G 2 . Similarly , if g = (g1 ,g2 ),
then
∑
χ∈ˆG
χ(g) =
⎛
⎝∑
χ1 ∈ˆG 1
χ1 (g1 )
⎞
⎠
⎛
⎝∑
χ1 ∈ˆG 2
χ2 (g2 )
⎞
⎠,
so that (4.13) holds for G if it holds for G 1 and G 2 . □
Theorem 4.4 Let G be a finite abelian group. Then ˆG is isomorphic to G ,
and (4.12) and (4.13) both hold.
Proof Any finite abelian group is isomorphic to a direct product of cyclic
groups, say
G ∼
=Cn1 ⊗ Cn2 ⊗···⊗ Cnr .
The result then follows immediately from the lemmas. □
Though G and ˆG are isomorphic, the isomorphism is not canonical. That is,
no particular one-to-one correspondence between the elements of G and those
of ˆG is naturally distinguished. | {
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4.2 Dirichlet characters 117
Corollary 4.5 The multiplicative group (Z/q Z)× of reduced residue classes
(mod q ) has ϕ(q ) Dirichlet characters. If χ is such a character , then
q∑
n=1
(n,q )=1
χ(n) =
{ ϕ(q ) if χ = χ0 ,
0 otherwise. (4.14)
If (n,q ) = 1, then
∑
χ
χ(n) =
{ ϕ(q ) if n ≡ 1 (mod q ),
0 otherwise, (4.15)
where the sum is extended over the ϕ(q ) Dirichlet characters χ (mod q ).
As we remarked at the outset, for our purposes it is convenient to define the
Dirichlet characters (mod q ) on all integers; we do this by setting χ(n) = 0
when ( n,q ) > 1. Thus χ is a totally multiplicative function with period q that
vanishes whenever ( n,q ) > 1, and any such function is a Dirichlet character
(mod q ). In this book a character is understood to be a Dirichlet character unless
the contrary is indicated.
Corollary 4.6If χi is a character (mod qi ) for i = 1,2, then χ1 (n)χ2 (n)
is a character (mod [ q1 ,q2 ]).I fq = q1 q2 , (q1 ,q2 ) = 1, and χ is a character
(mod q ), then there exist unique characters χi (mod q ),i = 1,2, such that
χ(n) = χ1 (n)χ2 (n) for all n.
Proof The first assertion follows immediately from the observations that
χ1 (n)χ2 (n) is totally multiplicative, that it vanishes if ( n,[q1 ,q2 ]) > 1, and
that it has period [ q1 ,q2 ]. As for the second assertion, we may suppose that
(n,q ) = 1. By the Chinese Remainder Theorem we see that
(Z/q Z)× ∼
=(Z/q1 Z)× ⊗ (Z/q2 Z)×
if ( q1 ,q2 ) = 1. Thus the result follows from Lemma 4.2. □
Our proof of Theorem 4.4 depends on Abel’s theorem that any finite abelian
group is isomorphic to the direct product of cyclic groups, but we can prove
Corollary 4.5 without appealing to this result, as follows. By the Chinese Re-
mainder Theorem we see that
(Z/q Z)× ∼
=
⨂
pα∥ q
(Z/pαZ)×.
If p is odd, then the reduced residue classes (mod pα) form a cyclic group; in
classical language we say there is a primitive root g. Thus if ( n, p) = 1, then
there is a unique ν (mod ϕ( pα)) such that gν ≡ n (mod pα). The number ν is | {
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118 Primes in arithmetic progressions: I
called the index of n, and is denoted ν = indg n. From Lemma 4.2 it follows
that the characters (mod pα), p > 2, are given by
χk (n) = e
(k indg n
ϕ( pα)
)
(4.16)
for ( n, p) = 1. W e obtain ϕ( pα) different characters by allowing k to assume
integral values in the range 1 ≤ k ≤ ϕ( pα). By Lemma 4.3 it follows that if q
is odd, then the general character (mod q )i sg i v e nb y
χ(n) = e
(∑
pα∥ q
k indg n
ϕ( pα)
)
(4.17)
for ( n,q ) = 1, where it is understood that k = k( pα) is determined (mod ϕ( pα))
and that g = g( pα) is a primitive root (mod pα).
The multiplicative structure of the reduced residues (mod 2 α) is more com-
plicated. For α = 1o r α = 2 the group is cyclic (of order 1 or 2, respectively),
and (4.16) holds as before. For α ≥ 3 the group is not cyclic, but if n is odd, then
there exist unique µ(mod 2) and ν(mod 2 α−2 ) such that n ≡ (−1)µ5ν (mod 2 α).
In group-theoretic terms this means that
(
Z/2αZ)× ∼
=C2 ⊗ C2α−2
when α ≥ 3. By Lemma 4.3 the characters in this case take the form
χ(n) = e
(j µ
2 + kν
2α−2
)
(4.18)
for odd n where j = 0o r1a n d1 ≤ k ≤ 2α−2 . Thus (4.17) holds if 8 ∤ q , but if
8|q , then the general character takes the form
χ(n) = e
⎛
⎜
⎝j µ
2 + kν
2α−2 +
∑
pα∥ q
p>2
ℓindg n
ϕ( pα)
⎞
⎟
⎠ (4.19)
when ( n,q ) = 1.
By definition, if f (n) is totally multiplicative, f (n) = 0 whenever ( n,q ) > 1,
and f (n) has period q , then f is a Dirichlet character (mod q ). It is useful to
note that the first condition can be relaxed.
Theorem 4.7If f is multiplicative, f (n) = 0 whenever (n,q ) > 1, and f has
period q , then f is a Dirichlet character modulo q .
ProofIt suffices to show that f is totally multiplicative. If ( mn ,q ) > 1, then
f (mn ) = f (m) f (n) since 0 = 0. Suppose that ( mn ,q ) = 1. Hence in partic-
ular ( m,q ) = 1, so that the map k ↦→n + kq (mod m) permutes the residue
classes (mod m). Thus there is a k for which n + kq ≡ 1 (mod m), and | {
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4.2 Dirichlet characters 119
consequently ( m,n + kq ) = 1. Then
f (mn ) = f (m(n + kq )) (by periodicity)
= f (m) f (n + kq ) (by multiplicativity)
= f (m) f (n) (by periodicity),
and the proof is complete. □
W e shall discuss further properties of Dirichlet characters in Chapter 9.
4.2.1 Exercises
1. Let G be a finite abelian group of order n. Let g1 ,g2 ,..., gn denote the
elements of G , and let χ1 (g),χ2 (g),...,χ n (g) denote the characters of G .
Let U = [uij ]b et h e n × n matrix with elements uij = χi (g j )/√n. Show
that UU ∗ = U ∗U = I , i.e., that U is unitary .
2. Show that for arbitrary real or complex numbers c1 ,..., cq ,
∑
χ
⏐
⏐
⏐
q∑
n=1
cn χ(n)
⏐
⏐
⏐
2
= ϕ(q )
q∑
n=1
(n,q )=1
|cn |2
where the sum on the left-hand side runs over all Dirichlet characters
χ(mod q ).
3. Show that for arbitrary real or complex numbers cχ,
q∑
n=1
⏐
⏐
⏐
∑
χ
cχχ(n)
⏐
⏐
⏐
2
= ϕ(q )
∑
χ
|cχ|2
where the sum over χ is extended over all Dirichlet characters (mod q ).
4. Let ( a,q ) = 1, and suppose that k is the order of a in the multiplicative group
of reduced residue classes (mod q ).
(a) Show that if χ is a Dirichlet character (mod q ), then χ(a)i sa kth root
of unity .
(b) Show that if z is a kth root of unity , then
1 + z +···+ zk−1 =
{ k if z = 1,
0 otherwise .
(c) Let ζ be a kth root of unity . By taking z = χ(a)/ζ, show that each kth
root of unity occurs precisely ϕ(q )/k times among the numbers χ(a)a s
χ runs over the ϕ(q ) Dirichlet characters (mod q ).
5. Let χ be a Dirichlet character (mod q ), and let k denote the order of χ in the
character group.
(a) Show that if (a,q ) = 1, then χ(a)i sa kth root of unity . | {
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120 Primes in arithmetic progressions: I
(b) Show that each kth root of unity occurs precisely ϕ(q )/k times among the
numbers χ(a)a s a runs over the ϕ(q ) reduced residue classes (mod q ).
6. Let χ be a character (mod q ) such that χ(a) =± 1 whenever ( a,q ) = 1, and
put S(χ) = ∑ q
n=1 nχ(n). Thus S(χ) is an integer.
(a) Show that if ( a,q ) = 1 then aχ(a)S(χ) ≡ S(χ) (mod q ).
(b) Show that there is an a such that ( a,q ) = 1 and ( aχ(a) − 1,q )|12.
(c) Deduce that 12 S(χ) ≡ 0 (mod q ).
In algebraic number fields we encounter not only Dirichlet characters, but
also characters of ideal class groups and of Galois groups. In addition, algebraic
number fields possessing one or more complex embeddings also have a further
kind of character, Hecke’sGr ¨ossencharaktere. In a sequence of exercises, be-
ginning with the one below , we develop the basic properties of these characters
for the Gaussian fieldQ(√
−1).
7. Let K be the Gaussian field,
K = Q
(√
−1
)
={ a + bi : a,b ∈ Q},
and let OK be the ring of algebraic integers in K ,
OK ={ a + bi : a,b ∈ Z}.
Elements α = a + bi ∈ K have a norm, N (α) = a2 + b2 , and we observe
that N (αβ) = N (α) N (β). An element αof a ring is a unit if αhas an inverse
in the ring. The ring OK has precisely four units, namely i k for k = 0,1,2,3.
T wo elements α,β ∈ OK are associates if α = uβ for some unit u. For each
integer m we define the Hecke Gr ¨ossencharakter
χm (α) =
{ e4mi arg α if α ̸=0,
0i f α = 0.
(a) Show that if α and β are associates then χm (α) = χm (β).
(b) Show that χm (αβ) = χm (α)χm (β) for all α and β in OK .
4.3 Dirichlet L -functions
Let χ be a character (mod q ). For σ> 1w ep u t
L (s,χ ) =
∞∑
n=1
χ(n)n−s . (4.20)
Since χ is totally multiplicative, by Theorem 1.9 we have
L (s,χ ) =
∏
p
(1 − χ( p) p−s )−1 (4.21) | {
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4.3 Dirichlet L -functions 121
for σ> 1. Thus we see that
L (s,χ0 ) =
∞∑
n=1
(n,q )=1
n−s = ζ(s)
∏
p|q
(
1 − p−s )
(4.22)
for σ> 1. By (4.14) we see that if χ ̸=χ0 , then
∑
1≤n≤kq
χ(n) = 0
for k = 1,2,3,... . Hence
⏐
⏐
⏐
⏐
⏐
∑
n≤x
χ(n)
⏐
⏐
⏐
⏐
⏐≤ q (4.23)
for any x , so that by Theorem 1.3, the series (4.20) converges for σ> 0. This
result is best possible since the terms in (4.20) do not tend to 0 when σ = 0. On
the other hand, we shall show in Chapter 10 that the function L (s,χ ) is entire
if χ ̸=χ0 .F o r σ> 1 we can take logarithms in (4.21), and differentiate, as in
Corollary 1.11, and thus we obtain
Theorem 4.8If χ ̸=χ0 , then L (s,χ ) is analytic for σ> 0. On the other
hand, the function L (s,χ0 ) is analytic in this half-plane except for a simple
pole at s = 1 with residue ϕ(q )/q . In either case,
log L (s,χ ) =
∞∑
n=2
/Lambda1 (n)
log n χ(n)n−s (4.24)
for σ> 1, and
− L ′
L (s,χ ) =
∞∑
n=1
/Lambda1 (n)χ(n)n−s . (4.25)
In these last formulæ we see how relations for L -functions parallel those
for the zeta functions. Indeed, when manipulating Dirichlet series formally , the
only property ofn−s that is used is that it is totally multiplicative. Hence all
such calculations can be made with n−s replaced by χ(n)n−s . For example, we
know that ∑ µ(n)2 n−s = ζ(s)/ζ(2s) for σ> 1. Hence formally
∞∑
n=1
µ(n)2 χ(n)n−s = L (s,χ )/L (2s,χ 2 ). (4.26)
Since |χ(n)n−s |≤ n−σ, this latter series is absolutely convergent whenever the
former one is, and by (4.21) we see that (4.26) holds for σ> 1. In fact, by a
theorem of Stieltjes (see Exercise 1.3.2), the identity (4.26) holds for σ> 1/2
if χ ̸=χ0 . | {
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122 Primes in arithmetic progressions: I
W e now use the identity (4.15) to capture a prescribed residue class. If
(a,q ) = 1, then
1
ϕ(q )
∑
χ
χ(a)χ(n) =
{ 1i f n ≡ a (mod q ),
0 otherwise (4.27)
where the sum is extended over all characters χ (mod q ). This is the multiplica-
tive analogue of (4.1). Hence if ( a,q ) = 1 then
∞∑
n=1
n≡a (q )
/Lambda1 (n)n−s = 1
ϕ(q )
∞∑
n=1
/Lambda1 (n)n−s ∑
χ
χ(a)χ(n)
= −1
ϕ(q )
∑
χ
χ(a) L ′
L (s,χ ) (4.28)
for σ> 1. As L (s,χ0 ) has a simple pole at s = 1, the function L ′
L (s,χ ) has a
simple pole at 1 with residue −1. Thus the term arising from χ0 on the right-hand
side above is
1
ϕ(q )(s − 1) + Oq (1) (4.29)
as s → 1+. This enables us to prove that there are infinitely many primes
p ≡ a (mod q ), provided that we can show that the terms from χ ̸=χ0 on the
right-hand side of (4.28) do not interfere with the main term (4.29). But L (s,χ )
is analytic for σ> 0, so that L ′
L (s,χ ) is analytic except at zeros of L (s,χ ).
Hence
lim
s→1+
L ′
L (s,χ ) = L ′
L (1,χ ) (4.30)
for χ ̸=χ0 , provided that L (1,χ ) ̸=0. Thus the following result lies at the
heart of the matter.
Theorem 4.9(Dirichlet) If χ is a character (mod q ) with χ ̸=χ0 , then
L (1,χ ) ̸=0.
Suppose that ( a,q ) = 1. Then the above, with (4.28), (4.29), and (4.30) give
the estimate
∞∑
n=1
n≡a (q )
/Lambda1 (n)n−s = 1
ϕ(q )(s − 1) + Oq (1) | {
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4.3 Dirichlet L -functions 123
as s → 1+. Consequently
∞∑
n=1
n≡a (q )
/Lambda1 (n)
n =∞ .
Here the contribution of the proper prime powers is
∑
pk ≡a (q )
k≥2
log p
pk ≤
∑
p
log p
∞∑
k=2
p−k =
∑
p
log p
p( p − 1) < ∞, (4.31)
and thus we have
Corollary 4.10(Dirichlet’s theorem) If (a,q ) = 1, then there are infinitely
many primes p ≡ a (mod q ), and indeed
∑
p≡a (q )
log p
p =∞ .
W e call a character real if all its values are real (i.e., χ(n) = 0o r ±1 for all
n). Otherwise a character is complex. A character is quadratic if it has order
2 in the character group: χ2 = χ0 but χ ̸=χ0 . Thus a quadratic character is
real, and a real character is either principal or quadratic. In Chapter 9 we shall
express quadratic characters in terms of the Kronecker symbol
(d
n
)
.
Proof of Theorem 4.9 W e treat quadratic and complex characters separately .
Case 1: Complex χ. From (4.24) we have
∏
χ
L (s,χ ) = exp
(∑
χ
∞∑
n=2
/Lambda1 (n)
log n χ(n)n−s
)
for σ> 1. By (4.15) this is
= exp
⎛
⎜
⎝ϕ(q )
∞∑
n=2
n≡1( q )
/Lambda1 (n)
log n n−s
⎞
⎟
⎠.
If we take s = σ> 1, then the sum above is a non-negative real number, and
hence we see that
∏
χ
L (σ,χ ) ≥ 1 (4.32)
for σ> 1. Now L (s,χ0 ) has a simple pole at s = 1, but the other L (s,χ )
are analytic at s = 1. Thus L (1,χ ) = 0 can hold for at most one χ, since
otherwise the product in (4.32) would tend to 0 as σ → 1+.I f χ is a character
(mod q ), then
χ is a character (mod q ), and χ ̸=χ if χ is complex. Moreover | {
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124 Primes in arithmetic progressions: I
L (s,χ ) = L (s,χ) by the Schwarz reflection principle, so that L (1,χ) = 0i f
L (1,χ ) = 0. Consequently L (1,χ ) ̸=0 for complex χ.
Case 2: Quadratic χ. Let r (n) = ∑
d |n χ(d ). Thus ∑ ∞
n=1 r (n)n−s =
ζ(s)L (s,χ ) for σ> 1, r (n) is multiplicative, and
r ( pα) =
⎧
⎪
⎪
⎨
⎪
⎪
⎩
1i f p | q ,
α + 1i f χ( p) = 1,
1i f χ( p) =− 1 and 2 | α,
0i f χ( p) =− 1 and 2 ∤ α.
Hence r (n) ≥ 0 for all n, and r (n2 ) ≥ 1 for all n. Suppose that L (1,χ ) = 0.
Then ζ(s)L (s,χ ) is analytic for σ> 0, and by Landau’s theorem (Theorem
1.7) the series ∑ r (n)n−s converges for σ> 0. But this is false, since
∞∑
n=1
r (n)n−1/2 ≥
∞∑
n=1
r (n2 )n−1 ≥
∞∑
n=1
n−1 =+ ∞ .
Hence L (1,χ ) ̸=0. Since L (σ,χ ) > 0 for σ> 1 when χ is quadratic, we see
in fact that L (1,χ ) > 0 in this case. □
By using the techniques of Chapter 2 we can prove more than the mere
divergence of the series in Corollary 4.10.
Theorem 4.11Suppose that χ is a non-principal Dirichlet character . Then
for x ≥ 2,
(a)
∑
n≤x
χ(n)/Lambda1 (n)
n ≪χ 1,
(b)
∑
p≤x
χ( p) log p
p ≪χ 1,
(c)
∑
p≤x
χ( p)
p = b(χ) + Oχ
(1
log x
)
,
(d)
∏
p≤x
(
1 − χ( p)
p
)−1
= L (1,χ ) + Oχ
(1
log x
)
where
b(χ) = log L (1,χ ) −
∑
pk
k>1
χ( pk )
kp k .
