Chapter 4

The Riemann zeta function and

L-functions

4.1 Basic facts

We prove some results that will be used in the proof of the Prime Number Theorem

(for arithmetic progressions). The L-function of a Dirichlet character χ modulo q is

defined by

L(s, χ) =∞∑n=1

χ(n)n−s.

We view ζ(s) =∑∞

n=1 n−s as the L-function of the principal character modulo 1,

more precisely, ζ(s) = L(s, χ(1)0 ), where χ

(1)0 (n) = 1 for all n ∈ Z.

We first prove that ζ(s) has an analytic continuation to {s ∈ C : Re s > 0}\{1}.We use an important summation formula, due to Euler.

Lemma 4.1 (Euler’s summation formula). Let a, b be integers with a < b and

f : [a, b]→ C a continuously differentiable function. Then

b∑n=a

f(n) =

∫ b

a

f(x)dx+ f(a) +

∫ b

a

(x− [x])f ′(x)dx.

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Remark. This result often occurs in the more symmetric form

b∑n=a

f(n) =

∫ b

a

f(x)dx+ 12(f(a) + f(b)) +

∫ b

a

(x− [x]− 12)f ′(x)dx.

Proof. Let n ∈ {a, a+ 1, . . . , b− 1}. Then∫ n+1

n

(x− [x]

)f ′(x)dx =

∫ n+1

n

(x− n)f ′(x)dx

=[(x− n)f(x)

]n+1

n−∫ n+1

n

f(x)dx = f(n+ 1)−∫ n+1

n

f(x)dx.

By summing over n we get∫ b

a

(x− [x])f ′(x)dx =b∑

n=a+1

f(n)−∫ b

a

f(x)dx,

which implies at once Lemma 4.1.

We also need a result from the Prerequisites on analytic functions defined by

integrals, which we recall here.

Theorem 0.42. Let D be a measurable subset of Rm, U an open subset of C and

f : D × U → C a function with the following properties:

(i) f is measurable on D × U (with U viewed as subset of R2);

(ii) for every fixed x ∈ D, the function z 7→ f(x, z) is analytic on U ;

(iii) for every compact subset K of U there is a measurable function MK : D → Rsuch that

|f(x, z)| 6MK(x) for x ∈ D, z ∈ K,∫D

MK(x)dx <∞.

Then the function F given by

F (z) :=

∫D

f(x, z)dx

is analytic on U , and for every k > 1,

F (k)(z) =

∫D

f (k)(x, z)dx,

106

where f (k)(x, z) denotes the k-th derivative with respect to z of the analytic function

z 7→ f(x, z).

We now state and prove our result on the Riemann zeta function.

Theorem 4.2. ζ(s) has a unique analytic continuation to the set

{s ∈ C : Re s > 0, s 6= 1}, with a simple pole with residue 1 at s = 1.

Proof. By Corollary 0.39 we know that an analytic continuation of ζ(s), if such

exists, is unique.

For the moment, let s ∈ C with Re s > 1. Then by Lemma 4.1, with f(x) = x−s,

N∑n=1

n−s =

∫ N

1

x−sdx+ 1 +

∫ N

1

(x− [x])(−sx−1−s)dx

=1−N1−s

s− 1+ 1− s

∫ N

1

(x− [x])x−1−sdx.

If we let N → ∞ then the left-hand side converges, and also the first term on the

right-hand side, since |N−1−s| = N−1−Re s → 0. Hence the integral on the right-hand

side must converge as well. Thus, letting N →∞, we get for Re s > 1,

(4.1) ζ(s) =1

s− 1+ 1− s

∫ ∞1

(x− [x])x−1−sdx.

We now show that the integral on the right-hand side defines an analytic function

on U := {s ∈ C : Re s > 0}, by means of Theorem 0.42.

The function F (x, s) := (x − [x])x−1−s is measurable on [1,∞) × U (for its set

of discontinuities has Lebesgue measure 0 since it is the countable union of the sets

{n} × U with n ∈ Z>2, which all have Lebesgue measure 0) and for every fixed x it

is analytic in s. So conditions (i) and (ii) of Theorem 0.42 are satisfied. We verify

(iii). Let K be a compact subset of U . Then there is σ > 0 such that Re s > σ for

all s ∈ K. Now for x > 1 and s ∈ K we have

|(x− [x])x−1−s| 6 x−1−σ

and∫∞1x−1−σdx < ∞. So (iii) is also satisfied, and we may indeed conclude that

the integral on the right-hand side of (4.1) defines an analytic function on U .

107

Consequently, the right-hand side of (4.1) is analytic on {s ∈ C : Re s > 0}\{1}and since it is of the shape 1

s−1+something analytic around s = 1 it has a simple

pole at s = 1 with residue 1.

We may take this as our analytic continuation of ζ(s).

Theorem 4.3. Let q ∈ Z>2, and let χ be a Dirichlet character mod q.