Proof W e show first that
∑
n≤x
χ(n) log n
n =− L ′(1,χ ) + Oq
(log x
x
)
. (4.33)
T o this end we put S(x ) = ∑
n≤x χ(n). Then from (4.23) we see that S(x ) ≪χ 1. | {
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4.3 Dirichlet L -functions 125
Thus the error term above is
∑
n>x
χ(n) log n
n =
∫ ∞
x
log u
u dS (u)
=− S(x ) log x
x −
∫ ∞
x
S(u)(1 − log u)u−2 du
≪χ
log x
x .
As log n = ∑
d |n /Lambda1 (d ), the left-hand side of (4.33) is
∑
md ≤x
/Lambda1 (d )χ(md )
md =
∑
d ≤x
/Lambda1 (d )χ(d )
d
∑
m≤x /d
χ(m)
m . (4.34)
Here the inner sum is of the form
∑
m≤y
χ(m)
m = L (1,χ ) −
∑
m>y
χ(m)
m ,
and this last sum is∫ ∞
y
u−1 dS (u) =− S( y)
y +
∫ ∞
y
S(u)u−2 du ≪χ y−1 .
Hence the right-hand side of (4.34) is
L (1,χ )
∑
d ≤x
/Lambda1 (d )χ(d )
d + Oχ
(
1
x
∑
d ≤x
/Lambda1 (d )
)
.
This last error term is ≪χ 1, and then (a) follows from (4.33) and the fact that
L (1,χ ) ̸=0. The derivation of (b) from (a), and of (c) from (b) proceeds as in
the proof of Theorem 2.7. Continuing as in that proof, we see from (c) that
∑
1<n≤x
/Lambda1 (n)χ(n)
n log n = c(χ) + Oχ
(1
log x
)
where
c(χ) = b(χ) +
∑
pk
k>1
χ( pk )
kp k .
W e let s → 1+ in (4.24), and deduce by Theorem 1.1 that c(χ) = log L (1,χ ).
T o complete the derivation of (d) it suffices to argue as in the proof of
Theorem 2.7. □
By forming a linear combination of these estimates as in (4.27) we obtain
Corollary 4.12 If (a,q ) = 1 and x ≥ 2, then
(a)
∑
n≤x
n≡a (q )
/Lambda1 (n)
n = 1
ϕ(q ) log x + Oq (1), | {
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126 Primes in arithmetic progressions: I
(b)
∑
p≤x
n≡a (q )
log p
p = 1
ϕ(q ) log x + Oq (1),
(c)
∑
p≤x
n≡a (q )
1
p = 1
ϕ(q ) log log x + b(q ,a) + Oq
(1
log x
)
,
(d)
∏
p≤x
n≡a (q )
(
1 − 1
p
)−1
= c(q ,a)(log x )1/ϕ(q )
(
1 + Oq
(1
log x
))
where
b(q ,a) = 1
ϕ(q )
(
C0 +
∑
p|q
log
(
1 − 1
p
)
+
∑
χ̸=χ0
χ(a) log L (1,χ )
)
−
∑
pk ≡a (q )
k>1
1
kp k
and
c(q ,a) =
(
eC0 ϕ(q )
q
∏
χ̸=χ0
(
L (1,χ )χ(a) ∏
p
(
1 − 1
p
)−χ( p) (
1 − χ( p)
p
)))1/ϕ(q )
.
Proof T o derive (a) from Theorem 4.11(a) we use (4.27) and the estimate
∑
n≤x
/Lambda1 (n)χ0 (n)
n = log x + Oq (1),
which follows from Theorem 2.7(a) since
∑
pk
p|q
log p
pk =
∑
p|q
log p
p − 1 ≪q 1.
W e derive (b) and (c) similarly from the corresponding parts of Theorem 4.11.
In the latter case we use the estimate
∑
p≤x
χ0 ( p)
p = log log x + b(χ0 ) + Oq
(1
log x
)
where
b(χ0 ) = C0 +
∑
p|q
log
(
1 − 1
p
)
−
∑
pk
k>1
χ0 ( pk )
kp k .
T o derive (d) we observe first that
∏
p≤x
(
1 − χ0 ( p)
p
)−1
=
∏
p≤x
p|q
(
1 − 1
p
)∏
p≤x
(
1 − 1
p
)−1
,
which by Theorem 2.7(e) is
= ϕ(q )
q
⎛
⎜
⎝
∏
p|q
p>x
(
1 − 1
p
)
⎞
⎟
⎠
−1
e−C0 (log x )
(
1 + O
(1
log x
))
. | {
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4.3 Dirichlet L -functions 127
Here each term in the product is 1 + O (1/x ), and the number of factors is
≤ ω(q ), so the product is 1 + Oq (1/x ), and hence the above is
= eC0 ϕ(q )
q (log x )
(
1 + Oq
(1
log x
))
.
T o complete the proof it suffices to combine this with Theorem 4.11(d)
in (4.27). □
4.3.1 Exercises
1. Let χ be a Dirichlet character (mod q ). Show that if σ> 1, then
(a)
∞∑
n=1
(−1)n−1 χ(n)n−s = (1 − χ(2)21−s )L (s,χ );
(b)
∞∑
n=1
d (n)2 χ(n)n−s = L (s,χ )4
L (2s,χ 2 ) .
2. (Mertens 1895a,b) Let r (n) = ∑
d |n χ(d ).
(a) Show that if χ is a non-principal character (mod q ), then
∑
n>x
χ(n)√n ≪χ
1√x .
(b) Show that if χ is a non-principal character (mod q ), then
∑
n≤x
r (n)
n1/2 = 2x 1/2 L (1,χ ) + Oχ(1).
(c) Recall that if χ is quadratic then r (n) ≥ 0 for all n, and that r (n2 ) ≥ 1.
Deduce that if χ is a quadratic character, then the left-hand side above
is ≫ log x .
(d) Conclude that if χ is a quadratic character, then L (1,χ ) > 0.
3. (Mertens 1897, 1899) For u ≥ 0, put f (u) = ∑
m≤u (1 − m/u).
(a) Show that f (u) ≥ 0, that f (u) is continuous, and that if u is not an
integer, then
f ′(u) = [u]([u] + 1)
2u2 ;
deduce that f is increasing.
(b) Show also that
f (u) = u
2 − 1
u
∫ u
0
{v} d v = u
2 − 1
2 + O (1/u) .
(c) Let r (n) = ∑
d |n χ(d ), and assume that χ is non-principal. Show that
∑
n≤x
r (n)(1 − n/x ) =
∑
d ≤x
χ(d ) f (x /d ) . | {
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128 Primes in arithmetic progressions: I
(d) Write ∑
d ≤x = ∑
d ≤y + ∑
y<d ≤x = S1 + S2 where 1 ≤ y ≤ x . Use
part (b) to show that S1 = 1
2 xL (1,χ ) + Oχ(x /y) + O ( y2 /x ).
(e) Use the results of part (a) to show that S2 ≪χ f (x /y).
(f) By making an appropriate choice of y, deduce that if χis a non-principal
character, then
∑
n≤x
r (n)(1 − n/x ) = x
2 L (1,χ ) + Oχ
(
x 1/3 )
.
(g) Argue that if χ is a quadratic character, then the left-hand side above
is ≫ x 1/2 ; deduce that L (1,χ ) > 0.
4. (Ingham 1929) Let f1 (n) and f2 (n) be totally multiplicative functions, and
suppose that | fi (n)|≤ 1 for all n.
(a) Show that if σ> 1, then
∞∑
n=1
(∑
d |n
f1 (d )
)(∑
d |n
f2 (d )
)
n−s
=
ζ(s)
(∞∑
n=1
f1 (n)n−s
)(∞∑
n=1
f2 (n)n−s
)(∞∑
n=1
f1 (n) f2 (n)n−s
)
∞∑
n=1
f1 (n) f2 (n)n−2s
=
∏
p
(
1 − f1 ( p) f2 ( p)
p2s
)
∏
p
(
1 − 1
ps
)(
1 − f1 ( p)
ps
)(
1 − f2 ( p)
ps
)(
1 − f1 ( p) f2 ( p)
ps
).
(b) By considering
F (s) =
∞∑
n=1
⏐
⏐
⏐
∑
d |n
χ(d )d −iu
⏐
⏐
⏐
2
n−s ,
show that L (1 + iu ,χ ) ̸=0.
5. Let π(x ; q ,a) denote the number of primes p ≡ a (mod q ) with p not
exceeding x . Similarly , let
ϑ(x ; q ,a) =
∑
p≤x
p≡a (q )
log p,ψ (x ; q ,a) =
∑
n≤x
n≡a (q )
/Lambda1 (n).
(a) Show that
ϑ(x ; q ,a) = ψ(x ; q ,a) + O
(
x 1/2 )
.
(b) Show that
π(x ; q ,a) = ϑ(x ; q ,a)
log x + O
( x
(log x )2
)
. | {
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4.3 Dirichlet L -functions 129
(c) Show that if x ≥ C , C ≥ 2, and ( a,q ) = 1, then
∑
x /C <p≤x
p≡a (q )
log p
p = log C
ϕ(q ) + Oq (1).
(d) Show that for any positive integer q there is a small number cq and a
large number Cq such that if x ≥ 2Cq and ( a,q ) = 1, then
∑
x /Cq <p≤x
p≡a (q )
log p
p > cq .
(e) Show that for any positive integer q there is a Cq such that if ( a,q ) = 1,
then
π(x ; q ,a) ≫q
x
log x
uniformly for x ≥ Cq .
(f) Show that if ( a,q ) = 1, then
lim inf
x →∞
π(x ; q ,a)
x /log x ≤ 1
ϕ(q ) , lim sup
x →∞
π(x ; q ,a)
x /log x ≥ 1
ϕ(q ) .
6. (a) Show that
ϑ(x ) ≤ π(x ) log x ≤ ϑ(x ) + O
(x
log x
)
for x ≥ 2.
(b) Let P denote a set of prime numbers, and put
πP(x ) =
∑
p≤x
p∈P
1,ϑ P(x ) =
∑
p≤x
p∈P
log p.
Show that
ϑP(x ) = πP(x ) log x + O
(x
log x
)
for x ≥ 2, where the implicit constant is absolute.
(c) Let
n =
∏
p≤y
p∈P
p .
Show that log n = ω(n) log y + O ( y/log y) for y ≥ 2.
(d) From now on, assume that ϑP(x ) ≫ x for all sufficiently large x , where
the implicit constant may depend on P. Show that log log n = log y +
OP(1). | {
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130 Primes in arithmetic progressions: I
(e) Deduce that
d (n) = n(log 2 +o(1))/log log n
as y →∞ .
7. Let R(n) denote the number of ordered pairs a,b such that a2 + b2 = n
with a ≥ 0 and b > 0. Also, let r (n) denote the number of such pairs for
which ( a,b) = 1. Finally , let χ−4 =
(−4
n
)
be the non-principal character
(mod 4). W e recall that if the prime factorization of n is written in the form
n = 2α ∏
pβ∥ n
p≡1 (4)
pβ ∏
q γ ∥ n
q ≡3 (4)
q γ,
then r (n) > 0 if and only if γ = 0 for all primes q and α ≤ 1. W e also
recall that
R(n) =
∑
d 2 |n
r (n/d 2 ) =
∑
d |n
χ−4 (d ) =
{ ∏
p (β + 1) if 2 |γ for all q ,
0 otherwise.
(a) Show that ∑ ∞
n=1 R(n)n−s = ζ(s)L (s,χ−4 ) for σ> 1.
(b) Show that ∑ ∞
n=1 r (n)n−s = ζ(s)L (s,χ−4 )/ζ(2s) for σ> 1.
(c) Show that if x ≥ 0 and y ≥ 2, then
card{n ∈ (x ,x + y]: r (n) > 0}≪ y
√log y .
(d) Show that
card{n ≤ x : R(n) > 0}≪ x√log x
for x ≥ 2.
(e) Suppose that n is of the form
n =
∏
p≤y
p≡1 (4)
p.
Thus log n = ϑ( y; ,4,1) ≍ y for y ≥ 5, and hence log y = log log n +
O (1). Show that for such n,
R(n) = n(log 2 +o(1))/log log n .
In the above it is noteworthy that although R(n) ≤ d (n) for all n, that
R(n) is usually 0 and has a smaller average value (cf. Exercise 2.1.9)
than d (n) (cf. Theorem 2.3), the maximum order of magnitude of R(n)
is the same as for d (n). | {
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4.3 Dirichlet L -functions 131
8. Let K = Q(
√−1) be the Gaussian field, OK ={ a + ib : a,b ∈ Z} the ring
of integers in K . Ideals a in OK are principal, a = (a + ib ), and have norm
N (a) = a2 + b2 .
(a) Explain why the number of ideals a with N (a) ≤ x is π
4 x + O (x 1/2 ).
(b) For σ> 1, let ζK (s) = ∑
a N (a)−s be the Dedekind zeta function of
K . Show that ζK (s) = ζ(s)L
(
s,χ−4
)
.
(c) For the Gaussian field K , show that N (ab) = N (a) N (b). (This is true
in any algebraic number field.)
(d) Assume that ideals in K factor uniquely into prime ideals. (This is true
in any algebraic number field, and is particularly easy to establish for
the Gaussian field since it has a division algorithm.) Deduce that if
σ>1, then
ζK (s) =
∏
p
(
1 − 1
N (p)
)−1
where the product runs over all prime ideals p in OK .
(e) Define a function µ(a) = µK (a) in such a way that
1
ζK (s) =
∑
a
µ(a)
N (a)s
for σ> 1.
(f) Let a and b be given ideals. Show that
∑
d|a
d|b
µ(d) =
{ 1 if gcd( a,b) = 1,
0 otherwise .
(g) Among pairs a, b of ideals with N (a) ≤ x , N (b) ≤ x , show that the
probability that gcd( a,b) = 1i s
1
ζK (2) + O
(
x −1/2 )
= 6
π2 L
(
2,χ−4
)+ O
(
x 1/2 )
.
9. (Erd ˝ os 1946, 1949, 1957, V aughan 1974, Saffari, unpublished, but see
Bateman, Pomerance & V aughan 1981; cf. Exercise 2.3.7) Let /Phi1 q (z) =∏
d |q (zd − 1)µ(q /d ) denote the q th cyclotomic polynomial. Suppose that
q =
∏
p≤y
p≡±2 (5)
p
where y is chosen so that ω(q ) is odd.
(a) Show that if d |q and ω(d ) is even, then |e(d /5) − 1|=| e(1/5) − 1|.
(b) Show that if d |q and ω(d ) is odd, then |e(d /5) − 1|=| e(2/5) − 1|.
(c) Deduce that |/Phi1 q (e(1/5))|=| e(1/5) + 1|d (q )/2 . | {
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132 Primes in arithmetic progressions: I
(d) Deduce that /Phi1 q (z) has a coefficient whose absolute value is at least
exp
(
q (log 2 −ε)/log log q )
if y > y0 (ε).
10. Gr ¨ossencharaktere for Q(√−1), continued from Exercise 4.2.7.
(a) For σ> 1 put
L (s,χm ) =
∑
α∈OK
′
χm (α) N (α)−s = 1
4
∑
a,b∈Z
(a,b)̸=(0,0)
χm (a + bi )(a2 + b2 )−s
where ∑ ′
α denotes a sum over unassociated members of OK . Show
that the above sum is absolutely convergent in this half-plane.
(b) W e recall that members of OK factor uniquely into Gaussian primes.
Also, the Gaussian primes are obtained by factoring the rational primes:
The prime 2 ramifies, 2= i 3 (1 + i )2 , the rational primes p ≡ 1 (mod 4)
split into two distinct Gaussian primes, p = (a + bi )(a − bi ), and the
rational primes q ≡ 3 (mod 4) are inert. Show that
L (s,χm ) =
∏
p
(1 − χm (p) N (p)−s )−1
for σ> 1 where the product is over an unassociated family of Gaussian
primes p.
(c) By grouping associates together, show that if 4 ∤ m, then the sum
∑
a,b∈Z
(a,b)̸=(0,0)
emi arg(a+bi ) (a2 + b2 )−s
vanishes identically for σ> 1.
(d) For 0 ≤ θ ≤ 2π, put N (x ; θ) = card{(a,b) ∈ Z2 : a2 + b2 ≤ x ,0 <
arg(a + bi ) ≤ θ}. Show that for x ≥ 1,
N (x ; θ) = θ
2 x + O
(
x 1/2 )
uniformly in θ.
(e) Show that if m ̸=0, then
∑
a2 +b2 ≤x
a>0,b≥0
χm (a + bi ) =
∫ π/2
0
e4mi θ dN (x ; θ) ≪| m|x 1/2 .
(f) Show that if m ̸=0, then the Dirichlet series L (s,χm ) is convergent for
σ> 1/2.
(g) Show that L (s,χm ) and L (s,χ−m ) are identically equal, and hence that
L (σ,χm ) ∈ R for σ> 1/2. | {
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4.4 Notes 133
4.4 Notes
Section 4.1. Ramanujan’s sum was introduced by Ramanujan (1918). Incredi-
bly , both Hardy and Ramanujan missed the fact thatcq (n) be written in closed
form: The formula on the extreme right of (4.7) is due to H ¨ older (1936). Nor-
mally one would say that a functionf is even if f (x ) = f (−x ). However, in
the present context, an arithmetic function f with period q is said to be even
if f (n) is a function only of ( n,q ). Thus cq (n) is an even function. The space
of almost-even functions is rather small, but includes several arithmetic func-
tions of interest. For such functions one may hope for a representation in the
formf (n) = ∑ ∞
q =1 aq cq (n), called a Ramanujan expansion . For a survey of the
theory of such expansions, see Schwarz (1988). Hildebrand (1984) established
definitive results concerning the pointwise convergence of Ramanujan expan-
sions. An appropriate Parseval identity has been established for mean-square
summable almost-even functions; see Hildebrand, Schwarz & Spilker (1988).
Section 4.2. The first instance of characters of a non-cyclic group occurs in
Gauss’s analysis of the genus structure of the class group of binary quadratic
forms. The quotient of the class group by the principal genus is isomorphic to
C2 ⊗ C2 ⊗···⊗ C2 , and the associated characters are given by Kronecker’s
symbol. Dirichlet (1839) defined the Dirichlet characters for the multiplicative
group (Z/q Z)× of reduced residues modulo q , and the same technique suffices
to construct the characters for any finite Abelian group. More generally , if
G is a group, then a homomorphism h : G −→ GL (n,C) is called a group
representation, and the trace of h(g)i sa group character . Note that if a and
b are conjugate elements of G , say a = gbg −1 , then h(a) and h(b) are similar
matrices. Hence they have the same eigenvalues, and in particular tr h(a) =
tr h(b). Thus a group character is constant on conjugacy classes. In the case of a
finite Abelian group it suffices to take n = 1, and in this case the representation
and its trace are essentially the same. For an introduction to characters in a
wider setting, see Serre (1977).
Section 4.3. Dirichlet (1837a,b,c) first proved Corollary 4.10 in the case that
q is prime. The definition of the Dirichlet characters is not difficult in that case,
since the multiplicative group ( Z/pZ)× of reduced residues is cyclic. The most
challenging part of the proof is to show that L (1,χ ) when χ is the Legendre
symbol (mod p). If p ≡ 3 (mod 4), then
p−1∑
a=1
a
(a
p
)
≡
p−1∑
a=1
a = p( p − 1)
2 ≡ 1 (mod 2) ,
and hence the sum on the left is non-zero. It follows by (9.9) that L (1,χp ) ̸=0
in this case. If p ≡ 1 (mod 4), then one has the identity of Exercise 9.3.7(c), | {
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134 Primes in arithmetic progressions: I
and thus to show that L (1,χp ) ̸=0 it suffices to show that Q ̸=1. Dirichlet
established this by means of Gauss’s theory of cyclotomy . Accounts of this are
found in Davenport (2000, Sections 1–3), and in Narkiewicz (2000, pp. 64–
65). An alternative proof thatQ ̸=1 was given more recently by Chowla &
Mordell (1961) (cf. Exercise 9.3.8). In order to prove that L (1,χ ) ̸=0 when χ
is quadratic, Dirichlet related L (1,χ ) to the class number of binary quadratic
forms. Suppose that d is a fundamental quadratic discriminant, and put χd (n) =(d
n
)
, the Kronecker symbol (as discussed in Section 9.3). Suppose first that
d > 0. Among the solutions of Pell’s equation x 2 − dy 2 = 4, let ( x0 ,y0 )b e
the solution with x0 > 0, y0 > 0, and y0 minimal, and put η = 1
2 (x0 + y0
√
d ).
Dirichlet showed that
L (1,χd ) = h log η√
d
(4.35)
where h is the number of equivalence classes of binary quadratic forms with
discriminant d . Since h ≥ 1 and y0 ≥ 1, it follows that L (1,χd ) ≫ (log d )/
√
d
in this case. Now suppose that d < 0 and that w denotes the number of auto-
morphs of the positive definite binary quadratic forms of discriminant d (i.e.,
w = 6i f d =− 3, w = 4i f d =− 4, and w = 2i f d < −4). Dirichlet showed
that
L (1,χd ) = 2πh
w√−d
. (4.36)
Thus L (1,χd ) ≥ π/√−d when d < −4.
Our treatment of quadratic characters in the proof of Theorem 4.9 is due
to Landau (1906). Mertens (1895a,b, 1897, 1899) gave two elementary proofs
thatL (1,χ ) > 0 when χ is quadratic; cf. Exercises 2.4.2 and 2.4.3. For a
definitive account of Mertens’ methods, see Bateman (1959). Other proofs
have been given by T eege (1901), Gel’fond & Linnik (1962, Chapter 3 Section
2), Bateman (1966, 1997), Pintz (1971), and Monsky (1993). See also Baker,
Birch & Wirsing (1973).
4.5 References
Baker, A., Birch, B. J., & Wirsing, E. A. (1973). On a problem of Chowla, J. Number
Theory 5, 224–236.
Bateman, P . T . (1959). Theorems implying the non-vanishing of ∑ χ(m)m−1 for real
residue-characters, J. Indian Math. Soc. 23, 101–115.
(1966). Lower bounds for ∑ h(m)/m for arithmetical function h similar to real
residue characters, J. Math. Anal. Appl. 15, 2–20. | {
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4.5 References 135
(1997). A theorem of Ingham implying that Dirichlet’s L -functions have no zeros
with real part one, Enseignement Math . (2) 43, 281–284.
Bateman, P . T ., Pomerance, C., & V aughan, R. C. (1981). On the size of the coefficients
of the cyclotomic polynomial, Coll. Math. Soc. J. Bolyai , pp. 171–202.
Carmichael, R. (1932). Expansions of arithmetical functions in infinite series, Proc.
London Math. Soc . (2) 34, 1–26.
Chowla, S. & Mordell, L. J. (1961). Note on the nonvanishing of L (1), Proc. Amer .
Math. Soc . 12, 283–284.
Davenport, H. (2000). Multiplicative Number Theory , Graduate T exts Math. 74. New
Y ork: Springer-V erlag.
Delange, H. (1976). On Ramanujan expansions of certain arithmetical functions, Acta
Arith. 31, 259–270.
Dirichlet, P . G. L. (1839a). Sur l’usage des int´ etrales d´ efinies dans la sommation des
s´eries finies ou infinies, J. Reine Angew . Math. 17, 57–67; W erke, V ol. 1, Berlin:
Reimer, 1889, pp. 237–256.
(1837b). Beweis eines Satzes ueber die arithmetische Progression, Ber V erhandl. Kgl.
Preuss. Akad. Wiss ., 108–110; W erke, V ol. 1, Berlin: Reimer, 1889, pp. 307–312.
(1837c). Beweis des Satzes, dass jede unbegrenzte arithmetische Progression, deren
erstes Glied und Differenz ganze Zahlen ohne gemeinschaftlichen Factor sind, un-
endlich viele Primzahlen enth¨ alt,Abhandl. Kgl. Preuss. Akad. Wiss . 45–81; W erke,
V ol. 1, Berlin: Reimer, 1889, pp. 313–342.
(1839). Recherches sur diverses applications de l’analyse infinit´ esimale a la th´ eorie
des nombres, J. Reine Angew . Math. 19, 324–369; W erke, V ol. 1, Berlin: Reimer,
1889, pp. 411–496.
Erd ˝ os, P . (1946). On the coefficients of the cyclotomic polynomial, Bull. Amer . Math.
Soc. 52, 179–184.
(1949). On the coefficients of the cyclotomic polynomial, P ortugal. Math. 8, 63–71.
(1957). On the growth of the cyclotomic polynomial in the interval (O, 1). Proc.
Glasgow Math. Assoc. 3, 102–104.
Friedman, A. (1957). Mean-values and polyharmonic polynomials, Michigan Math. J .
4, 67–74.
Gel’fond, A. O. & Linnik, Ju. V . (1962). Elementary Methods in Analytic Number
Theory. Moscow: Gosudarstv . Izdat. Fiz.-Mat. Lit.; English translation, Chicago:
Rand McNally , 1965; English translation, Cambridge: M. I. T . Press, 1966.
Grytczuk, A. (1981). An identity involving Ramanujan’s sum, Elem. Math. 36, 16–17.
Hildebrand, A. (1984). ¨Uber die punkweise Konvergenz von Ramanujan-Entwicklungen
zahlentheoretischer Funktionen, Acta Arith. 44, 108–140.
Hildebrand, A., Schwarz, W ., & Spilker, J. (1988). Still another proof of Parseval’s
equation for almost-even arithmetical functions, Aequationes Math. 35, 132–139.
H¨older, O. (1936). Zur Theorie der Kreisteilungsgleichung, Prace Mat. –Fiz. 43, 13–23.
Ingham, A. E. (1929). Note on Riemann’s ζ-function and Dirichlet’s L -functions,
J. London Math. Soc. 5, 107–112.
Landau, E. (1906). ¨Uber das Nichtverschwinden einer Dirichletschen Reihe, Sitzungsber .
Akad. Wiss. Berlin 11, 314–320; Collected W orks, V ol. 2. Essen: Thales, 1986, pp.
230–236.
Mertens, F . (1895a). ¨Uber Dirichletsche Reihen, Sitzungsber . Kais. Akad. Wiss. Wien
104, 2a, 1093–1153. | {
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136 Primes in arithmetic progressions: I
(1895b). ¨Uber das Nichtverschwinden Dirichletscher Reihen mit reelen Gliedern,
Sitzber . Kais. Akad. Wiss. Wien 104, 2a, 1158–1166.
(1897). ¨Uber Multiplikation und Nichtverschwinden Dirichlet’scher Reihen, J. Reine
Angew . Math. 117, 169–184.
(1899). Eine asymptotische Aufgabe, Sitzber . Kais. Akad. Wiss. Wien 108, 2a, 32–37.
Monsky , P . (1993). Simplifying the proof of Dirichlet’s theorem, Amer . Math. Monthly
100, 861–862.
Narkiewicz, W . (2000). The Development of Prime Number Theory , Berlin: Springer-
V erlag.
Pintz, J. (1971). On a certain point in the theory of Dirichlet’s L -functions, I,II, Mat.
Lapok 22, 143–148; 331–335.
Ramanujan, S. (1918). On certain trigonometrical sums and their applications in the
theory of numbers, Trans. Cambridge Philos. Soc. 22, 259–276; Collected papers .
Cambridge: Cambridge University Press, 1927, pp. 179–199.
Redmond, D. (1983). A remark on a paper: “ An identity involving Ramanujan’s sum”
by A. Grytczuk, Elem. Math. 38, 17–20.
Reznick, B. (1995). Some constructions of spherical 5-designs, Linear Algebra Appl. ,
226/228, 163–196.
Schwarz, W . (1988). Ramanujan expansions of arithmetical functions, Ramanujan revis-
ited, Proc. Centenary Conference (Urbana, June 1987). Boston: Academic Press,
pp. 187–214.
Serre, J.–P . (1977). Linear representation of finite groups , Graduate T exts Math. 42.
New Y ork: Springer-V erlag.
T eege, H. (1901). Beweis, daß die unendliche Reihe ∑ n=∞
n=1
(
p
n
)
1
n einen positiven von
Null verschiedenen W ert hat, Mitt. Math. Ges. Hamburg 4, 1–11.
V aughan, R. C. (1974). Bounds for the coefficients of cyclotomic polynomials, Michigan
Math. J. 21, 289–295.
Wintner, A. (1943). Eratosthenian averages . Baltimore: W averly Press. | {
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5
Dirichlet series: II
5.1 The inverse Mellin transform
In Chapter 1 we saw that we can express a Dirichlet series α(s) = ∑ ∞
n=1 an n−s
in terms of the coefficient sum A(x ) = ∑
n≤x an , by means of the formula
α(s) = s
∫ ∞
1
A(x )x −s−1 dx , (5.1)
which holds for σ> max(0,σc ). This is an example of a Mellin transform. In
the reverse direction, Perron’s formula asserts that
A(x ) = 1
2πi
∫ σ0 +i ∞
σ0 −i ∞
α(s) x s
s ds (5.2)
for σ0 > max(0,σc ). This is an example of an inverse Mellin transform.
T o understand why we might expect that (2) should be true, note that if
σ0 > 0, then by the calculus of residues
1
2πi
∫ σ0 +i ∞
σ0 −i ∞
ys ds
s =
{ 1i f y > 1,
0i f 0 < y < 1. (5.3)
Thus we would expect that
1
2πi
∫ σ0 +i ∞
σ0 −i ∞
α(s) x s
s ds =
∑
n
an
2πi
∫ σ0 +i ∞
σ0 −i ∞
(x
n
)s ds
s =
∑
n≤x
an . (5.4)
The interchange of limits here is difficult to justify , since α(s) may not be
uniformly convergent, and because the integral in (5.3) is neither uniformly nor
absolutely convergent. Moreover, ifx is an integer, then the term n = x in (5.4)
gives rise to the integral (5.3) with y = 1, and this integral does not converge,
although its Cauchy principal value exists:
lim
T →∞
1
2πi
∫ σ0 +iT
σ0 −iT
ds
s = 1
2 (5.5)
for σ0 > 0. W e now give a rigorous form of Perron’s formula.
137 | {
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138 Dirichlet series: II
Theorem 5.1 (Perron’s formula) If σ0 > max(0,σc ) and x > 0, then
∑
n≤x
′
an = lim
T →∞
1
2πi
∫ σ0 +iT
σ0 −iT
α(s) x s
s ds .
Here ∑ ′ indicates that if x is an integer , then the last term is to be counted with
weight 1/2.
Proof Choose N so large that N > 2x + 2, and write
α(s) =
∑
n≤N
an n−s +
∑
n>N
an n−s = α1 (s) + α2 (s),
say . By (5.4), modified in recognition of (5.5), we see that
∑
n≤x
′
an = lim
T →∞
1
2πi
∫ σ0 +iT
σ0 −iT
α1 (s) x s
s ds ;
here the justification is trivial since there are only finitely many terms. As for
α2 (s), we observe that
α2 (s) =
∫ ∞
N
u−s d ( A(u) − A( N )) = s
∫ ∞
N
( A(u) − A( N ))u−s−1 du .
But A(u) − A( N ) ≪ uθ for θ> max(0,σc ), and hence
α2 (s) ≪
(
1 + |s|
σ − θ
)
N θ−σ
for σ>θ> max(0,σc ). Implicit constants here and in the rest of this proof
may depend on the an . Hence
∫ T ±iT
σ0 ±iT
α2 (s) x s
s ds ≪ N θ
σ0 − θ
∫ ∞
σ0
(x
N
)σ
d σ ≪ N θ
σ0 − θ
(x /N )σ0
log N /x ,
and ∫ T +iT
T −iT
α2 (s) x s
s ds ≪ N θ(x /N )σ0
for large T . W e take θ so that σ0 >θ> max(0,σc ). Hence by Cauchy’s theorem
∫ σ0 +iT
σ0 −iT
=
∫ T −iT
σ0 −iT
+
∫ T +iT
T −iT
+
∫ σ0 +iT
T +iT
≪ x σ0 N θ−σ0 .
On combining our estimates, we see that
lim sup
T →∞
⏐
⏐
⏐
⏐
∑
n≤x
′
an − 1
2πi
∫ σ0 +iT
σ0 −iT
α(s) x s
s ds
⏐
⏐
⏐
⏐≪ x σ
0 N θ−σ0 .
Since this holds for arbitrarily large N , it follows that the lim sup is 0, and the
proof is complete. □ | {
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5.1 The inverse Mellin transform 139
W e have now established a precise relationship between (5.1) and (5.2), but
Theorem 5.1 is not sufficiently quantitative to be useful in practice. W e express
the error term more explicitly in terms of thesine integral
si(x ) =−
∫ ∞
x
sin u
u du .
By integration by parts we see that si( x ) ≪ 1/x for x ≥ 1, and hence that
si(x ) ≪ min(1,1/x ) (5.6)
for x > 0. W e also note that
si(x ) + si(−x ) =−
∫ +∞
−∞
sin u
u du =− π. (5.7)
Theorem 5.2 If σ0 > max(0,σa ) and x > 0, then
∑
n≤x
′
an = 1
2πi
∫ σ0 +iT
σ0 −iT
α(s) x s
s ds + R (5.8)
where
R = 1
π
∑
x /2<n<x
an si
(
T log x
n
)
− 1
π
∑
x <n<2x
an si
(
T log n
x
)
+ O
(
4σ0 + x σ0
T
∑
n
|an |
nσ0
)
.
Proof Since the series α(s) is absolutely convergent on the interval [ σ0 −
iT ,σ0 + iT ], we see that
1
2πi
∫ σ0 +iT
σ0 −iT
α(s) x s
s ds =
∑
n
an
1
2πi
∫ σ0 +iT
σ0 −iT
(x
n
)s ds
s .
Thus it suffices to show that
1
2πi
∫ σ0 +iT
σ0 −iT
ys ds
s =
⎧
⎪
⎪
⎨
⎪
⎪
⎩
1 + O ( yσ0 /T )i f y ≥ 2,
1 + 1
π si(T log y) + O (2σ0 /T )i f 1 ≤ y ≤ 2,
− 1
π si(T log 1 /y) + O (2σ0 /T )i f 1 /2 ≤ y ≤ 1,
O ( yσ0 /T )i f y ≤ 1/2
(5.9)
for σ0 > 0.
T o establish the first part of this formula, suppose that y ≥ 2, and let C be
the piecewise linear path from −∞ − iT to σ0 − iT to σ0 + iT to −∞ + iT .
Then by the calculus of residues we see that
1
2πi
∫
C
ys ds
s = 1, | {
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140 Dirichlet series: II
since the integrand has a pole with residue 1 at s = 0. In addition,
∫ σ0 ±iT
−∞±iT
ys ds
s =
∫ σ0
−∞
yσ±iT
σ ± iT d σ ≪ 1
T
∫ σ0
−∞
yσ d σ = yσ0
T log y ≪ yσ0
T ,
so we have (5.9) in the case y ≥ 2. The case y ≤ 1/2 is treated similarly , but
the contour is taken to the right, and there is no residue.
Suppose now that 1 ≤ y ≤ 2, and take C to be the closed rectangular path
from σ0 − iT to σ0 + iT to iT to −iT to σ0 − iT , with a semicircular inden-
tation of radius ε at s = 0. Then by Cauchy’s theorem
1
2πi
∫
C
ys ds
s = 0.
W e note that∫ σ0 ±iT
±iT
ys ds
s ≪ 1
T
∫ σ0
0
yσ d σ ≤ 1
T
∫ σ0
0
2σ d σ ≪ 2σ0
T .
The integral around the semicircle tends to 1 /2a s ε → 0, and the remaining
integral is
1
2πi lim
ε→0
(∫ iT
i ε
+
∫ −i ε
−iT
)
ys ds
s = 1
2πi lim
ε→0
∫ T
ε
(
yit − y−it )dt
t
= 1
π
∫ T log y
0
sin v d v
v
= 1
2 + 1
πsi(T log y)
by (5.7). This gives (5.9) when 1 ≤ y ≤ 2 and the case 1 /2 ≤ y ≤ 1 is treated
similarly . □
In many situations, Theorem 5.2 contains more information than is really
needed – it is often more convenient to appeal to the following less precise result.