(i) L(s, χ) =∏

p(1− χ(p)p−s)−1 for s ∈ C, Re s > 1.

(ii) If χ 6= χ(q)0 , then L(s, χ) converges, and is analytic on {s ∈ C : Re s > 0}.

(iii) L(s, χ(q)0 ) can be continued to an analytic function on {s ∈ C : Re s > 0, s 6= 1},

and for s in this set we have

L(s, χ(q)0 ) = ζ(s) ·

∏p|q

(1− p−s).

Hence L(s, χ(q)0 ) has a simple pole at s = 1.

Proof. (i) χ is a strongly multiplicative function, and L(s, χ) converges absolutely

for Re s > 1. Apply Corollary 2.14.

(ii) Let N be any positive integer. Then N = tq + r for certain integers t, r

with t > 0 and 0 6 r < q. By one of the orthogonality relations for characters (see

Theorem 3.8), we have∑q

m=1 χ(m) = 0,∑2q

m=q+1 χ(m) = 0, etc. Hence∣∣∣∣∣N∑n=1

χ(n)

∣∣∣∣∣ =∣∣∣χ(tq + 1) + · · ·+ χ(tq + r)

∣∣∣ 6 r < q.

This last upper bound is independent of N . Now Theorem 2.2 implies that the

L-series L(s, χ) converges and is analytic on Re s > 0.

(iii) By (i) we have for Re s > 1,

L(s, χ(q)0 ) =

∏p-q

(1− p−s)−1 = ζ(s)∏p|q

(1− p−s).

The right-hand side is defined and analytic on {s ∈ C : Re s > 0, s 6= 1}, and so it

can be taken as an analytic continuation of L(s, χ(q)0 ) on this set.

Corollary 4.4. Both ζ(s) and L(s, χ) for any character χ modulo an integer q > 2

are 6= 0 on {s ∈ C : Re s > 1}.

Proof. Use part (i) of the above theorem, together with Corollary 2.14.

108

4.2 Non-vanishing on the line Re s = 1

We prove that ζ(s) 6= 0 if Re s = 1 and s 6= 1, and L(s, χ) 6= 0 for any s ∈ Cwith Re s = 1 and any non-principal character χ modulo an integer q > 2. We

have to distinguish two cases, which are treated quite differently. We interpret ζ(s)

as L(s, χ(1)0 ). Recall that a character χ mod q is called real if χ(a) ∈ R for every

a ∈ Z. Clearly, real characters modulo q assume only the values −1, 1 on the integers

coprime with q and so, χ is real if and only if χ2 = χ(q)0 .

Theorem 4.5. Let q ∈ Z>1, χ a character mod q, and t a real. Assume that either

t 6= 0, or t = 0 but χ is not real. Then L(1 + it, χ) 6= 0.

Proof. We use a famous idea, due to Hadamard. It is based on the inequality

(4.2) 3 + 4 cos θ + cos 2θ = 2(1 + cos θ)2 > 0 for θ ∈ R.

Fix a real t such that t 6= 0 if χ is real. Suppose that L(1 + it, χ) = 0. Consider the

function in s,

F (s) := L(s, χ(q)0 )3 · L(s+ it, χ)4 · L(s+ 2it, χ2).

By the conditions imposed on χ and t, the function L(s+ 2it, χ2) is analytic around

s = 1. Further, L(s, χ(q)0 ) has a simple pole at s = 1, while L(s + it, χ) has by

assumption a zero at s = 1. Recall that for a function f meromorphic around

s = s0, ords=s0(f) is the integer n0 such that f(s) has a Laurent series expansion∑∞n=n0

an(s− s0)n around s = s0 with an0 6= 0. This ords=s0-function is additive on

products. Thus,

ords=1(F ) = 3 · ords=1

(L(s, χ

(q)0 ))

+ 4 · ords=1

(L(s+ it, χ)

)+ ords=1

(L(s+ 2it, χ2)

)> 3 · (−1) + 4 + 0 = 1.

This shows that F is analytic around s = 1, and has a zero at s = 1. We now prove

that |F (σ)| > 1 (or rather, log |F (σ)| > 0) for σ > 1. This gives a contradiction since

by continuity, limσ↓1 |F (σ)| should be 0. So our assumption that L(1 + it, χ) = 0

must be false.

109

From the definition of the function F we obtain that for σ > 1 we have

log |F (σ)| = log∏p

∣∣∣∣∣ 1

1− χ(q)0 (p)p−σ

∣∣∣∣∣3

·∣∣∣∣ 1

1− χ(p)p−σ−it

∣∣∣∣4 · ∣∣∣∣ 1

1− χ(p)2p−σ−2it

∣∣∣∣

=∑p-q

(3 log

∣∣∣∣ 1

1− p−σ

∣∣∣∣+ 4 log

∣∣∣∣ 1

1− χ(p)p−σ−it

∣∣∣∣+ log

∣∣∣∣ 1

1− χ(p)2p−σ−2it

∣∣∣∣) .Note that if p - q then χ(p) is a root of unity. Hence |χ(p)p−it| = |χ(p)e−it log p| = 1.