Corollary 5.3In the situation of Theorem 5.2,
R ≪
∑
x /2<n<2x
n̸=x
|an | min
(
1, x
T |x − n|
)
+ 4σ0 + x σ0
T
∞∑
n=1
|an |
nσ0
.
Proof From (5.6) we see that
si(T | log n/x |) ≪ min
(
1, 1
T | log n/x |
)
.
But n/x = 1 + (n − x )/x and | log(1 + δ)|≍| δ| uniformly for −1/2 ≤ δ ≤ 1,
so the above is
≍ min
(
1, x
T |x − n|
)
if x /2 ≤ n ≤ 2x . Thus the stated bound follows from Theorem 5.2. □ | {
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5.1 The inverse Mellin transform 141
In classical harmonic analysis, for f ∈ L1 (T) we define Fourier coefficients
ˆf (k) =
∫1
0 f (x )e(−kα) d α, and we expect that the Fourier series ∑ ˆf (k)e(kα)
provides a useful formula for f (α). As it happens, the Fourier series may
diverge, or converge to a value other than f (α), but for most f a satisfactory
alternative can be found. For example, if f is of bounded variation, then
f (α−) + f (α+)
2 = lim
K →∞
K∑
−K
ˆf (k)e(kα).
A sharp quantitative form of this is established in Appendix D.1. Analogously ,
iff ∈ L 1 (R), then we can define the Fourier transform of f ,
ˆf (t ) =
∫ +∞
−∞
f (x )e(−tx ) dx , (5.10)
and we expect that
f (x ) =
∫ +∞
−∞
ˆf (t )e(tx ) dt . (5.11)
As in the case of Fourier series, this may fail, but it is not difficult to show that
iff is of bounded variation on [ − A, A] for every A, then
f (α−) + f (α+)
2 = lim
T →∞
∫ T
−T
ˆf (t )e(tx ) dt . (5.12)
The relationship between (5.1) and (5.2) is precisely the same as between
(5.10) and (5.11). Indeed, if we takef (x ) = A(e2πx )e−2πσx , then f ∈ L 1 (R)b y
Theorem 1.3, and by changing variables in (5.1) we find that
ˆf (t ) = α(σ + it )
2π(σ + it ) .
Thus (5.2) is equivalent to (5.11), and an appeal to (5.12) provides a second
(real variable) proof of Theorem 5.1.
In general, if
F (s) =
∫ ∞
0
f (x )x s−1 dx , (5.13)
then we say that F (s) is the Mellin transform of f (x ). By (5.10) and (5.11) we
expect that
f (x ) = 1
2πi
∫ σ0 +i ∞
σ0 −i ∞
F (s)x −s ds , (5.14)
and when this latter formula holds we say that f is the inverse Mellin transform
of F . Thus if A(x ) is the summatory function of a Dirichlet series α(s), then
α(s)/s is the Mellin transform of A(1/x ) for σ> max(0,σc ), and Perron’s
formula (Theorem 5.1) asserts that if σ0 > max(0,σc ), then A(1/x ) is the inverse | {
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142 Dirichlet series: II
Mellin transform of α(s)/s. Further instances of this pairing arise if we take a
weight function w(x ), and form a weighted summatory function
Aw(x ) =
∞∑
n=1
an w(n/x ).
Let K (s) denote the Mellin transform of w(x ),
K (s) =
∫ ∞
0
w(x )x s−1 dx .
Then we expect that
α(s)K (s) =
∫ ∞
0
Aw(x )x −s−1 dx , (5.15)
and that
Aw(x ) = 1
2πi
∫ σ0 +i ∞
σ0 −i ∞
α(s)K (s)x s ds . (5.16)
Alternatively , we may start with a kernel K (s), and define the weight w(x )
to be its inverse Mellin transform. The precise conditions under which these
identities hold depends on the weight or kernel; we mention several important
examples.
1. Ces ` aro weights.For a positive integer k, put
Ck (x ) = 1
k!
∑
n≤x
an (x − n)k . (5.17)
Then Ck (x ) =
∫x
0 Ck−1 (u) du for k ≥ 1 where C0 (x ) = A(x ), and hence
Ck (x ) ≪ x θ for θ> k + max(0,σc ). (The implicit constant here may depend
on k,o n θ, and on the an .) By integrating (5.1) by parts repeatedly , we see
that
α(s) = s(s + 1) ··· (s + k)
∫ ∞
1
Ck (x )x −s−k−1 dx (5.18)
for σ> max(0,σc ). By following the method used to prove Theorem 5.1, it
may also be shown that
Ck (x ) = 1
2πi
∫ σ0 +i ∞
σ0 −i ∞
α(s) x s+k
s(s + 1) ··· (s + k) ds (5.19)
when x > 0 and σ0 > max(0,σc ). Here the critical step is to show that if y ≥ 1
and σ0 > 0, then
1
2πi
∫ σ0 +i ∞
σ0 −i ∞
ys
s(s + 1) ··· (s + k) ds =
k∑
j =0
Res
( ys
s(s + 1) ··· (s + k)
⏐
⏐
⏐
⏐
s=− j | {
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5.1 The inverse Mellin transform 143
by the calculus of residues; this is
=
k∑
j =0
(−1) j y− j
j !(k − j )! = 1
k! (1 − 1/y)k
by the binomial theorem.
2. Riesz typical means.For positive integers k and positive real x put
Rk (x ) = 1
k!
∑
n≤x
an (log x /n)k . (5.20)
Then Rk (x ) =
∫x
0 Rk−1 (u)/ud u where R0 (x ) = A(x ), so that Rk (x ) ≪ x θ for
θ> max(0,σc ). (The implicit constant here may depend on k,o n θ, and on the
an .) By integrating (5.1) by parts repeatedly we see that
α(s) = sk+1
∫ ∞
1
Rk (x )x −s−1 dx (5.21)
for σ> max(0,σc ). By following the method used to prove Theorem 5.1 we
also find that
Rk (x ) = 1
2πi
∫ σ0 +i ∞
σ0 −i ∞
α(s) x s
sk+1 ds (5.22)
when x > 0 and σ0 > max(0,σc ). Here the critical observation is that if y ≥ 1
and σ0 > 0, then
1
2πi
∫ σ0 +i ∞
σ0 −i ∞
ys
sk+1 ds = Res
(ys
sk+1
⏐
⏐
⏐
⏐
s=0
= 1
k! (log y)k .
3. Abelian weights. For σ> 0w eh a v e
Ŵ(s) =
∫ ∞
0
e−u us−1 du = ns
∫ ∞
0
e−nx x s−1 dx .
W e multiply by an n−s and sum, to find that
α(s)Ŵ(s) =
∫ ∞
0
P (x )x s−1 dx (5.23)
where
P (x ) =
∞∑
n=1
an e−nx . (5.24)
These operations are valid for σ> max(0,σa ), but by partial summation
P (x ) ≪ x −θ as x → 0+ for θ> max(0,σc ), so that the integral in (5.23) is
absolutely convergent in the half-plane σ> max(0,σc ). Hence the integral is
an analytic function in this half-plane, so that by the principle of uniqueness | {
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144 Dirichlet series: II
of analytic continuation it follows that (5.23) holds for σ> max(0,σc ). In the
opposite direction,
P (x ) = 1
2πi
∫ σ0 +i ∞
σ0 −i ∞
α(s)Ŵ(s)x −s ds (5.25)
for x > 0, σ> max(0,σc ). T o prove this we recall from Theorem 1.5 that
α(s) ≪ τ uniformly for σ ≥ ε+ max(0,σc ), and from Stirling’s formula
(Theorem C.1) we see that |Ŵ(s)|≍ e− π
2 |t ||t |σ−1/2 as |t |→∞ with σ bounded.
Thus the value of the integral is independent of σ0 , and in particular we may
assume that σ0 > max(0,σa ). Consequently the terms in α(s) can be integrated
individually , and it suffices to appeal to Theorem C.4.
The formulæ (5.23) and (5.25) provide an important link between the Dirich-
let series α(s) and the power series generating function P (x ). Indeed, these
formulæ hold for complex x , provided that ℜx > 0. In particular, by taking
x = δ − 2πi α we find that
∞∑
n=1
an e(nα)e−nδ = 1
2πi
∫ σ0 +i ∞
σ0 −i ∞
α(s)Ŵ(s)(δ − 2πi α)−s ds .
It may be noted in the above examples that smoother weights w(x ) give rise
to kernels K (s) that tend to 0 rapidly as |t |→∞ . Further useful kernels can
be constructed as linear combinations of the above kernels.
Since the Mellin transform is a Fourier transform with altered variables, all
results pertaining to Fourier transforms can be reformulated in terms of Mellin
transforms. Particularly useful is Plancherel’s identity , which asserts that iff ∈
L 1 (R) ∩ L 2 (R), then ∥ f ∥ 2 =∥ ˆf ∥ 2 . This is the analogue for Fourier transforms
of Parseval’s identity for Fourier series, which asserts that ∑
k | ˆf (k)|2 =∥ f ∥ 2
2.
By the changes of variables we noted before, we obtain
Theorem 5.4(Plancherel’s identity) Suppose that
∫∞
0 |w(x )|x −σ−1 dx < ∞,
and also that
∫∞
0 |w(x )|2 x −2σ−1 dx < ∞. Put K (s) =
∫∞
0 w(x )x −s−1 d x . Then
2π
∫ ∞
0
|w(x )|2 x −2σ−1 dx =
∫ +∞
−∞
|K (σ + it )|2 dt .
Among the many possible applications of this theorem, we note in particular
that
2π
∫ ∞
0
| A(x )|2 x −2σ−1 dx =
∫ +∞
−∞
⏐
⏐
⏐α(σ + it )
σ + it
⏐
⏐
⏐
2
dt (5.26)
for σ> max(0,σc ). | {
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5.1 The inverse Mellin transform 145
5.1.1 Exercises
1. Show that if σc <σ 0 < 0, then
lim
T →∞
1
2πi
∫ σ0 +iT
σ0 −iT
α(s) x s
s ds =
∑ ′
n>x an .
2. (a) Show that if y ≥ 0, then
− π
2 = si(0) ≤ si( y) ≤ si(π) = 0.28114 ....
(b) Show that if y ≥ 0, then
ℑ
∫ ∞
y
eiu
u du = ℑ
∫ y+i ∞
y
eiz
z dz .
(c) Deduce that if y ≥ 0, then |si( y)| < 1/y.
3. (a) Let β> 0 be fixed. Show that if σ0 > 0, then
1
2πi
∫ σ0 +i ∞
σ0 −i ∞
Ŵ(s/β) ys ds = βe−y−β
.
(b) Let β> 0 be fixed. Show that if x > 0 and σ0 > max(0,σc ), then
1
2πi
∫ σ0 +i ∞
σ0 −i ∞
α(s)Ŵ(s/β)x s ds = β
∞∑
n=1
an e−(n/x )β
.
4. (a) Suppose that a > 0 and that b is real. Explain why
1
2πi
∫ σ0 +i ∞
σ0 −i ∞
ea2 s2/2+bs ds = e−b2/(2a2 )
2πi
∫ σ0 +i ∞
σ0 −i ∞
ea2 (s+b/a2 )2/2 ds .
(b) Explain why the values of the integrals above are independent of the
value of σ0 . Hence show that if σ0 =− b/a2 , then the above is
= e−b2/(2a2 )
2π
∫ +∞
−∞
e−a2 t 2/2 dt = 1√
2π a
e−b2/a2
.
(c) Show that if a > 0, x > 0 and σ0 >σ c , then
1
2πi
∫ σ0 +i ∞
σ0 −i ∞
α(s)ea2 s2/2 x s ds = 1√
2π a
∞∑
n=1
an exp
(
− (log x /n)2
2a2
)
.
5. T ake k = 1 in (5.22) for several different values of x , and form a suitable
linear combination, to show that if x ≥ 0 and and σc < 0, then
2
π
∫ +∞
−∞
α(it )
(
sin 1
2 t log x
t
)2
dt =
∑
n≤x
an log x /n. | {
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146 Dirichlet series: II
6. Let w(x ) ր, and suppose that w(x ) ≪ x σ as x →∞ for some fixed σ.
Let σw be the infimum of those σ such that
∫∞
0 w(x )x −σ−1 dx < ∞, and
put
K (s) =
∫ ∞
0
w(x )x −s−1 dx
for σ>σ w.
(a) Show that Aw(x ) = ∑ ∞
n=1 an w(x /n) satisfies Aw(x ) ≪ x θ for θ>
max(σw,σc ).
(b) Show that
K (s)α(s) =
∫ ∞
0
Aw(x )x −s−1 dx
for σ> max(σw,σc ).
(c) Show that
1
2 ( Aw(x −) + Aw(x +)) = 1
2πi lim
T →∞
∫ σ0 +iT
σ0 −iT
α(s)K (s)x s ds
for σ0 > max(σw,σc ), x > 0.
7. Show that
ζ(s) =− s
∫ ∞
0
{x }
x s+1 dx
for 0 <σ< 1, and that
2π
∫ ∞
0
{x }2 x −2σ−1 dx =
∫ +∞
−∞
⏐
⏐
⏐ζ(σ + it )
σ + it
⏐
⏐
⏐
2
dt
for 0 <σ< 1.
8. (a) Show that if f ∈ L 1 (R) and f ′ ∈ L 1 (R), then ˆf ′(t ) = 2πit ˆf (t ).
(b) Suppose that f is a function such that f ∈ L 1 (R), that xf (x ) ∈ L 2 (R),
and that f ′ ∈ L 1 (R) ∩ L 2 (R). Show that
∫ +∞
−∞
| f (x )|2 dx =−
∫ +∞
−∞
x
(
f ′(x )
f (x ) + f (x ) f ′(x )
)
dx .
The Cauchy–Schwarz inequality asserts that
⏐
⏐
⏐
⏐
∫ +∞
−∞
a(x )b(x ) dx
⏐
⏐
⏐
⏐
2
≤
(∫ +∞
−∞
|a(x )|2 dx
)(∫ +∞
−∞
|b(x )|2 dx
)
.
By means of this inequality , or otherwise, show that
(∫ +∞
−∞
|xf (x )|2 dx
)(∫ +∞
−∞
|t ˆf (t )|2 dt
)
≥ 1
16π2
(∫ +∞
−∞
| f (x )|2 dx
)2
. | {
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5.2 Summability 147
This is a form of the Heisenberg uncertainty principle. From it we see that
iff tends to 0 rapidly outside [ − A, A], and if ˆf tends to 0 rapidly outside
[−B,B ], then AB ≫ 1.
9. (a) Note the identity
f
g = 1
2 | f + g|2 − 1
2 | f − g|2 + i
2 | f + ig |2 − i
2 | f − ig |2 .
(b) Show that if f ∈ L 1 (R) ∩ L 2 (R) and if g ∈ L 1 (R) ∩ L 2 (R), then
∫ +∞
−∞
f (x )g(x ) dx =
∫ +∞
−∞
ˆf (t )ˆg(t ) dt .
10. Suppose that F is strictly increasing, and that for i = 1,2 the functions fi
are real-valued with fi ∈ L 1 (R) ∩ L 2 (R) and F ( fi ) ∈ L 1 (R) ∩ L 2 (R).
(a) Show that
∫ +∞
−∞
( f1 (x ) − f2 (x ))( F ( f1 (x )) − F ( f2 (x ))) dx
=
∫ +∞
−∞
(ˆf1 (t ) − ˆf2 (t )
)(ˆF ( f1 )(t ) − ˆF ( f2 )(t )
)
dt .
(b) Suppose additionally that ˆfi (t ) = 0 for |t |≥ T , and that ˆF ( f1 )(t ) =
ˆF ( f2 )(t ) for −T ≤ t ≤ T . Show that f1 = f2 a.e.
5.2 Summability
W e say that an infinite series ∑ an is Abel summable to a, and write ∑ an = a
(A) if
lim
r →1−
∞∑
n=0
an r n = a.
Abel proved that if a series converges, then it is A-summable to the same value.
Because of this historical antecedent, we call a theorem ‘ Abelian’ if it states
that one kind of summability implies another. Perhaps the simplest Abelian
theorem asserts that if∑ ∞
n=1 an converges to a, then
lim
N →∞
N∑
n=1
(
1 − n
N
)
an = a. (5.27)
This is the Ces` aro method of summability of order 1, and so we abbreviate the
relation above as∑ an = a (C, 1). On putting sN = ∑ N
n=1 an , we reformulate | {
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148 Dirichlet series: II
the above by saying that if lim N →∞ sN = a, then
lim
N →∞
1
N
N∑
n=1
sn = a. (5.28)
Here, as in Abel summability and in most other summabilities, each term in
the second limit is a linear function of the terms in the first limit. Following
T oeplitz and Schur, we characterize those linear transformationsT = [tmn ] that
preserves limits of sequences. W e call T regular if the following three conditions
are satisfied:
There is a C = C (T ) such that
∞∑
n=1
|tmn |≤ C for all m; (5.29)
lim
m→∞
tmn = 0 for all n; (5.30)
lim
m→∞
∞∑
n=1
tmn = 1. (5.31)
W e now show that regular transformations preserve limits, and relegate the
verification of the converse to exercises.
Theorem 5.5Suppose that T satisfies (5.29) above. If {an } is a bounded
sequence, then the sequence
bm =
∞∑
n=1
tmn an (5.32)
is also bounded. If T satisfies (5.29) and (5.30), and if limn→∞ an = 0,
then limm→∞ bm = 0. Finally, if T is regular and limn→∞ an = a, then
limm→∞ bm = a.
The important special case (5.28) is obtained by noting that the (semi-infinite)
matrix [ tmn ] with
tmn =
{
1/m if 1 ≤ n ≤ m,
0i f n > m
is regular. Moreover, the proof of Theorem 5.5 requires only a straightforward
elaboration of the usual proof of (5.28).
ProofIf |an |≤ A and (5.29) holds, then
|bm |≤
∞∑
n=1
|tmn an |≤ A
∞∑
n=1
|tmn |≤ CA . | {
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5.2 Summability 149
T o establish the second assertion, suppose that ε> 0 and that |an | <ε for
n > N = N (ε). Now
|bm |≤
N∑
n=1
|tmn an |+
∑
n>N
|tmn an |= /Sigma1 1 + /Sigma1 2 ,
say . From (5.29) and the argument above with A = ε we see that /Sigma1 2 ≤ C ε.
From (5.30) we see that lim m→∞ /Sigma1 1 = 0. Hence lim sup m→∞ |bm |≤ C ε, and
we have the desired conclusion since ε is arbitrary . Finally , suppose that T is
regular and that lim n→∞ an = a. W e write an = a + αn , so that
bm = a
∞∑
n=1
tmn +
∞∑
n=1
tmn αn .
Since lim n→∞ αn = 0, we may appeal to the preceding case to see that
the second sum tends to 0 as m →∞ . Hence by (5.31) we conclude that
limm→∞ bm = a, and the proof is complete. □
In Chapter 1 we used Theorem 1.1 to show that if S is a sector of the
form S ={ s : σ>σ 0 , |t − t0 |≤ H (σ − σ0 )} where H is an arbitrary positive
constant, and if the Dirichlet series α(s) converges at the point s0 , then
lims→s0
s∈S
α(s) = α(s0 ).
T o see how this may also be derived from Theorem 5.5, let {sm } be an arbitrary
sequence of points of S for which lim m→∞ sm = s0 . It suffices to show that
limm→∞ α(sm ) = α(s0 ). T ake
tmn = ns0 −sm − (n + 1)s0 −sm ,
so that
α(sm ) =
∞∑
n=1
tmn
(n∑
k=1
ak k−s0
)
.
In view of Theorem 5.5, it suffices to show that [ tmn ] is regular. The conditions
(5.30) and (5.31) are clearly satisfied, and (5.29) follows on observing that if
s∈ S, then s − s0 ≪H σ − σ0 , so that
⏐
⏐ns0 −s − (n + 1)s0 −s ⏐
⏐=
⏐
⏐
⏐
⏐(s − s0 )
∫ n+1
n
us0 −s−1 du
⏐
⏐
⏐
⏐
≪H (σ − σ0 )
∫ n+1
n
uσ0 −σ−1 du
= nσ0 −σ − (n + 1)σ0 −σ.
Thus we have the result. Abel’s analogous theorem on the convergence of power
series can be derived similarly from Theorem 5.5. | {
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150 Dirichlet series: II
The converse of Abel’s theorem on power series is false, but T auber (1897)
proved a partial converse: If an = o(1/n) and ∑ an = a (A), then ∑ an = a.
Following Hardy and Littlewood, we call a theorem ‘T auberian’ if it provides
a partial converse of an Abelian theorem. The qualifying hypothesis (‘an =
o(1/n)’ in the above) is the ’T auberian hypothesis’. For simplicity we begin
with partial converses of (5.27).
Theorem 5.6If ∑ ∞
n=1 an = a (C, 1) , then ∑ an = a provided that one of the
following hypotheses holds :
(a) an ≥ 0 for n ≥ 1;
(b) an = O (1/n) for n ≥ 1;
(c) There is a constant A such that a n ≥− A/n for all n ≥ 1.
Proof Clearly (a) implies (c). If (b) holds, then both ℜan and ℑan satisfy (c).
Thus it suffices to prove that ∑ an = a when (c) holds. W e observe that if H
is a positive integer, then
N∑
n=1
an = N + H
H
N +H∑
n=1
an
(
1 − n
N + H
)
− N
H
N∑
n=1
an
(
1 − n
N
)
− 1
H
∑
N <n<N +H
an ( N + H − n) (5.33)
= T1 − T2 − T3 ,
say . T ake H = [εN ] for some ε> 0. By hypothesis, lim N →∞ T1 = a(1 + ε)/ε,
and lim N →∞ T2 = a/ε. From (c) we see that
T3 ≥− A
∑
N <n<N +H
1
n ≥− AH
N ≥− Aε.
Hence on combining these estimates in (5.33) we see that
lim sup
N →∞
N∑
n=1
an ≤ a + Aε.
Since ε can be taken arbitrarily small, it follows that
lim sup
N →∞
N∑
n=1
an ≤ a.
T o obtain a corresponding lower bound we note that
N∑
n=1
an = N
H
N∑
n=1
an
(
1 − n
N
)
− N − H
H
N −H∑
n=1
an
(
1 − n
N − H
)
(5.34)
+ 1
H
∑
N −H <n<N
an (n + H − N ). | {
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5.2 Summability 151
Arguing as we did before, we find that
lim inf
N →∞
N∑
n=1
an ≥ a − Aε/(1 − ε),
so that
lim inf
N →∞
N∑
n=1
an ≥ a,
and the proof is complete. □
If we had argued from (a) or (b), then the treatment of the term T3 above
would have been simpler, since from (a) it follows that T3 ≥ 0, while from
(b) we have T3 ≪ ε.
Our next objective is to generalize and strengthen Theorem 5.6. The type of
generalization we have in mind is exhibited in the following result, which can
be established by adapting the above proof: Letβ be fixed, β ≥ 0. If
N∑
n=1
an
(
1 − n
N
)
= (a + o(1)) N β,
and if an ≥− An β−1 , then
N∑
n=1
an = (a(β + 1) + o(1)) N β.
Concerning the possibility of strengthening Theorem 5.6, we note that by an
Abelian argument (or by an application of Theorem 5.5) it may be shown that
∑an = a (C, 1) implies that ∑ an = a (A). Thus if we replace (C, 1) by (A)
in Theorem 5.6, then we have weakened the hypothesis, and the result would
therefore be stronger. Indeed, Hardy (1910) conjectured and Littlewood (1911)
proved that if∑ an = a (A) and an = O (1/n), then ∑ an = a. That is, the
condition ‘ an = o(1/n)’ in T auber’s theorem can be replaced by the condition
(b) above. In fact the still weaker condition (c) suffices, as will be seen by
takingβ = 0 in Corollary 5.9 below . W e now formulate a general result for the
Laplace transform, from which the analogues for power series and Dirichlet
series follow easily .
Theorem 5.7(Hardy–Littlewood) Suppose that a (u) is Riemann-integrable
over [0,U ] for every U > 0, and that the integral
I (δ) =
∫ ∞
0
a(u)e−uδ du | {
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152 Dirichlet series: II
converges for every δ> 0. Let β be fixed, β ≥ 0, and suppose that
I (δ) = (α + o(1))δ−β (5.35)
as δ → 0+. If, moreover , there is a constant A ≥ 0 such that
a(u) ≥− A(u + 1)β−1 (5.36)
for all u ≥ 0, then
∫ U
0
a(u) du =
( α
Ŵ(β + 1) + o(1)
)
U β. (5.37)
The basic properties of the gamma function are developed in Appendix C,
but for our present purposes it suffices to put
Ŵ(β) =
∫ ∞
0
uβ−1 e−u du
for β> 0. From this it follows by integration by parts that
βŴ(β) = Ŵ(β + 1) (5.38)
when β> 0.
The amount of unsmoothing required in deriving (5.37) from (5.35) is now
much greater than it was in the proof of Theorem 5.6. Nevertheless we follow
the same line of attack. T o obtain the proper perspective we review the preceding
proof. LetJ = [0,1], let χJ (u) be its characteristic function, and put K (u) =
max(0,1 − u) for u ≥ 0. Thus ∑ N
n=1 an = ∑
n an χJ (n/N ), and ∑ N
n=1 an (1 −
n/N ) = ∑
n an K (n/N ). Our strategy was to approximate to χJ (u) by linear
combinations of K (κu) for various values of κ, κ> 0. The relation underlying
(5.33) and (5.34) is both simple and explicit:
1
ε
(
K (u) − (1− ε)K (u/(1 − ε))
)
≤ χJ (u) ≤ 1
ε((1+ ε)K (u/(1+ ε)) − K (u));
(5.39)
we took ε = H /N . In the present situation we wish to approximate to χJ (u)b y
linear combinations of e−κu , κ> 0. W e make the change of variable x = e−u ,
so that 0 ≤ x ≤ 1, and we put J = [1/e,1]. Then we want to approximate to
χJ (x ) by a linear combination P (x ) of the functions x κ, κ> 0. In fact it suffices
to use only integral values of κ, so that P (x ) is a polynomial that vanishes at
the origin. In place of (5.33), (5.34) and (5.39) we shall substitute
Lemma 5.8Let ε be given, 0 <ε< 1/4, and put J = [1/e,1], K =
[e−1−ε,e−1+ε]. There exist polynomials P ±(x ) such that for 0 ≤ x ≤ 1 we have
P−(x ) ≤ χJ (x ) ≤ P+(x ) (5.40) | {
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5.2 Summability 153
and
| P±(x ) − χJ (x )|≤ εx (1 − x ) + 5χK(x ). (5.41)
Proof Let g(x ) = (χJ (x ) − x )/(x (1 − x )). Then g is continuous in [0 ,1]
apart from a jump discontinuity at x = 1/e of height e2 /(e − 1) < 5. Hence
by W eierstrass’s theorem on the uniform approximation of continuous func-
tions by polynomials we see that there are polynomialsQ±(x ) such that
Q−(x ) ≤ g(x ) ≤ Q+(x ) for 0 ≤ x ≤ 1, and for which
|g(x ) − Q±(x )|≤ ε+ 5χK(x ) (5.42)
for 0 ≤ x ≤ 1. Then the polynomials P±(x ) = x + x (1 − x ) Q±(x ) have the
desired properties. □
Proof of Theorem 5.7 W e suppose first that α = 0. W e note that if P (x )i sa
polynomial such that P (0) = 0, say P (x ) = ∑ R
r =1 cr x r , then by (5.35) we see
that
∫ ∞
0
a(u) P (e−uδ) du =
R∑
r =1
cr I (r δ) = o(δ−β) (5.43)
as δ → 0+. In the notation of the above lemma,
∫ U
0
a(u) du =
∫ ∞
0
a(u)χJ (e−u/U ) du .
If (5.40) holds, then by (5.36) we see that
∫ ∞
0
a(u)
(
P+
(
e−u/U )
− χJ
(
e−u/U ))
du
≥− A
∫ ∞
0
(u + 1)β−1 (
P+
(
e−u/U )
− χJ
(
e−u/U ))
du .
By (5.41) this latter integral is
≪ ε
∫ ∞
0
(u + 1)β−1 e−u/U (1 − e−u/U ) du +
∫ (1+ε)U
(1−ε)U
(u + 1)β−1 du .
In the first term, the integrand is ≪ (u + 1)βU −1 for 0 ≤ u ≤ U ;i ti s ≪
uβ−1 e−u/U for u ≥ U . Hence the first integral is ≪ U β. The second integral is
≪ εU β. On taking δ = 1/U , P = P+ in (5.43) and combining our results, we
find that
∫ U
0
a(u) du ≤ A1 εU β + o(U β). | {
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154 Dirichlet series: II
Since ε can be arbitrarily small, we deduce that
lim sup
U →∞
U −β
∫ U
0
a(u) du ≤ 0.
By arguing similarly with P− instead of P+, we see that the corresponding
liminf is ≥ 0, and so we have (5.37) in the case α = 0.
Suppose now that α ̸=0, β> 0. W e note first that
∫ ∞
0
(u + 1)β−1 e−uδ du = eδ
∫ ∞
1
vβ−1 e−vδ d v = eδ
∫ ∞
0
vβ−1 e−vδ d v + O (eδ),
and that ∫ ∞
0
vβ−1 e−vδ d v = δ−β
∫ ∞
0
wβ−1 e−w d w = δ−βŴ(β).
Hence if b(u) = a(u) − α(u + 1)β−1 /Ŵ(β), then b(u) ≥− B (u + 1)β−1 , and
∫ ∞
0
b(u)e−uδ du = o(δ−β).
Thus
∫U
0 b(u) du = o(U β), so that
∫ U
0
a(u) du = α
βŴ(β) U β + o(U β),
and we have (5.37), in view of (5.38).
For the remaining case, β = 0, it suffices to consider b(u) = a(u) −
αχ[0,1] (u). □
Corollary 5.9 Suppose that p (z) = ∑ ∞
n=0 an zn converges for |z| < 1, and
that β ≥ 0.I f p (x ) = (α + o(1))(1 − x )−β as x → 1−, and if a n ≥− An β−1
for n ≥ 1, then
N∑
n=0
an =
( α
Ŵ(β + 1) + o(1)
)
N β.
Proof Put a(u) = an for n ≤ u < n + 1. Then (5.36) holds, and
I (δ) =
∞∑
n=0
an
∫ n+1
n
e−uδ du = 1 − e−δ
δ p(e−δ).
But 1 − e−δ ∼ δ as δ → 0+, so that (5.35) holds. The result now follows by
taking U = N + 1 in (5.37). □
Corollary 5.10 If ∑ an = α (A), and if the sequence s N = ∑ N
n=0 an is
bounded, then ∑ an = α (C, 1) . | {
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5.2 Summability 155
Proof Ta k e β = 1, p(z) = ∑ ∞
n=0 sn zn = (1 − z)−1 ∑ ∞
n=0 an zn in Corollary
5.9. Then ∑ N
n=0 sn = (α + o(1)) N , which is the desired result. □
For Dirichlet series we have similarly
Theorem 5.11 Suppose that α(s) = ∑ ∞
n=1 an n−s converges for σ> 1, and
that β ≥ 0.I f α(σ) = (α + o(1))(σ − 1)−β as σ → 1+, and if a n ≥− A(1 +
log n)β−1 , then
N∑
n=1
an
n =
( α
Ŵ(β + 1) + o(1)
)
(log N )β.
Proof Ta k e a(u) = ∑
u−1≤log n<u an /n. Then I (δ) converges for δ> 0, and
moreover
I (δ) =
∞∑
n=1
an
n
∫ 1+log n
log n
e−uδ du = 1 − e−δ
δ α(1 + δ),
so that (5.37) follows. T o obtain the desired conclusion we require a further
appeal to our T auberian hypothesis. W e note that
∫ log N
0
a(u) du =
∑
n≤N
an
n −
∑
N /e<n≤N
an
n log ne
N .
By our T auberian hypothesis this is
≤
∑
n≤N
an
n + A1 (log N )β−1 ,
so that
∑
n≤N
an
n ≥
( α
Ŵ(β + 1) + o(1)
)
(log N )β − A1 (log N )β−1 .
On taking U = 1 + log N in (5.37) we may derive a corresponding upper bound
to complete the proof. □
The qualitative arguments we have given can be put in quantitative form as
the need arises. For example, it is easy to see that if
N∑
n=1
an = N + O
(√
N
)
, (5.44)
then
N∑
n=1
an ( N − n) = 1
2 N 2 + O
(
N 3/2 )
. (5.45) | {
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156 Dirichlet series: II
This is best possible (take an = 1 + n−1/2 ), but if the error term is oscilla-
tory , then smoothing may reduce its size (consider an = cos √n). Conversely if
(5.45) holds and if the sequence an is bounded, then the method used to prove
Theorem 5.6 can be used to show that
N∑
n=1
an = N + O
(
N 3/4 )
. (5.46)
This conclusion, though it falls short of (5.44), is best possible (take an =
1 + cos n1/4 ). W e can also put Theorem 5.7 in quantitative form, but here
the loss in precision is much greater, and in general the importance of The-
orem 5.7 and its corollaries lies in its versatility . For example, it can be
shown that if∑ ∞
n=0 an r n = (1 − r )−1 + O (1) as r → 1−, and if an = O (1),
then
N∑
n=0
an = N + O
(N
log N
)
.
This error term, though weak, is best possible (take an = 1 + cos(log n)2 ).
For Dirichlet series it can be shown that if
α(s) =
∞∑
n=1
an n−s = 1
s − 1 + O (1)
as s → 1+, and if the sequence an is bounded, then
N∑
n=1
an
n = log N + O
(log N
log log N
)
.
This is also best possible (take an = 1 + cos(log log n)2 ), but we can obtain a
sharper result by strengthening our analytic hypothesis. For example, it can be
shown that ifα(s) is analytic in a neighbourhood of 1 and if the sequence an is
bounded, then
N∑
n=1
an
n = O (1).
However, even this stronger assumption does not allow us to deduce that
N∑
n=1
an = o( N ),
as we see by considering an = cos log n. In Chapter 8 we shall encounter further
T auberian theorems in which the above conclusion is derived from hypotheses
concerning the behaviour ofα(s) throughout the half-plane σ ≥ 1. | {
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5.2 Summability 157
5.2.1 Exercises
1. Let T be a regular matrix such that tmn ≥ 0 for all m,n. Show that if
limn→∞ an =+ ∞ , then lim m→∞ bm =+ ∞ .
2. Show that if T = [tmn ] and U = [umn ] are regular matrices, then so is
TU = V = [vmn ] where
vmn =
∞∑
k=1
tmk ukn .
3. Show that if b = T a and lim m→∞ bm = a whenever lim n→∞ an = a, then
T is regular.
4. For n = 0,1,2,... let tn (x ) be defined on [0 ,1), and suppose that the tn
satisfy the following conditions:
(i) There is a constant C such that if x ∈ [0,1), then ∑ ∞
n=0 |tn (x )|≤ C .
(ii) For all n, lim x →1− tn (x ) = 0.
(iii) lim x →1−
∑ ∞
n=0 tn (x ) = 1.
Show that if lim n→∞ an = a and if b(x ) = ∑ ∞
n=0 an tn (x ), then
limx →1− b(x ) = a.
5. (Kojima 1917) Suppose that the numbers tmn satisfy the following
conditions:
(i) There is a constant C such that ∑ ∞
n=1 |tmn |≤ C for all m.
(ii) For all n, lim m→∞ tmn exists.
(iii) lim m→∞
∑ ∞
n=1 tmn exists.
Show that if lim n→∞ an exists and if bm = ∑ ∞
n=1 tmn an , then lim m→∞ bm
exists.
6. For positive integers n let Kn (x ) be a function defined on [0 ,∞) such that
(i)
∫∞
0 Kn (x ) dx → 1a s n →∞ ;
(ii)
∫∞
0 |Kn (x )| dx ≤ C for all n;
(iii) lim n→∞ Kn (x ) = 0 uniformly for 0 ≤ x ≤ X .
Suppose that a(x ) is a bounded function, and that bn =
∫∞
0 a(x )Kn (x ) dx .