So we have χ(p)p−it = eiϕp with ϕp ∈ R. Hence

log |F (σ)| =∑p-q

(3 log

∣∣∣∣ 1

1− p−σ

∣∣∣∣+ 4 log

∣∣∣∣ 1

1− p−σeiϕp

∣∣∣∣+ log

∣∣∣∣ 1

1− p−σe2iϕp

∣∣∣∣) .Recall that

log1

1− z=∞∑n=1

zn

n, log

∣∣∣∣ 1

1− z

∣∣∣∣ = Re log1

1− zfor z ∈ C with |z| < 1.

Hence for r, ϕ ∈ R with 0 < r < 1,

log

∣∣∣∣ 1

1− reiϕ

∣∣∣∣ = Re

(log

1

1− reiϕ

)= Re

(∞∑n=1

(reiϕ)n

n

)

=∞∑n=1

rn

nRe (einϕ) =

∞∑n=1

rn

n· cosnϕ.

This leads to

log |F (σ)| =∑p-q

(3∞∑n=1

p−nσ

n+ 4

∞∑n=1

p−nσ

n· cosnϕp +

∞∑n=1

p−nσ

ncos 2nϕp

)

=∑p-q

∞∑n=1

p−nσ

n(3 + 4 cosnϕp + cos 2nϕp) > 0,

using (4.2). This shows that indeed, |F (σ)| > 1 for σ > 1, giving us the contradiction

we want.

110

It remains to prove that L(1, χ) 6= 0 for any real character χ mod q that is

not principal. Dirichlet needed this fact already in his proof that for every pair of

integers q, a with q > 3 and gcd(a, q) = 1 there are infinitely many primes p with

p ≡ a (mod q). Dirichlet had a rather complicated proof that L(1, χ) 6= 0, based

on Dirichlet series associated with quadratic forms (in modern language: Dedekind

zeta functions for quadratic number fields) and class number formulas.

Landau found a much more direct proof, which we give here, based on a simple

result for Dirichlet series, which more or less asserts that a Dirichlet series with non-

negative real coefficients can not be continued analytically beyond the boundary of

its half plane of convergence.

Lemma 4.6 (Landau). Let f : Z>0 → R be an arithmetic function with f(n) > 0

for all n. Suppose that Lf (s) =∑∞

n=1 f(n)n−s has abscissa of convergence σ0.

Then Lf (s) cannot be continued analytically to any open set containing {s ∈ C :

Re s > σ0} ∪ {σ0}.

Proof.Suppose Lf (s) can be continued to an ana-

lytic function g(s) on an open set containing

{s ∈ C : Re s > σ0} ∪ {σ0}. Then there is

δ > 0 such that g(s) is analytic on the open

disk D(σ0, δ) with center σ0 and radius δ. Let

σ1 := σ0 + δ/3. Then D(σ1, 2δ/3) ⊂ D(σ0, δ),

so g(s) is analytic and has a Taylor series ex-

pansion around σ1 converging on D(σ1, 2δ/3).

Now take σ with σ0 − δ/3 < σ < σ0, so that

σ ∈ D(σ1, 2δ/3). Using the Taylor series ex-

pansion of g(s) around σ1, we get

g(σ) =∞∑k=0

g(k)(σ1)

k!· (σ − σ1)k.

Since σ1 is larger than the abscissa of conver-

gence σ0 of Lf (s), we have

111

g(k)(σ1) = L(k)f (σ1) =

∞∑n=1

f(n)(− log n)kn−σ1 for k > 0.

Hence

g(σ) =∞∑k=0

1

k!

(∞∑n=1

f(n)(− log n)kn−σ1

)(σ − σ1)k

=∞∑k=0

1

k!

(∞∑n=1

f(n)(log n)kn−σ1

)(σ1 − σ)k.

Now all terms are non-negative, hence it is allowed to interchange the summations.

Thus,

g(σ) =∞∑n=1

f(n)n−σ1

(∞∑k=0

1

k!(log n)k(σ1 − σ)k

)

=∞∑n=1

f(n)n−σ1e(logn)(σ1−σ)(

using ez =∞∑k=0

zk/k!)

=∞∑n=1

f(n)n−σ1nσ1−σ =∞∑n=1

f(n)n−σ.

We see that Lf (s) converges for s = σ. But this is impossible, since σ is smaller

than the abscissa of convergence σ0 of Lf (s). So our initial assumption that Lf (s)

has an analytic continuation to an open set containing {s ∈ C : Re s > σ0} ∪ {σ0}is false.

Remark. Lemma 4.6 becomes false if we drop the condition that f(n) > 0 for all n.