Show that if lim x →∞ a(x ) = a, then lim n→∞ bn = a.
7. Let rm be a sequence of positive real numbers with rm → 1− as m →∞ .
For m ≥ 1, n ≥ 1, put tmn = nr n−1
m (1 − rm )2 .
(a) Show that [ tmn ] is regular.
(b) Show that if an = ∑ n−1
k=0 ck (1 − k/n) and bm is defined by (5.32), then
bm = ∑ ∞
k=0 ck r k
m .
(c) Show that if ∑ cn = c (C, 1), then ∑ cn = c (A).
8. Suppose that T = [tmn ]i sg i v e nb y
tmn =
⎧
⎪
⎪
⎨
⎪
⎪
⎩
0i f n = 0,
m!n
mn+1 (m − n)! if m ≥ n > 0,
0i f m < n. | {
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158 Dirichlet series: II
(a) Show that
m∑
n=k
tmn = m!
mk (m − k)!
for 1 ≤ k ≤ m .
(b) V erify that T is regular.
(c) Show that if an = ∑ n
k=0 x k /k! for n ≥ 0, then bm = (1 + x /m)m for
m ≥ 1.
9. (Mercer’s theorem) Suppose that
bm = 1
2 am + 1
2 · a1 + a2 +···+ am
m
for m ≥ 1. Show that
an = 2n
n + 1 bn − 2
n(n + 1)
n−1∑
m=1
mbm .
Conclude that lim n→∞ an = a if and only if lim m→∞ bm = a.
10. For a non-negative integer k we say that ∑ an = a (C, k)i f
lim
x →∞
∑
n≤x
an
(
1 − n
x
)k
= a.
This is Ces `aro summability of order k .
(a) Show that if ∑ an = a (C, j ), then ∑ an = a (C, k) for all k ≥ j .
(b) Show that if ∑ an = a (C, k) for some k, then ∑ an = a (A).
11. Show that if ∑ an = a (A), then lim s→0+
∑ an n−s = a. (See Wintner 1943
for T auberian converses.)
12. For a non-negative integer k we say that ∑ an = a (R, k)i f
lim
x →∞
∑
n≤x
an
(
1 − log n
log x
)k
= a.
This is Riesz summability of order k .
(a) Show that if ∑ an = a (R, j ), then ∑ an = a (R, k) for all k ≥ j .
(b) Show that if ∑ an = a (R, k) for some k, then ∑
s→0+ α(s) = a.
13. Put tmn = 0 for n > m, set
tmm = m + 1
log(m + 1) (log(m + 1) − log m),
while for 1 ≤ n < m put
tmn = n + 1
log(m + 1) (− log n + 2 log( n + 1) − log(n + 2)) . | {
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5.2 Summability 159
(a) Show that if
an =
n∑
k=1
ck
(
1 − k
n + 1
)
for n ≥ 1, then the bm given in (5.32) satisfies
bm =
m∑
k=1
ck
(
1 − log k
log(n + 1)
)
.
(b) Show that tmn ≥ 0 for all m,n.
(c) Show that
∞∑
n=1
tmn = 1 + log 2
log(m + 1) .
(d) Show that lim m→∞ tmn = 0.
(e) Conclude that if ∑ ck = c (C, 1), then ∑ ck = c (R, 1) .
14. Let A(x ) = ∑
0<n≤x an .
(a) Show that
N∑
n=1
an
(
1 − n
N
)
= 1
N
∫ N
0
A(x ) dx .
(b) Show that
N∑
n=1
an
(
1 − log n
log N
)
= 1
log N
∫ N
1
A(x )
x dx .
(c) Suppose that t is a fixed non-zero real number. By Corollary 1.15, or
otherwise, show that
N∑
n=1
n−1−it
(
1 − n
N
)
= N −it
(1 − it )2 + ζ(1 + it ) + O
(log N
N
)
.
(d) Similarly , show that
N∑
n=1
n−1−it
(
1 − log n
log N
)
= ζ(1 + it ) + O
( 1
log N
)
.
(e) Conclude that ∑ ∞
n=1 n−1−it is not summable (C, 1), but that it is
summable (R, 1) to ζ(1 + it ).
15. W e say that a series is Lambert summable , and write ∑ an = a (L), if
lim
r →1−
(1 − r )
∞∑
n=1
nan r n
1 − r n = a.
(a) Show that if ∑ an = a, then ∑ an = a (L). | {
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160 Dirichlet series: II
(b) Show that if an is a bounded sequence and |z| < 1, then
∞∑
n=1
nan zn
1 − zn =
∞∑
n=1
(∑
d |n
da d
)
zn .
(c) Show that ∑ ∞
n=1 µ(n)/n = 0 (L).
(d) Deduce that if ∑ ∞
n=1 µ(n)/n converges, then its value is 0. (See (6.18)
and (8.6).)
(e) Show that ∑ ∞
n=1 (/Lambda1 (n) − 1)/n =− 2C0 (L).
(f) Deduce that if ∑
n≤x /Lambda1 (n)/n = log x + c + o(1) then c =− C0 . (See
Exercise 8.1.1.)
16. (Bohr 1909; Riesz 1909; Phragm´ en (cf. Landau 1909, pp. 762, 904))
Let α(s) = ∑ an n−s , β(s) = ∑ bn n−s , and γ(s) = α(s)β(s) = ∑ cn n−s
where cn = ∑
d |n ad bn/d . Further, put A(x ) = ∑
n≤x an and B (x ) =∑
n≤x bn .
(a) Show that
∫ x
1
A( y) B (x /y) dy
y =
∑
n≤x
cn log x /n.
(b) Show that if ∑ an converges and ∑ bn converges, then ∑ cn =
α(0)β(0) (R, 1).
(c) (Landau 1907) By taking j = 0 in Exercise 12(a), or otherwise, show
that if the three series ∑ an , ∑ bn , ∑ cn all converge, then ∑ cn =(∑ an
)(∑ bn
)
.
17. Suppose that f (n) ր∞ . Construct an so that |an |≤ f (n)/n for all n,
lim sup
N →∞
N∑
n=1
an = 1, lim inf
N →∞
N∑
n=1
an =− 1,
but
lim
N →∞
N∑
n=1
an (1 − n/N ) = 0.
18. (Landau 1908) Show that if f (x ) ∼ x as x →∞ and xf ′(x ) is increasing,
then lim x →∞ f ′(x ) = 1.
19. (Landau (1913); cf. Littlewood (1986, p. 54–55); Schoenberg 1973) Show
that if f (x ) → 0a s x →∞ , and if f ′′(x ) = O (1), then f ′(x ) → 0a s
x →∞ .
20. (T auber’s ‘second theorem’) Suppose that P (δ) = ∑ ∞
n=0 an e−nδ for δ> 0,
and put sN = ∑ N
n=0 an .
(a) Show that if an = O (1/n), then sN = P (1/N ) + O (1).
(b) Show that if an = o(1/n), then sN = P (1/N ) + o(1). | {
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5.2 Summability 161
(c) Let B ( N ) = ∑ N
n=1 nan . Show that if ∑ an converges, then B ( N ) =
o( N )a s N →∞ .
(d) Show that if P (δ) converges for δ> 0, then
sN − P (1/N ) = B ( N )
N +
∫ N
1
B (u)
(1
u2 − e−u/N
u2 − e−u/N
uN
)
du
+
∫ ∞
N
B (u)e−u/N
(u
N − 1
)du
u2 .
(e) Show that if B ( N ) = o( N ), then sN − P (1/N ) = o(1).
(f) Show that if ∑ an = a (A), then ∑ an = a if and only if B ( N ) = o( N ).
21. (a) Using Ramanujan’s identity ∑ ∞
n=1 d (n)2 n−s = ζ(s)4 /ζ(2s) and Theo-
rem 5.11, show that ∑
n≤x d (n)2 /n ∼ (4π2 )−1 (log x )4 .
(b) Show that if ∑
n≤x d (n)2 ∼ cx (log x )3 as x →∞ , then c = 1/π2 .
22. Show that ∑ ∞
n=1 1/(d (n)ns ) ∼ c(s − 1)−1/2 as s → 1+ where
c =
∏
p
(
( p2 − p)1/2 log
( p
p − 1
))
.
Deduce that
∑
n≤x
1
nd (n) ∼ 2c√π(log x )1/2
as x →∞ .
23. Show that if ∑
n≤N an /n = O (1) and lim s→1+
∑ ∞
n=1 an n−s = a, then
lim
x →∞
∑
n≤x
an
n
(
1 − log n
log x
)
= a.
24. Show that ∫ ∞
0
sin x
x e−sx dx = arctan 1 /s
for s > 0. Using Theorem 5.7, deduce that
∫ ∞
0
sin x
x dx = π
2 .
25. Suppose that f (u) ≥ 0, that
∫∞
0 f (u) du < ∞, and that
∫∞
0 (1 −
e−δu ) du ∼ δ1/2 as δ → 0+. Show that
∫∞
U f (u) du ∼ (πU )−1/2 as U →
∞.
26. Show that ∑ ∞
n=1 an = a if and only if
lim
r →1−
∞∑
n=0
an r 2n
= a. | {
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162 Dirichlet series: II
27. Suppose that for every ε> 0 there is an η> 0 such that∑
N <n≤(1+η) N |an | <ε whenever N > 1/η. Show that if ∑ an = a (A),
then ∑ an = a.
28. Show that if ∑ an = a (C, 1) and if an+1 − an = O (|an |/n), then ∑ an = a.
29. (Hardy & Littlewood 1913, Theorem 27) Show that if ∑ an = a (A) and if
an+1 − an = O (|an |/n), then ∑ an = a.
30. (Hardy 1907) Show that
lim
x →1−
∞∑
k=0
(−1)k x 2k
does not exist.
5.3 Notes
Section 5.1. Theorem 5.1 and the more general (5.22) were first proved rig-
orously by Perron (1908). Although the Mellin transform had been used by
Riemann and Cahen, it was Mellin (1902) who first described a general class
of functions for which the inversion succeeds. Hjalmar Mellin was Finnish, but
his family name is of Swedish origin, so it is properly pronounced m˘e · l¯en′.
However, in English-speaking countries the uncultured pronunciation m ˘el′· ˘ın
is universal.
In connection with Theorem 5.4, it should be noted that Plancherel’s formula
∥ f ∥ 2 =∥ ˆf ∥ 2 holds not just for all f ∈ L 1 (R) ∩ L 2 (R) but actually for all
f ∈ L 2 (R). However, in this wider setting one must adopt a new definition for
ˆf , since the definition we have taken is valid only for f ∈ L 1 (R). See Goldberg
(1961, pp. 46–47) for a resolution of this issue.
For further material concerning properties of Dirichlet series, one should
consult Hardy & Riesz (1915), Titchmarsh (1939, Chapter 9), or Widder (1971,
Chapter 2). Beyond the theory developed in these sources, we call attention to
two further topics of importance in number theory . Wiener (1932, p. 91) proved
that if the Fourier series off ∈ L 1 (T) is absolutely convergent and is never zero,
then the Fourier series of 1 /f is also absolutely convergent. Wiener’s proof was
rather difficult, but Gel’fand (1941) devised a simpler proof depending on his
theory of normed rings. L´ evy (1934) proved more generally that the Fourier
series ofF ( f ) is absolutely convergent provided that F is analytic at all points
in the range of f . Elementary proofs of these theorems have been given by
Zygmund (1968, pp. 245–246) and Newman (1975). These theorems were
generalized to absolutely convergent Dirichlet series by Hewitt & Williamson
(1957), who showed that ifα(s) = ∑ an n−s is absolutely convergent for σ ≥
σ0 , then 1 /α(s) is represented by an absolutely convergent Dirichlet series | {
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5.3 Notes 163
in the same half-plane, if and only if the values taken by α(s) in this half-
plane are bounded away from 0. Ingham (1962) noted a fallacy in Zygmund’s
account of L´ evy’s theorem, corrected it, and gave an elementary proof of the
generalization to absolutely convergent Dirichlet series. See also Goodman &
Newman (1984). Secondly , Bohr (1919) developed a theory concerning the
values taken on by an absolutely convergent Dirichlet series. This is described
by Titchmarsh (1986, Chapter 11), and in greater detail by Apostol (1976,
Chapter 8). For a small footnote to this theory , see Montgomery & Schinzel
(1977).
Section 5.2. That conditions (5.29)–(5.31) are necessary and sufficient for
the transformation T to preserve limits was proved by T oeplitz (1911) for upper
triangular matrices, and by Steinhaus (1911) in general. See also Kojima (1917)
and Schur (1921). For more on the T oeplitz matrix theorem and various aspects
of T auberian theorems, see Peyerimhoff (1969).
Theorem 5.6 under the hypothesis (a) is trivial by dominated convergence.
Theorem 5.6(b) is a special case of a theorem of Hardy (1910), who considered
the more general (C,k) convergence, and Theorem 5.6(c) is similarly a special
case of a theorem of Landau (1910, pp. 103–113).
T auber (1897) proved two theorems, the second of which is found in Exer-
cise 5.2.18. Littlewood (1911) derived his strengthening of T auber’s first theo-
rem by using high-order derivatives. Subsequently Hardy & Littlewood (1913,
1914a, b, 1926, 1930) used the same technique to obtain Theorem 5.8 and
its corollaries. Karamata (1930, 1931a, b) introduced the use of W eierstrass’s
approximation theorem. Karamata also considered a more general situation,
in which the right-hand sides of (5.35) and (5.36) are multiplied by a slowly
oscillating functionL (1/δ), and the right-hand side of (5.37) is multiplied by
L (U ). Our exposition employs a further simplification due to Wielandt (1952).
Other proofs of Littlewood’s theorem have been given by Delange (1952) and
by Eggleston (1951). Ingham (1965) observed that a peak function similar
to Littlewood’s can be constructed by using high-order differencing instead
of differentiation. Since many proofs of the W eierstrass theorem involve con-
structing a peak function, the two methods are not materially different. Sharp
quantitative T auberian theorems have been given by Postnikov (1951), Kore-
vaar (1951, 1953, 1954a–d), Freud (1952, 1953, 1954), Ingham (1965), and
Ganelius (1971).
For other accounts of the Hardy–Littlewood theorem, see Hardy (1949) or
Widder (1946, 1971). For a brief survey of applications of summability to
classical analysis, see Rubel (1989).
Wiener (1932, 1933) invented a general T auberian theory that contains the
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164 Dirichlet series: II
as a special case. Wiener’s theory is discussed by Hardy (1949), Pitt (1958), and
Widder (1946). Among the longer expositions of T auberian theory , the recent
accounts of Korevaar (2002, 2004) are especially recommended.
5.4 References
Apostol, T . (1976). Modular Functions and Dirichlet Series in Number Theory , Graduate
T exts Math. 41. New Y ork: Springer-V erlag.
Bohr, H. (1909). ¨Uber die Summabilit¨ at Dirichletscher Reihen, Nachr . K ¨onig. Gesell.
Wiss. G ¨ottingen Math.-Phys. Kl. , 247–262; Collected Mathematical W orks , V ol. I.
København: Dansk Mat. Forening, 1952, A2.
(1919). Zur Theorie algemeinen Dirichletschen Reihen, Math. Ann. 79, 136–156;
Collected Mathematical W orks , V ol. I. København: Dansk Mat. Forening, 1952,
A13.
Delange, H. (1952). Encore une nouvelle d´ emonstration du th´ eor` eme taub´ erien de Lit-
tlewood, Bull. Sci. Math . (2) 76, 179–189.
Edwards, D. A. (1957). On absolutely convergent Dirichlet series, Proc. Amer . Math.
Soc. 8, 1067–1074.
Eggleston, H. G. (1951). A T auberian lemma, Proc. London Math. Soc. (3) 1, 28–45.
Freud, G. (1952). Restglied eines T auberschen Satzes, I, Acta Math. Acad. Sci. Hungar .
2, 299–308.
(1953). Restglied eines T auberschen Satzes, II, Acta Math. Acad. Sci. Hungar . 3,
299–307.
(1954). Restglied eines T auberschen Satzes, III, Acta Math. Acad. Sci. Hungar . 5,
275–289.
Ganelius, T . (1971). T auberian Remainder Theorems , Lecture Notes Math. 232. Berlin:
Springer-V erlag.
Gel’fand, I. M. (1941). ¨Uber absolut konvergente trigonometrische Reihen und Integrale,
Mat. Sb. N. S . 9, 51–66.
Goldberg R. R. (1961). F ourier Transforms, Cambridge Tract 52. Cambridge: Cambridge
University Press.
Goodman, A. & Newman, D. J. (1984). A Wiener type theorem for Dirichlet series,
Proc. Amer . Math. Soc . 92, 521–527.
Hardy , G. H. (1907). On certain oscillating series, Quart. J. Math . 38, 269–288; Collected
P apers, V ol. 6. Oxford: Clarendon Press, 1974, pp. 146–167.
(1910). Theorems relating to the summability and convergence of slowly oscillating
series, Proc. London Math. Soc . (2) 8, 301–320; Collected P apers, V ol. 6. Oxford:
Clarendon Press, 1974, pp. 291–310.
(1949). Divergent Series , Oxford: Oxford University Press.
Hardy , G. H. & Littlewood, J. E. (1913). Contributions to the arithmetic theory of
series, Proc. London Math. Soc . (2) 11, 411–478; Collected P apers, V ol. 6. Oxford:
Clarendon Press, 1974, pp. 428–495.
(1914a). T auberian theorems concerning power series and Dirichlet series whose co-
efficients are positive, Proc. London Math. Soc . (2) 13, 174–191; Collected P apers,
V ol. 6. Oxford: Clarendon Press, 1974, pp. 510–527. | {
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5.4 References 165
(1914b). Some theorems concerning Dirichlet’s series, Messenger Math. 43, 134–147;
Collected P apers, V ol. 6. Oxford: Clarendon Press, 1974, pp. 542–555.
(1926). A further note on the converse of Abel’s theorem, Proc. London Math.
Soc. (2) 25, 219–236; Collected P apers , V ol. 6. Oxford: Clarendon Press, 1974,
pp. 699–716.
(1930). Notes on the theory of series XI: On T auberian theorems, Proc. London
Math. Soc . (2) 30, 23–37; Collected P apers, V ol. 6. Oxford: Clarendon Press, 1974,
pp. 745–759.
Hardy , G. H. & Riesz, M. (1915). The General Theory of Dirichlet’s Series , Cambridge
Tract No. 18. Cambridge: Cambridge University Press. Reprint: Stechert–Hafner
(1964).
Hewitt, E. & Williamson, H. (1957). Note on absolutely convergent Dirichlet series,
Proc. Amer . Math. Soc . 8, 863–868.
Ingham, A. E. (1962). On absolutely convergent Dirichlet series . Studies in Mathemati-
cal Analysis and Related T opics. Stanford: Stanford University Press, pp. 156–164.
(1965). On tauberian theorems, Proc. London Math. Soc . (3) 14A, 157–173.
Karamata, J. (1930). ¨Uber die Hardy–Littlewoodschen Umkehrungen des Abelschen
Stetigkeitssatzes, Math. Z . 32, 319–320.
(1931a). Neuer Beweis und V erallgemeinerung einiger T auberian-S¨ atze, Math. Z . 33,
294–300.
(1931b). Neuer Beweis und V erallgemeinerung der T auberschen S¨ atze, welche die
Laplacesche und Stieltjessche Transformation betreffen, J. Reine Angew . Math .
164, 27–40.
Kojima, T . (1917). On generalized T oeplitz’s theorems on limit and their application,
T ˆohoku Math. J . 12, 291–326.
Korevaar, J. (1951). An estimate of the error in T auberian theorems for power series,
Duke Math. J . 18, 723–734.
(1953). Best L 1 approximation and the remainder in Littlewood’s theorem, Proc.
Nederl. Akad. W etensch. Ser . A 56 (= Indagationes Math. 15), 281–293.
(1954a). A very general form of Littlewood’s theorem, Proc. Nederl. Akad. W etensch.
Ser . A 57 (= Indagationes Math. 16), 36–45.
(1954b). Another numerical T auberian theorem for power series, Proc. Nederl. Akad.
W etensch. Ser . A 57 (= Indagationes Math. 16), 46–56.
(1954c). Numerical T auberian theorems for Dirichlet and Lambert series, Proc.
Nederl. Akad. W etensch. Ser .A 57 (= Indagationes Math. 16), 152–160.
(1954d). Numerical T auberian theorems for power series and Dirichlet series, I, II,
Proc. Nederl. Akad. W etensch. Ser .A 57 (= Indagationes Math. 16), 432–443,
444–455.
(2001). T auberian theory , approximation, and lacunary series of powers, Trends in
approximation theory (Nashville, 2000), Innov . Appl. Math. Nashville: V anderbilt
University Press, pp. 169–189.
(2002). A century of complex T auberian theory , Bull. Amer . Math. Soc. (N.S.) 39,
475–531.
(2004). T auberian Theory . A Century of Developments. Grundl. Math. Wiss . 329.
Berlin: Springer-V erlag.
Landau, E. (1907). ¨Uber die Multiplikation Dirichletscher Reihen, Rend. Circ. Mat.
P alermo 24, 81–160. | {
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166 Dirichlet series: II
(1908). Zwei neue Herleitungen f ¨ ur die asymptotische Anzahl der Primzahlen unter
einer gegebenen Grenze, Sitzungsberichte Akad. Wiss . Berlin 746–764; Collected
W orks, V ol.4. Essen: Thales V erlag, 1986, pp. 21–39.
(1909). Handbuch der Lehre von der V erteilung der Primzahlen , Leipzig: T eubner.
Reprint: Chelsea (New Y ork), 1953.
(1910). ¨Uber die Bedeutung einiger neuerer Grenzwerts¨ atze der Herren Hardy und
Axer, Prace mat.-fiz. (W arsaw)21, 97–177; Collected W orks, V ol. 4. Essen: Thales
V erlag, 1986, pp. 267–347.
(1913). Einige Ungleichungen f ¨ ur zweimal differentiierbare Funktionen, Proc. Lon-
don Math. Soc . (2) 13, 43–49; Collected W orks, V ol. 6. Essen: Thales V erlag, 1986,
pp. 49–55.
L´evy , P . (1934). Sur la convergence absolue des s´ eries de Fourier, Compositio Math . 1,
1–14.
Littlewood, J. E. (1911). The converse of Abel’s theorem on power series, Proc. London
Math. Soc . (2) 9, 434–448; Collected P apers , V ol. 1. Oxford: Oxford University
Press, 1982, pp. 757–773.
(1986). Littlewood’s Miscellany , Bollobas, B. Ed., Cambridge: Cambridge University
Press.
van de Lune, J. (1986). An Introduction to T auberian Theory: From T auber to Wiener .
CWI Syllabus 12. Amsterdam: Mathematisch Centrum.
Mellin, H. (1902). ¨Uber den Zusammenhang zwischen den linearen Differential- und
Differenzengleichungen, Acta Math . 25, 139–164.
Montgomery , H. L. & Schinzel, A. (1977). Some arithmetic properties of polynomials in
several variables. Transcendence Theory: Advances and Applications (Cambridge,
1976). London: Academic Press, pp. 195–203.
Newman, D. J. (1975). A simple proof of Wiener’s 1 /f theorem, Proc. Amer . Math. Soc .
48, 264–265.
Perron, O. (1908). Zur Theorie der Dirichletschen Reihen, J. Reine Angew . Math . 134,
95–143.
Peyerimhoff, A. (1969). Lectures on summability , Lecture Notes Math. 107. Berlin:
Springer-V erlag.
Pitt, H. R. (1958). T auberian Theorems. T ata Monographs. London: Oxford University
Press.
Postnikov , A. G. (1951). The remainder term in the T auberian theorem of Hardy and
Littlewood, Dokl. Akad. Nauk SSSR N. S . 77, 193–196.
Riesz, M. (1909). Sur la sommation des s´ eries de Dirichlet, C. R. Acad. Sci. Paris 149,
18–21.
Rubel, L. (1989). Summability theory: a neglected tool of analysis, Amer . Math. Monthly
96, 421–423.
Schoenberg, I. J. (1973). The elementary cases of Landau’s problem of inequalities
between derivatives, Amer . Math. Monthly 80, 121–158.
Schur, I. (1921). ¨Uber lineare Transformationen in der Theorie der unendlichen Reihen,
J. Reine Angew . Math. 151, 79–111.
Steinhaus, H. (1911). Kilka sl ´ow o uog ´olnieniu poje ¸cia granicy, W arsaw: Prace mat-fiz
22, 121–134.
T auber, A. (1897). Ein Satz aus der Theorie der unendlichen Reihen, Monat. Math . 8,
273–277. | {
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5.4 References 167
Titchmarsh, E. C. (1939). The Theory of Functions , Second Edition. Oxford: Oxford
University Press.
(1986). The Theory of the Riemann Zeta-function , Second Edition. Oxford: Oxford
University Press.
T oeplitz, O. (1911). ¨Uber algemeine lineare Mittelbildungen , W arsaw: Prace mat–fiz
22, 113–119.
Widder, D. V . (1946). The Laplace transform , Princeton: Princeton University Press.
(1971). An Introduction to Transform Theory . New Y ork: Academic Press.
Wielandt, H. (1952). Zur Umkehrung des Abelschen Stetigkeitssatzes, Math Z . 56, 206–
207.
Wiener, N. (1932). T auberian theorems, Ann. of Math . (2) 33, 1–100.
(1933). The F ourier Integral, and Certain of its Applications . Cambridge: Cambridge
University Press.
Wintner, A. (1943). Eratosthenian averages . Baltimore: W averly Press.
Zygmund, A. (1968). Trigonometric series , V ol. 1, Second Edition. Cambridge: Cam-
bridge University Press. | {
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6
The Prime Number Theorem
6.1 A zero-free region
The Prime Number Theorem (PNT) asserts that
π(x ) ∼ x
log x
as x tends to infinity . W e shall prove this by using Perron’s formula, but in
the course of our arguments it will be important to know that ζ(s) ̸=0 for
σ ≥ 1. In Chapter 1 we saw that ζ(s) ̸=0 for σ> 1, but it remains to show
that ζ(1 + it ) ̸=0. T o obtain a quantitative form of the Prime Number The-
orem we take some care to show that ζ(s) ̸=0 for σ ≥ 1 − δ(t ) where δ(t )
is some function of t . W e would like the width δ(t ) of the zero-free region
to be as large as possible, as the rate at which δ(t ) tends to 0 determines the
size of the estimate we can derive for the error term in the Prime Number
Theorem.
W e begin by reviewing some basic facts concerning functions of a complex
variable. If P (z) is a polynomial, then the rate of growth of | P (z)| as |z|→
∞ reflects the number of zeros of P (z). This is generalized to other analytic
functions by Jensen’s formula. For our purposes we are content to establish the
following simple consequence of Jensen’s formula.
Lemma 6.1(Jensen’s inequality) If f (z) is analytic in a domain containing
the disc |z|≤ R, if | f (z)|≤ M in this disc, and if f (0) ̸=0, then for r < R the
number of zeros of f in the disc |z|≤ r does not exceed
log M/| f (0)|
log R/r .
Proof Let z1 ,z2 ,..., z K denote the zeros of f in the disc |z|≤ R, and
168 | {
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6.1 A zero-free region 169
put
g(z) = f (z)
K∏
k=1
R2 − z zk
R(z − zk ) .
The kth factor of the product has been constructed so that it has a pole at zk , and
so that it has modulus 1 on the circle |z|= R. Hence g is an analytic function
in the disc |z|≤ R, and if |z|= R, then |g(z)|=| f (z)|≤ M . Hence by the
maximum modulus principle, |g(0)|≤ M . But
|g(0)|=| f (0)|
K∏
k=1
R
|zk | .
Each factor in the product is ≥ 1, and if |zk |≤ r , then the factor is ≥ R/r .I f
there are L such zeros, then the above is ≥| f (0)|( R/r )L , which gives the stated
upper bound for L . □
W e now show that a bound for the modulus of an analytic function can be
derived from a one-sided bound for its real part in a slightly larger region.
Lemma 6.2(The Borel–Carath´ eodory Lemma) Suppose that h (z) is analytic
in a domain containing the disc |z|≤ R, that h (0) = 0, and that ℜh(z) ≤ M
for |z|≤ R. If |z|≤ r < R, then
|h(z)|≤ 2 Mr
R − r
and
|h′(z)|≤ 2 MR
( R − r )2 .
Proof It suffices to show that
⏐
⏐
⏐
⏐
h(k) (0)
k!
⏐
⏐
⏐
⏐≤ 2 M
Rk (6.1)
for all k ≥ 1, for then
|h(z)|≤
∞∑
k=1
⏐
⏐
⏐
⏐
h(k) (0)
k!
⏐
⏐
⏐
⏐r k ≤ 2 M
∞∑
k=1
(r
R
)k
= 2 Mr
R − r ,
and
|h′(z)|≤
∞∑
k=1
|h(k) (0)|kr k−1
k! ≤ 2 M
R
∞∑
k=1
k
(r
R
)k−1
= 2 MR
( R − r )2 .
T o prove (6.1) we first note that
∫ 1
0
h( Re (θ)) d θ = 1
2πi
∮
|z|=R
h(z) dz
z = h(0) = 0. | {
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170 The Prime Number Theorem
Moreover, if k > 0, then
∫ 1
0
h( Re (θ))e(kθ) d θ = R−k
2πi
∮
|z|=R
h(z)zk−1 dz = 0,
and
∫ 1
0
h( Re (θ))e(−kθ) d θ = Rk
2πi
∮
|z|=R
h(z)z−k−1 dz = Rk h(k) (0)
k! .
By forming a linear combination of these identities we see that if k > 0, then
∫ 1
0
h( Re (θ))(1 + cos 2 π(kθ + φ)) d θ = Rk e(−φ)h(k) (0)
2 · k! .
By taking real parts it follows that
ℜ
(1
2 Rk e(−φ)h(k) (0)/k!
)
≤ M
∫ 1
0
(1 + cos 2 π(kθ + φ)) d θ = M
for k > 0. Since this holds for any real φ, we are free to choose φ so that
e(−φ)h(k) (0) =| h(k) (0)|. Then the above inequality gives (6.1), and the proof
is complete. □
If P (z) = c ∏ K
k=1 (z − zk ), then
P ′
P (z) =
K∑
k=1
1
z − zk
.
W e now generalize this to analytic functions f (z), to the extent that f ′/f can
be approximated by a sum over its nearby zeros.
Lemma 6.3Suppose that f (z) is analytic in a domain containing the disc
|z|≤ 1, that | f (z)|≤ M in this disc, and that f (0) ̸=0. Let r and R be fixed,
0 <r < R < 1. Then for |z|≤ r we have
f ′
f (z) =
K∑
k=1
1
z − zk
+ O
(
log M
| f (0)|
)
where the sum is extended over all zeros z k of f for which |zk |≤ R. (The implicit
constant depends on r and R, but is otherwise absolute. )
Proof If f (z) has zeros on the circle |z|= R, then we replace R by a very
slightly larger value. Thus we may assume that f (z) ̸=0 for |z|= R. Set
g(z) = f (z)
K∏
k=1
R2 − z zk
R(z − zk ) . | {
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6.1 A zero-free region 171
By Lemma 6.1 we know that
K ≤ log M/| f (0)|
log 1 /R ≪ log M
| f (0)| . (6.2)
If |z|= R, then each factor in the product has modulus 1. Consequently |g(z)|≤
M when |z|= R, and by the maximum modulus principle |g(z)|≤ M for |z|≤
R. W e also note that
|g(0)|=| f (0)|
K∏
k=1
R
|zk | ≥| f (0)|.
Since g(z) has no zeros in the disc |z|≤ R, we may put h(z) = log(g(z)/g(0)).
Then h(0) = 0, and
ℜh(z) = log |g(z)|− log |g(0)|≤ log M − log | f (0)|
for |z|≤ R. Hence by the Borel–Carath´ eodory lemma we see that
h′(z) ≪ log M
| f (0)| (6.3)
for |z|≤ r . But
h′(z) = g′
g (z) = f ′
f (z) −
K∑
k=1
1
z − zk
+
K∑
k=1
1
z − R2/zk
. (6.4)
Now |R2/zk |≥ R, so that if |z|≤ r then |z − R2/zk |≥ R − r . Hence for |z|≤ r
the last sum above has modulus
≤ K
R − r ≪ log M
| f (0)|
by (6.2). T o obtain the stated result it suffices to combine this estimate and (6.3)
in (6.4).□
W e now apply these general principles to the zeta function.
Lemma 6.4 If |t |≥ 7/8 and 5/6 ≤ σ ≤ 2, then
ζ′
ζ (s) =
∑
ρ
1
s − ρ + O (log τ)
where τ =| t |+ 4 and the sum is extended over all zeros ρ of ζ(s) for which
|ρ − (3/2 + it )|≤ 5/6.
Proof W e apply Lemma 6.3 to the function f (z) = ζ(z + (3/2 + it )), with
R = 5/6 and r = 2/3. T o complete the proof it suffices to note that | f (0)|≫ 1
by the (absolutely convergent) Euler product formula (1.17), and that f (z) ≪ τ
for |z|≤ 1 by Corollary 1.17. □ | {
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172 The Prime Number Theorem
If the zeta function were to have a zero of multiplicity m at 1 + i γ, then we
would have
ζ′
ζ (1 + δ + i γ) ∼ m
δ
as δ → 0+. But
ℜ ζ′
ζ (1 + δ + i γ) =−
∞∑
n=1
/Lambda1 (n)n−1−δ cos(γ log n),
and in the very worst case this could be no larger than
∞∑
n=1
/Lambda1 (n)n−1−δ =− ζ′
ζ (1 + δ) ∼ 1
δ.
Thus m is at most 1, and even in this case ζ′/ζ would be essentially as large as
it could possibly be. Roughly speaking, this would imply that pi γ is near −1
for most primes. But then it would follow that p2i γ is near 1 for most primes,
so that
ζ′
ζ (1 + δ + 2i γ) ∼− 1
δ
as δ → 0+. Then ζ(s) would have a pole at 1 + 2i γ, contrary to Corollary
1.13. The essence of this informal argument is captured very effectively by the
following elementary inequality .
Lemma 6.5If σ> 1, then
ℜ
(
−3 ζ′
ζ (σ) − 4 ζ′
ζ (σ + it ) − ζ′
ζ (σ + 2it )
)
≥ 0.
Proof From Corollary 1.11 we see that the left-hand side above is
∞∑
n=1
/Lambda1 (n)n−1−δ(
3 + 4 cos( t log n) + cos(2t log n)
)
.
It now suffices to note that 3 + 4 cos θ + cos 2 θ = 2(1 + cos θ)2 ≥ 0 for
all θ. □
W e now use Lemmas 6.4 and 6.5 to establish the existence of a zero-free
region for the zeta function.
Theorem 6.6There is an absolute constant c > 0 such that ζ(s) ̸=0 for
σ ≥ 1 − c/log τ.
This is the classical zero-free region for the zeta function. | {
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6.1 A zero-free region 173
Proof Since ζ(s) is given by the absolutely convergent product (1.17) for
σ> 1, it suffices to consider σ ≤ 1. From (1.24) we see that
⏐
⏐
⏐
⏐ζ(s) − s
s − 1
⏐
⏐
⏐
⏐≤| s|
∫ ∞
1
u−σ−1 du = |s|
σ (6.5)
for σ> 0. From this we see that ζ(s) ̸=0 when σ> |s − 1|, i.e., in the parabolic
region σ> (1 + t 2 )/2. In particular, ζ(s) ̸=0 in the rectangle 8 /9 ≤ σ ≤ 1,
|t |≤ 7/8. Now suppose that ρ0 = β0 + i γ0 is a zero of the zeta function with
5/6 ≤ β0 ≤ 1, |γ0 |≥ 7/8. Since ℜρ ≤ 1 for all zeros ρ of ζ(s), it follows that
ℜ1/(s − ρ) > 0 whenever σ> 1. Hence by Lemma 6.4 with s = 1 + δ + i γ0
we see that
−ℜ ζ′
ζ (1 + δ + i γ0 ) ≤− 1
1 + δ − β0
+ c1 log(|γ0 |+ 4).
Similarly , by Lemma 6.4 with s = 1 + δ + 2i γ0 we find that
ℜ− ζ′
ζ (1 + δ + 2i γ0 ) ≤ c1 log(|2γ0 |+ 4).