For instance, if χ is a non-principal character mod q, then L(s, χ) =∑∞

n=1 χ(n)n−s

diverges if Re s < 0, but one can show that L(s, χ) has an analytic continuation to

the whole of C.

Theorem 4.7. Let q ∈ Z>2, and let χ be a real, non-principal character mod q.

Then L(1, χ) 6= 0.

Proof. Assume that L(1, χ) = 0. Consider the function

F (s) := L(s, χ)ζ(s).

112

By Theorems 4.2, 4.3, this function is analytic at least on {s ∈ C : Re s > 0, s 6= 1}.Further, ζ(s) has a simple pole at s = 1, and since L(s, χ) has by assumption a zero

at s = 1, we get

ords=1(F ) > 1 + (−1) = 0,

i.e., F is analytic at s = 1 as well. Hence F is analytic for all s with Re s > 0. We

show that for s ∈ C with Re s > 1, F (s) is expressable as a Dirichlet series with

non-negative coefficients. By Lemma 4.6, this Dirichlet series should have abscissa

of convergence 6 0. But we show that the abscissa of convergence of this series is

> 12

and derive a contradiction.

The series ζ(s) =∑∞

n=1 n−s and

∑∞n=1 χ(n)n−s converge absolutely if Re s > 1.

So by Theorem 2.12,

F (s) = Lf (s) =∞∑n=1

f(n)n−s for s ∈ C, Re s > 1,

where f = E ∗ χ, i.e.,

f(n) =∑d|n

χ(d) for n ∈ Z>0.

Hence f is a multiplicative function. We compute f in the prime powers. We have

χ(n) = ±1 for all n ∈ Z with gcd(n, q) = 1, while χ(n) = 0 if gcd(n, q) > 1. Hence,

if p is a prime and k a non-negative integer, we have

f(pk) =k∑j=0

χ(p)j =

1 if p|q,

k + 1 if p - q, χ(p) = 1,

1 if p - q, χ(p) = −1, k even,

0 if p - q, χ(p) = −1, k odd.

Therefore, f(pk) > 0 for all prime powers pk. Since f is multiplicative, it follows

that f(n) > 0 for all n ∈ Z>0.

The series Lf (s) has an analytic continuation to {s ∈ C : Re s > 0}, which is

F (s). So by Lemma 4.6, Lf (s) has abscissa of convergence σ0(f) 6 0. On the other

hand, from the above table and from the fact that f is multiplicative, it follows that

if n = m2 is a square, then f(n) > 1. Hence

Lf (σ) =∞∑n=1

f(n)n−σ >∞∑m=1

m−2σ =∞ if σ 6 12.

So σ0(f) > 12. This gives a contradiction, and so our assumption that L(1, χ) = 0

has to be false.

113

4.3 Functional equations

Euler’s Gamma function plays an important role in the functional equations of ζ(s),

the L-functions, and various generalizations thereof. Euler’s Gamma function is

given by

Γ(z) :=

∫ ∞0

e−ttz−1dt for z ∈ C with Re z > 0,

where tz := ez log t for z ∈ C, with log t the ordinary real natural logarithm of t. We

collect here some properties:

Lemma 4.8. (i) Γ(z) defines an analytic function on {z ∈ C : Re z > 0};(ii) Γ(z + 1) = zΓ(z) for z ∈ C, Re z > 0;

(iii) Γ(n) = (n− 1)! for n = 1, 2, . . . and Γ(12) =√π;

(iv) Γ(z) has an analytic continuation to C\{0,−1,−2, . . .} with a simple pole with

residue (−1)n/n! at z = −n, for n = 0, 1, 2, . . .;

(v) Γ(z) 6= 0 for z ∈ C \ {0,−1,−2, . . .};

(vi) Γ(z)Γ(1− z) =π

sin πzfor z ∈ C \ Z;

(vii) Γ(2z) =22z−1√π· Γ(z)Γ(z + 1

2) for z ∈ C, z 6= 0,−1

2,−1,−3

2,−2, . . . (duplication

formula).

Property (i) is an easy consequence of Theorem 0.42, and property (ii) an easy

application of integration by parts. The interested reader may consult Section 4.4

for the proofs of (i)–(vii).

Now define

(4.3) ξ(s) := 12s(s− 1)π−s/2Γ(1

2s)ζ(s) = (s− 1)π−s/2Γ(1

2s+ 1)ζ(s),

where we have used the identity 12sΓ(1

2s) = Γ(1

2s + 1). In his memoir from 1859,

Riemann proved the famous identity

(4.4) ξ(s) = 12

+ 12s(s− 1)

∫ ∞1

ω(t)(t(s/2)−1 + t−(s+1)/2

)dt if Re s > 1,

where ω(t) :=∑∞

m=1 e−πm2t. Using Theorem 0.42 one shows that the integral is

defined and analytic for all s ∈ C. This immediately leads to

114

Theorem 4.9. The function ξ has an analytic continuation to C.

For this continuation we have ξ(1−s) = ξ(s) for s ∈ C and moreover, ξ(0) = ξ(1) =12.

For the interested reader we have included a proof of (4.4) and Theorem 4.9 in

Section 4.5. See also H. Davenport, Multiplicative Number Theory, Chapter 8.

We deduce some consequences.

Corollary 4.10. ζ has an analytic continuation to C \ {1}, with a simple pole with

residue 1 at s = 1.

Proof. For ζ(s) with Re s > 1 we have an expression

(4.5) ζ(s) =ξ(s)πs/2 · Γ(1

2s+ 1)−1

s− 1.

By Theorem 4.9, ξ(s) has an analytic continuation to C. Further, πs/2 is analytic

on C, and moreover, Γ(12s+ 1)−1 defines an analytic function on C, since Γ has only

poles and no zeros. Hence (4.5) gives an analytic continuation of ζ(s) to C \ {1}.Using Γ(3

2) = 1

2Γ(1

2) = 1

2

√π, we get

lims→1

(s− 1)ζ(s) = ξ(1)π1/2Γ(32)−1 = 1,

which shows that the analytic continuation of ζ(s) defined by (4.5) has a simple pole

with residue 1 at s = 1.

Remark. On {s ∈ C : Re s > 0} \ {1} the right-hand side of (4.5) coincides with

the analytic continuation defined in (4.1) since analytic continuations to connected

sets are uniquely determined.

Corollary 4.11. ζ has simple zeros at s = −2,−4,−6, . . ..

ζ has no other zeros outside the critical strip {s ∈ C : 0 < Re s < 1}.

Proof. By Corollary 4.4 and Theorem 4.5, we know that ζ(s) 6= 0 for s ∈ C with

Re s > 1, s 6= 1. Further, lims→1(s−1)ζ(s) = 1, hence (s−1)ζ(s) 6= 0 if Re s > 1. We

know also that πs/2 6= 0 for s ∈ C, that Γ(12s+1) has a simple pole at s = −2,−4, . . .

and Γ(12s + 1) 6= 0 for s 6= −2,−4, . . .. Using the second expression of (4.3) this

implies ξ(s) 6= 0 for s ∈ C with Re s > 1, and then also ξ(s) 6= 0 if Re s 6 0 by

115

Theorem 4.9. Together with what we observed above about πs/2 and Γ(12s + 1),

using (4.5), we obtain that for s = s0 with Re s0 6 0,

ords=s0ζ(s) = −ords=s0Γ(12s+ 1) =

{1 if s0 ∈ {−2,−4,−6, . . .},0 if s0 6∈ {−2,−4,−6, . . .}.

This proves Corollary 4.11.

Corollary 4.12. Suppose ζ has a zero s0 with 0 < Re s0 < 1. Then 1− s0, s0 and

1− s0 are also zeros of ζ.

Proof. We have ξ(s0) = 0, and so ξ(1− s0) = 0 by Theorem 4.9. But π−s/2Γ(12s) is

non-zero and has no poles in the critical strip 0 < Re s < 1, hence ζ(1−s0) = 0. The

function ζ assumes real values on R>1, so by Schwarz’ reflection principle (Corollary

0.41 from the Prerequisites), ζ(s) = ζ(s) for s ∈ C\{1}, and in particular ζ(s0) = 0,

ζ(1− s0) = 0.

In Exercise 4.3 you will be asked to prove that ζ(s) < 0 for s ∈ R, 0 < s < 1.

Inserting ξ(0) = 12

and Γ(1) = 1 in (4.5) one easily shows that ζ(0) = −12.

There are also functional equations for L-functions L(s, χ), in the case that χ is

a primitive character modulo an integer q > 2 (that is to say, χ is not induced by a

character modulo d for any proper divisor d of q).

Notice that for any character χ modulo q we have χ(−1)2 = χ(1) = 1, hence

χ(−1) ∈ {−1, 1}. A character χ is called even if χ(−1) = 1, and odd if χ(−1) = −1.

There will be different functional equations for even and odd characters.

In Chapter 3 we defined the Gauss sum related to a character χ mod q by

τ(1, χ) =

q−1∑a=0

χ(a)e2πia/q.

By χ we denote the complex conjugate of a character χ.

Theorem 4.13. Let q be an integer with q > 2, and χ a primitive character mod q.

Put

ξ(s, χ) :=( qπ

)s/2Γ(1

2s)L(s, χ), c(χ) :=

√q

τ(1, χ)if χ is even,

ξ(s, χ) :=( qπ

)(s+1)/2

Γ(12(s+ 1)

)L(s, χ), c(χ) :=

i√q

τ(1, χ)if χ is odd.

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Then ξ(s, χ) has an analytic continuation to C, and

ξ(1− s, χ) = c(χ)ξ(s, χ) for s ∈ C.