From Corollary 1.13 we see that
− ζ′
ζ (1 + δ) = 1
δ + O (1).
On combining these estimates in Lemma 6.5 we conclude that
3
δ − 4
1 + δ − β0
+ c2 log(|γ0 |+ 4) ≥ 0.
W e take δ = 1/(2c2 log(|γ0 |+ 4)). Thus the above gives
7c2 log(|γ0 |+ 4) ≥ 4
1 + δ − β0
,
which is to say that
1 + 1
2c2 log(|γ0 |+ 4) − β0 ≥ 4
7c2 log(|γ0 |+ 4) .
Hence
1 − β0 ≥ 1
14c2 log(|γ0 |+ 4) ,
so the proof is complete. □
In the above argument it is essential that the coefficient of ζ(s) is larger
than the coefficient of ζ(σ). Among non-negative cosine polynomials T (θ) = | {
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174 The Prime Number Theorem
a0 + a1 cos 2 πθ +···+ aN cos 2 πN θ, the ratio a1 /a0 can be arbitrarily close
to 2, as we see in the Fej´ er kernel
/Delta1 N (θ) = 1 + 2
N −1∑
n=1
(
1 − n
N
)
cos 2 nπθ = 1
N
(sin πN θ
sin πθ
)2
≥ 0,
but it must be strictly less than 2 since
a0 − 1
2 a1 =
∫ 1
0
T (θ)(1 − cos 2 πθ) d θ> 0.
It is useful to have bounds for the zeta function and its logarithmic derivative
in the zero-free region.
Theorem 6.7Let c be the constant in Theorem 6.6 .I f σ> 1 − c/(2 log τ)
and |t |≥ 7/8, then
ζ′
ζ (s) ≪ log τ, (6.6)
| log ζ(s)|≤ log log τ + O (1) , (6.7)
and 1
ζ(s) ≪ log τ. (6.8)
On the other hand, if 1 − c/(2 log τ) <σ ≤ 2 and |t |≤ 7/8, then ζ′
ζ (s) =
−1/(s − 1) + O (1), log
(
ζ(s)(s − 1)
)
≪ 1, and 1/ζ(s) ≪| s − 1|.
Proof If σ> 1, then by Corollary 1.11 and the triangle inequality we see that
⏐
⏐
⏐
⏐
ζ′
ζ (s)
⏐
⏐
⏐
⏐≤
∞∑
n=1
/Lambda1 (n)n−σ =− ζ′
ζ (σ) ≪ 1
σ − 1 .
Hence (6.6) is obvious if σ ≥ 1 + 1/log τ. Let s1 = 1 + 1/log τ + it . In par-
ticular we have
ζ′
ζ (s1 ) ≪ log τ. (6.9)
From this estimate and Lemma 6.4 we deduce that
∑
ρ
ℜ 1
s1 − ρ ≪ log τ (6.10)
where the sum is over those zeros ρfor which |ρ − (3/2 + it )|≤ 5/6. Suppose
that 1 − c/(2 log τ) ≤ σ ≤ 1 + 1/log τ. Then by Lemma 6.4 we see that
ζ′
ζ (s) − ζ′
ζ (s1 ) =
∑
ρ
( 1
s − ρ − 1
s1 − ρ
)
+ O (log τ). (6.11) | {
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6.1 A zero-free region 175
Since |s − ρ|≍| s1 − ρ| for all zeros ρ in the sum, it follows that
1
s − ρ − 1
s1 − ρ ≪ 1
|s1 − ρ|2 log τ ≪ℜ 1
s1 − ρ.
Now (6.6) follows on combining this with (6.9) and (6.10) in (6.11).
T o derive (6.7) we begin as in our proof of (6.6). From Corollary 1.11 and
the triangle inequality we see that if σ> 1, then
| log ζ(s)|≤
∞∑
n=2
/Lambda1 (n)
log n n−σ = log ζ(σ).
But by Theorem 1.14 we know that ζ(σ) < 1 + 1/(σ − 1), so that (6.7)
holds when σ ≥ 1 + 1/log τ. In particular (6.7) holds at the point s1 =
1 + 1/log τ + it , so that to treat the remaining s it suffices to bound the
difference
log ζ(s) − log ζ(s1 ) =
∫ s
s1
ζ′
ζ (w) d w.
W e take the path of integration to be the line segment joining the endpoints.
Then the length of this interval multiplied by the bound (6.6) gives the error
term O(1) in (6.7).
The estimate (6.8) follows directly from (6.7), since log 1 /|ζ|=− ℜ log ζ.
The remaining estimates follow trivially from (6.5). □
The ideas we have used enable us not only to derive a zero-free region but
also to place a bound on the number of zeros ρ that might lie near the point
1 + it .
Theorem 6.8 Let n (r ; t ) denote the number of zeros ρ of ζ(s) in the disc
|ρ − (1 + it )|≤ r . Then n (r ; t ) ≪ r log τ, uniformly for r ≤ 3/4.
Proof If c1 is a small positive constant and r < c1 /log τ, then n(r ; t ) = 0b y
Theorem 6.6. Suppose that c1 /log τ ≤ r ≤ 1/6, |t |≥ 7/8. As in the proof of
Theorem 6.7, the estimate (6.10) holds when we take s1 = 1 + r + it . In the sum
over ρ, each term is non-negative, and those zeros ρcounted in n(r ; t ) contribute
at least 1 /(2r ) apiece. Hence their number is ≪ r log τ.I f1 /6 <r ≤ 3/4 and
|t |≥ 3, then the desired bound follows at once by applying Jensen’s inequality
(Lemma 6.1 above) to the function f (z) = ζ(z + 2 + it ), with R = 11/6, in
view of the bounds provided by Corollary 1.17. Note that | f (0)|≫ 1 because
of the absolute convergence of the Euler product. If 1 /6 <r ≤ 3/4 and |t |≤ 3,
then we apply Jensen’s inequality to the function f (z) = (z + 1 + it )ζ(z + 2 +
it ). □ | {
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176 The Prime Number Theorem
6.1.1 Exercises
1. (a) Show that if |z| < R, |w|≤ R, and z ̸=w, then
⏐
⏐
⏐
⏐
z
w− R2
(z − w) R
⏐
⏐
⏐
⏐≥ 1.
(b) Show that if |w|≤ ρ< R, |z|= r < R, and z ̸=w, then
⏐
⏐
⏐
⏐
z
w− R2
(z − w) R
⏐
⏐
⏐
⏐≥ rρ + R2
(r + ρ) R .
(c) Suppose that f is analytic in the disc |z|≤ R.F o r r ≤ R put M (r ) =
max|z|≤r | f (z)|. Show that if 0 <r < R and 0 <ρ< R, then the num-
ber of zeros of f in the disc |z|≤ ρ does not exceed
log M ( R)
M (r )
log rρ + R2
(r + ρ) R
.
2. Suppose that R, M , and ε are positive real numbers, and set h(z) =
2 Mz /(z + R + ε).
(a) Show that h(0) = 0, that h(z) is analytic for |z| < R + ε, and that
ℜh(z) ≤ M for |z|≤ R + ε.
(b) Show that if 0 <r < R, then
max
|z|≤r
|h(z)|=− h(−r ) = 2 Mr
R + ε− r .
(c) Show that if 0 <r < R, then
max
|z|≤r
|h′(z)|= h′(−r ) = 2 M ( R + ε)
( R + ε− r )2 .
3. Show that, in the situation of the Borel–Carath´ eodory lemma (Lemma 6.2),
if |z|≤ r < R, then
|h′′(z)|≤ 4 MR
( R − r )3 .
4. (Mertens 1898) Use the Dirichlet series expansion of log ζ(s) to show that
if σ> 1, then
|ζ(σ)3 ζ(σ + it )4 ζ(σ + 2it )|≥ 1.
The method used to establish a zero-free region for the zeta function can be
applied to any particular Dirichlet L -function, though the constants involved
may depend on the function. W e shall pursue this systematically in Chapter 11,
but in the exercise below we treat one interesting example. | {
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6.1 A zero-free region 177
5. Let χ0 denote the principal character (mod 4), and χ1 the non-principal
character (mod 4).
(a) Show thatL (1,χ1 ) = π/4, and hence that there is a neighbourhood of
1 in which L (s,χ1 ) ̸=0.
(b) Show that if σ> 1, then
ℜ
(
−3 L ′
L (σ,χ0 ) − 4 L ′
L (σ + it ,χ1 ) − L ′
L (σ + 2it ,χ0 )
)
≥ 0.
(c) Show that there is a constant c > 0 such that L (s,χ1 ) ̸=0 for σ>
1 − c/log τ.
(d) Show that there is a constant c > 0 such that if σ> 1 − c/log τ, then
L ′
L (s,χ1 ) ≪ log τ,
| log L (s,χ1 )|≤ log log τ + O (1),
1
L (s,χ1 ) ≪ log τ.
6. (a) Show that if 1 <σ 1 ≤ σ2 , then
ζ(σ2 )
ζ(σ1 ) ≤
⏐
⏐
⏐
⏐
ζ(σ2 + it )
ζ(σ1 + it )
⏐
⏐
⏐
⏐≤ ζ(σ1 )
ζ(σ2 )
for all real t .
(b) Show that if 1 <σ 1 ≤ σ2 ≤ 2, then
σ1 − 1
σ2 − 1 ≪
⏐
⏐
⏐
⏐
ζ(σ2 + it )
ζ(σ1 + it )
⏐
⏐
⏐
⏐≪ σ2 − 1
σ1 − 1
uniformly in t .
7. (Montgomery & V aughan 2001)
(a) Show that if σ> 1, then
⏐
⏐
⏐
⏐
ζ(σ + i (t + 1))
ζ(σ + it )
⏐
⏐
⏐
⏐≤ exp
(
2
∞∑
n=1
/Lambda1 (n)
nσ log n
⏐
⏐sin
(1
2 log n
)⏐
⏐
)
uniformly for all real t .
(b) Put f (θ) =| sin πθ|, and for integers k set ˆf (k) =
∫1
0 f (θ)e(−kθ) d θ
where e(θ) = e2πi θ. Show that ˆf (k) =− 2/(π(4k2 − 1)).
(c) By Corollary D.3, or otherwise, show that
| sin πθ|=
∞∑
k=−∞
ˆf (k)e(kθ) . | {
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178 The Prime Number Theorem
(d) Show that if 1 <σ ≤ 2, then
⏐
⏐
⏐
⏐
ζ(σ + i (t + 1))
ζ(σ + it )
⏐
⏐
⏐
⏐≤
∞∏
k=−∞
|ζ(σ + ik )|2 ˆf (k)
uniformly for all real t .
(e) Show that if σ> 1, then
(σ − 1)4/π ≪
⏐
⏐
⏐
⏐
ζ(σ + i (t + 1))
ζ(σ + it )
⏐
⏐
⏐
⏐≪ (σ − 1)−4/π
uniformly in t .
(f) Show that
(log t )−4/π ≪
⏐
⏐
⏐
⏐
ζ(1 + i (t + 1))
ζ(1 + it )
⏐
⏐
⏐
⏐≪ (log t )4/π
uniformly for t ≥ 2.
8. Suppose that a and b are fixed, 0 < a < b < 1. Suppose that f is analytic
in a domain containing the disc |z|≤ R, that f (0) ̸=0, and that | f (z)|≤ M
for |z|≤ R. Show that
f ′
f (z) =
K∑
k=1
1
z − zk
+ O
(1
R log M
| f (0)|
)
for |z|≤ aR where the sum is over those zeros zk of f (z) for which
|zk |≤ bR .
9. (Landau 1924a) Suppose that θ(t ) and φ(t ) are functions with the following
properties: φ(t ) > 0, φ(t ) ր, e−φ(t ) ≤ θ(t ) ≤ 1/2, θ(t ) ց. Suppose also
that
ζ(s) ≪ eφ(t )
for σ ≥ 1 − θ(t ), t ≥ 2.
(a) Show that
ζ′
ζ (s) =
∑
ρ
1
s − ρ + O
(φ(t + 1)
θ(t + 1)
)
for σ ≥ 1 − θ(t + 1)/3 where the sum is over zeros ρ for which |ρ −
(1 + θ(t + 1) + it )|≤ 5θ(t + 1)/3.
(b) Show that there is an absolute constant c > 0 such that ζ(s) ̸=0 for
σ ≥ 1 − c θ(2t + 1)
φ(2t + 1) .
(c) Show that the zero-free region (6.26) follows from the estimate (6.25). | {
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6.2 The Prime Number Theorem 179
(d) By mimicking the proof of Theorem 6.7, but with s1 = 1 +
θ(2t + 1)/φ(2t + 1) + it , show that
ζ′
ζ (s) ≪ φ(2t + 2)
θ(2t + 2) ,
| log ζ(s)|≤ log φ(2t + 2)
θ(2t + 2) + O (1),
1
ζ(s) ≪ φ(2t + 2)
θ(2t + 2)
for σ ≥ 1 − 1
2 cθ(2t + 2)/φ(2t + 2).
10. Suppose that ζ(s) ̸=0 for σ ≥ η(t ), t ≥ 2, where η(t ) ց, η(t ) ≫ 1/log t .
Show that
ζ′
ζ (s) ≪ log t
for σ ≥ 1 − 1
2 η(t + 1), t ≥ 2.
6.2 The Prime Number Theorem
W e are now in a position to prove the Prime Number Theorem in a quantitative
form. W e apply Perron’s formula toζ′
ζ (s) to obtain an asymptotic estimate for
ψ(x ) =
∑
n≤x
/Lambda1 (n),
and then use partial summation to derive an estimate for π(x ). It would be more
direct to apply Perron’s formula to log ζ(s), but our approach is technically
simpler since log ζ(s) has a logarithmic singularity at s = 1 while ζ′
ζ (s) has
only a simple pole there.
Theorem 6.9There is a constant c > 0 such that
ψ(x ) = x + O
( x
exp(c√ log x )
)
, (6.12)
ϑ(x ) = x + O
( x
exp(c√ log x )
)
, (6.13)
and
π(x ) = li(x ) + O
( x
exp(c√ log x )
)
(6.14)
uniformly for x ≥ 2. | {
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180 The Prime Number Theorem
Here li( x )i st h e logarithmic integral,
li(x ) =
∫ x
2
1
log u du .
By integrating this integral by parts K times we see that
li(x ) = x
K −1∑
k=1
(k − 1)!
(log x )k + OK
( x
(log x )K
)
. (6.15)
On combining this with (6.14) we see that
π(x ) = x
log x + O
( x
(log x )2
)
.
This is a quantitative form of the Prime Number Theorem. When this main term
is used, the error term is genuinely of the indicated size, since by (6.14) and
(6.15) again we see that
π(x ) = x
log x + x
(log x )2 + O
( x
(log x )3
)
.
Thus we see that in order to obtain a precise estimate of π(x ), it is essential
to use the logarithmic integral (or some similar function) to express the main
term.
ProofFrom Corollary 1.11 and Theorem 5.2 we see that
ψ(x ) = −1
2πi
∫ σ0 +iT
σ0 −iT
ζ′
ζ (s) x s
s ds + R (6.16)
for σ0 > 1, where by Corollary 5.3 we see that
R ≪
∑
x /2<n<2x
/Lambda1 (n) min
(
1, x
T |x − n|
)
+ (4x )σ0
T
∞∑
n=1
/Lambda1 (n)
nσ0
.
Here the second sum is − ζ′
ζ (σ0 ), which is ≍ 1/(σ0 − 1) for 1 <σ 0 ≤ 2. T o
estimate the first sum we note that /Lambda1 (n) ≤ log n ≪ log x . For the n that is
nearest to x we replace the minimum by its first member, and for all other
values of n we replace it by its second member. Thus the first sum is
≪ (log x )
(
1 + x
T
∑
1≤k≤x
1
k
)
≪ log x + x
T (log x )2 .
Suppose that 2 ≤ T ≤ x and that σ0 = 1 + 1/log x . Then
R ≪ x
T (log x )2 . | {
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6.2 The Prime Number Theorem 181
Put σ1 = 1 − c/log T where c is a small positive constant, and let C denote
the closed contour that consists of line segments joining the points σ0 − iT ,
σ0 + iT , σ1 + iT , σ1 − iT . From Theorem 6.6 we know that ζ′
ζ (s) has a simple
pole with residue −1a t s = 1, but that it is otherwise analytic within C. Hence
by the calculus of residues,
−1
2πi
∫
C
ζ′
ζ (s) x s
s ds = x .
If c is small, then the estimate (6.6) of Theorem 6.7 applies on this contour.
Hence
−
∫ σ1 +iT
σ0 +iT
ζ′
ζ (s) x s
s ds ≪ log T
T x σ0 (σ0 − σ1 ) ≪ x
T ,
and similarly for the integral from σ1 − iT to σ0 − iT . Using (6.6) again, we
also see that
−
∫ σ1 −iT
σ1 +iT
ζ′
ζ (s) x s
s ds ≪ x σ1 (log T )
∫ T
−T
dt
1 +| t | + x σ1
∫ 1
−1
dt
|σ1 + it − 1|
≪ x σ1 (log T )2 + x σ1
1 − σ1
≪ x σ1 (log T )2 .
On combining these estimates we conclude that
ψ(x ) = x + O
(
x (log x )2
(1
T + x −c/log T
))
.
W e choose T so that the two terms in the last factor of the error term are equal,
i.e., T = exp
(√c log x
)
. With this choice of T , the error term above is
≪ x (log x )2 exp
(
−
√
c log x
)
≪ x exp
(
− c
√
log x
)
since we may suppose that 0 < c < 1. Thus the proof of (6.12) is complete.
T o derive (6.13) it suffices to combine (6.12) with the first estimate of Corol-
lary 2.5. As for (6.14), we note that
π(x ) =
∫ x
2−
1
log u d ϑ(u) = li(x ) +
∫ x
2−
1
log u d (ϑ(u) − u).
By integrating by parts we see that this last integral is
ϑ(u) − u
log u
⏐
⏐
⏐
x
2−
+
∫ x
2
ϑ(u) − u
u(log u)2 du ,
and by (6.13) it follows that this is ≪ x exp(−c√ log x ). Thus we have (6.14),
and the proof is complete. □ | {
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Subsets and Splits