Remark. According to Theorem 3.19, if χ is primitive then |τ(1, χ)| =√q, hence

|c(χ)| = 1. In general, it is a difficult problem to compute c(χ) for large values of q.

The proof of Theorem 4.13 is similar to that of that of the functional equation

for ζ(s), but with some additional technicalities, see H. Davenport, Multiplicative

Number Theory, Chapter 9.

In Exercise 4.6 you will be asked to deduce some consequences.

4.4 Properties of Euler’s Gamma function

Euler’s Gamma function is defined by the integral

Γ(z) :=

∫ ∞0

e−ttz−1dt (z ∈ C, Re z > 0).

Lemma 4.14. Γ(z) defines an analytic function on {z ∈ C : Re z > 0}.

Proof. This is standard using Theorem 0.42. Let U := {z ∈ C : Re z > 0}. First,

the function F (t, z) := e−ttz−1 is continuous, hence measurable on R>0×U . Second,

for each fixed t > 0, z 7→ e−ttz−1 is analytic on U . Third, let K be a compact subset

of U . Then there exist δ, R > 0 such that δ 6 Re z 6 R for z ∈ K. This implies

that for z ∈ K, t > 0,

|e−ttz−1| 6M(t) :=

{tδ−1 for 0 6 t 6 1,

e−ttR−1 6 Ce−t/2 for t > 1,

where C is some constant. Now we have∫ ∞0

M(t)dt =

∫ 1

0

tδ−1dt+ C

∫ ∞1

e−t/2dt = δ−1 + 2C · e−1/2 <∞.

Hence all conditions of Theorem 0.42 are satisfied, and thus, Γ(z) is analytic on

U .

117

Using integration by parts, one easily shows that for z ∈ C with Re z > 0,

Γ(z) = z−1∫ ∞0

e−tdtz

= z−1([e−ttz

]t=∞t=0

+

∫ ∞0

e−ttzdt

)= z−1Γ(z + 1),

that is,

(4.6) Γ(z + 1) = zΓ(z) if Re z > 0.

One easily shows that Γ(1) = 1 and then by induction, Γ(n) = (n− 1)! for n ∈ Z>0.

We now show that Γ has a meromorphic continuation to C.

Theorem 4.15. The function Γ has a unique analytic continuation to C\{0,−1,−2, . . .}.If we denote this continuation also by Γ, it has the following properties:

(i) Γ has a simple pole with residue (−1)n/n! at z = −n for n = 0, 1, 2, . . .;

(ii) Γ(z + 1) = zΓ(z) for z ∈ C \ {0,−1,−2, . . .}.

Proof. Corollary 0.39 from the Prerequisites and the analyticity of the integral imply

that the analytic continuation, if it exists, is unique. We proceed to construct this

analytic continuation.

First let z ∈ C with Re z > 0. By repeatedly applying (4.6) we get

(4.7) Γ(z) =1

z(z + 1) · · · (z + n− 1)· Γ(z + n) for Re z > 0, n = 1, 2, . . . .

The right-hand side clearly defines an analytic continuation of Γ to Bn := {z ∈ C :

Re z > −n, z 6= 0, 1, . . . , 1 − n}, for n = 1, 2, . . .. According to Theorem 0.40 from

the Prerequisites, Γ has an analytic continuation to B := C \{0,−1,−2, . . .}, which

coincides on Bn with the right-hand side of (4.7), for n = 1, 2, . . ..

By (4.7) we have

limz→−n

(z + n)Γ(z) = limz→−n

(z + n)1

z(z + 1) · · · (z + n)Γ(z + n+ 1)

=1

(−n)(−n+ 1) · · · (−1)Γ(1) =

(−1)n

n!.

Hence Γ has a simple pole at z = −n of residue (−1)n/n!.

118

Both functions Γ(z + 1) and zΓ(z) are analytic on B, and by (4.6), they are

equal on the set {z ∈ C : Re z > 0} which has limit points in B. So by Corollary

0.39, Γ(z + 1) = zΓ(z) for z ∈ B.

Theorem 4.16. We have Γ(z)Γ(1− z) =π

sinπzfor z ∈ C \ Z.

Proof. We prove that zΓ(z)Γ(1− z) = πz/ sin πz, or equivalently,

(4.8) Γ(1 + z)Γ(1− z) =πz

sin πzfor z ∈ A := (C \ Z) ∪ {0},

which implies Theorem 4.16. Notice that by Theorem 4.15 the left-hand side is

analytic on A, while by limz→0 πz/ sin πz = 1 the right-hand side is also analytic

on A. By Corollary 0.37, it suffices to prove that if S is any infinite subset of A

with a limit point in A, then (4.8) holds for every z ∈ S. For this infinite set we

take S := { 12n : n ∈ Z>0}; this set has limit point 0 in A. Thus, (4.8), and hence

Theorem 4.16, follows once we have proved that

(4.9) Γ(1 + 1

2n

)· Γ(1− 1

2n

)=

π/2n

sin π/2n(n = 1, 2, . . .).

Notice that

Γ(1 + 1

2n

)· Γ(1− 1

2n

)=

∫ ∞0

e−ss1/2nds ·∫ ∞0

e−tt−1/2ndt

=

∫ ∞0

∫ ∞0

e−s−t(s/t)1/2ndsdt.

Define new variables u = s + t, v = s/t. Then s = uv/(v + 1), t = u/(v + 1). The

Jacobian of the substitution (s, t) 7→ (u, v) is

∂(s, t)

∂(u, v)=

∣∣∣∣∣∣∣∂s

∂u

∂s

∂v∂t

∂u

∂t

∂v

∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣v

v + 1

u

(v + 1)2

1

v + 1− u

(v + 1)2

∣∣∣∣∣∣∣=−uv − u(v + 1)3

=−u

(v + 1)2.

It follows that

Γ(1 + 1

2n

)· Γ(1− 1

2n

)=

∫ ∞0

∫ ∞0

e−uv1/2n∣∣∣∣ ∂(s, t)

∂(u, v)

∣∣∣∣ · dudv=

∫ ∞0

∫ ∞0

e−uv1/2nu

(v + 1)2· dudv =

∫ ∞0

e−uudu ·∫ ∞0

v1/2n

(v + 1)2dv.

119

In the last product, the first integral is equal to 1, while, as we will show below, the

second integral isπ/2n

sinπ/2n . This proves (4.9).

Indeed, we have∫ ∞0

v1/2n

(v + 1)2dv = −

∫ ∞0

v1/2nd( 1

v + 1

)= −

[v1/2n

v + 1

]∞0

+

∫ ∞0

1

v + 1· dv1/2n

=

∫ ∞0

dw

w2n + 1= 1

2

∫ ∞−∞

dw

w2n + 1.

We apply Lemma 0.32 and Theorem 0.33 from the Prerequisites. The zeros of

g(z) := z2n + 1 in the upper half plane are zk := e(πi/2n)+(kπi/n) (k = 0, . . . , n − 1)

and these are all simple, and by Lemma 0.32, the function 1z2n+1 has at zk a simple

pole with residue

1

g′(zk)= 1

2nz−(2n−1)k = 1

2ne−(2n−1)((πi/2n)+(kπi/n)) = − 1

2ne(πi/2n)+(kπi/n).

So by Theorem 0.33, on applying the sum formula for geometric sequences,∫ ∞0

v1/2n

(v + 1)2dv = 1

2

∫ ∞−∞

dw

w2n + 1= −πi · e

πi/2n

2n

n−1∑k=0

ekπi/n

= −πi · eπi/2n

2n

eπi − 1

eπi/n − 1=

π

2n· 2i

eπi/2n − e−πi/2n=

π/2n

sin π/2n.

This proves Theorem 4.16.

Corollary 4.17. Γ(12) =√π.

Proof. Substitute z = 12

in Theorem 4.16, and use Γ(12) > 0.

Corollary 4.18. (i) Γ(z) 6= 0 for z ∈ C \ {0,−1,−2, . . .}.(ii) 1/Γ is analytic on C, and 1/Γ has simple zeros at z = 0,−1,−2, . . . and is

non-zero elesewhere.

Proof. (i) Recall that Γ(n) = (n − 1)! 6= 0 for n = 1, 2, . . .. Further, by Theorem

4.16 we have Γ(z)Γ(1− z) sinπz = π 6= 0 for z ∈ C \ Z.

(ii) By (i), the function 1/Γ is analytic and non-zero on C \ {0,−1,−2, . . .}.Further, at z = 0,−1,−2, . . ., Γ has a simple pole, hence 1/Γ is analytic and has a

simple zero.

120

We give another expression for the Gamma function.

Theorem 4.19. For z ∈ C \ {0,−1,−2, . . .} we have

Γ(z) = limn→∞

n! · nz

z(z + 1) · · · (z + n).

Proof. Define

Fn(z) :=n! · nz

z(z + 1) · · · (z + n).

We prove by induction that for every non-negative integer m we have Γ(z) =

limn→∞ Fn(z) for z ∈ C with Re z > −m and (if m > 0) z 6= 0,−1, . . . , 1 − m.

We first consider the case m = 0, i.e., z ∈ C and Re z > 0. We skip a few details,

which are left to the reader. Using repeated integration by parts, one shows that∫ 1

0

(1− u)nuz−1du =n!

z(z + 1) · · · (z + n),

so

Fn(z) = nz∫ 1

0

(1− u)nuz−1du =

∫ n

0

(1− t

n

)ntz−1dt =

∫ ∞0

gn(t)dt,

where gn(t) =(1 − t

n

)ntz−1 if 0 < t 6 n and gn(t) = 0 if t > n. Now we have

(1− x−1)x 6 e−1 for all x > 1, and limx→∞(1− x−1)x = e−1. This implies

limn→∞

gn(t) = e−ttz−1, |gn(t)| 6 e−ttRe z−1 for t > 0, n > 0,

while moreover,∫∞0e−ttRe z−1dt = Γ(Re z) <∞. So by the Dominated Convergence

Theorem,

limn→∞

Fn(z) = limn→∞

∫ ∞0

gn(t)dt =

∫ ∞0

e−ttz−1dt = Γ(z).

We now do the induction step. Let m > 0 be an integer and suppose that

Γ(z) = limn→∞ Fn(z) for z ∈ C with Re z > −m and (if m > 0) z 6= 0,−1, . . . , 1−m.

Let z ∈ C with Re z > −m− 1 and z 6= 0, . . . ,m. Then

limn→∞

Fn(z + 1)

zFn(z)= lim

n→∞

n! · nz+1

(z + 1) · · · (z + n+ 1)· (z + 1) · · · (z + n+ 1)

(n+ 1)!(n+ 1)z

= limn→∞

( n

n+ 1

)z+1

= 1.

121

By the induction hypothesis we know that Γ(z + 1) = limn→∞ Fn(z + 1), and so

limn→∞

Fn(z) = limn→∞

Fn(z + 1)

z=

Γ(z + 1)

z= Γ(z).

This completes the induction step and thus the proof of Theorem 4.19.

We deduce some consequences. Recall that the Euler-Mascheroni constant γ is

given by

γ := limN→∞

(N∑n=1

1

n

)− logN.

Corollary 4.20. We have

Γ(z) = e−γzz−1∞∏n=1

ez/n

1 + z/nfor z ∈ C \ {0,−1,−2, . . .}.

Proof. Let z ∈ C \ {0,−1,−2, . . .}. Then for N ∈ Z>0 we have

Γ(z) = limN→∞

N z ·N !

z(z + 1) · · · (z +N)= z−1 lim

N→∞

ez logN

(1 + z)(1 + z/2) · · · (1 + z/N)

= z−1 limN→∞

e(logN−1−12−···− 1

N)z

N∏n=1

ez/n

1 + z/n

= e−γzz−1∞∏n=1

ez/n

1 + z/n.

Corollary 4.21. For the logarithmic derivative of Γ we have

Γ′(z)

Γ(z)= −γ − 1

z+∞∑n=1

z

n(z + n)for z ∈ C \ {0,−1,−2, . . .}.

Proof. Apply Corollary 0.47 from the Prerequisites.

As another consequence, we derive an infinite product expansion for sinπz.

Corollary 4.22. We have

sin πz = πz ·∞∏n=1

(1− z2

n2

)for z ∈ C.

122

Proof. For z ∈ C we have by Theorem 4.16, Corollary 4.18 and Corollary 4.20,

sinπz =π

Γ(z)Γ(1− z)=

π

Γ(z)(−z)Γ(−z)

= π(−z)−1eγzz∞∏n=1

(e−z/n

(1 + z

n

))· (−z)e−γz

∞∏n=1

(e−z/n

(1− z

n

))= πz

∞∏n=1

(1− z

n

)(1 + z

n

)= πz ·

∞∏n=1

(1− z2

n2).

We finish with another important consequence of Theorem 4.19, the so-called

duplication formula.

Corollary 4.23. We have

Γ(2z) =22z−1√π· Γ(z)Γ(z + 1

2) for z ∈ C, z 6= 0,−1

2,−1,−3

2,−2, . . . .

Proof. Let A be the set of z indicated in the lemma. We show that the function

F (z) := 22zΓ(z)Γ(z + 12)/Γ(2z) is constant on A. Substituting z = 1

2gives that the

constant is 2√π, and then Corollary 4.23 follows.

Let z ∈ A. To get nice cancellations in the numerator and denominator, we use

the expressions

Γ(z) = limn→∞

n! · nz

z(z + 1) · · · (z + n)= lim

n→∞

2n+1 · n! · nz

2z(2z + 2) · · · (2z + 2n),

Γ(z + 12) = lim

n→∞

n! · nz+1/2

(z + 1/2)(z + 3/2) · · · (z + n+ 1/2)

= limn→∞

2n+1 · n! · nz+1/2

(2z + 1)(2z + 3) · · · (2z + 2n+ 1),

Γ(2z) = limn→∞

(2n+ 1)! · (2n+ 1)2z

2z(2z + 1) · · · (2z + 2n+ 1)

(i.e., in Theorem 4.19 we substitute 2z for z and take the limit over the odd integers).

123

Thus,

F (z) =22zΓ(z)Γ(z + 1

2)

Γ(2z)

= 22z limn→∞

{22n+2(n!)2n2z+1/2

2z(2z + 1) · · · (2z + 2n+ 1)· 2z(2z + 1) · · · (2z + 2n+ 1)

(2n+ 1)! · (2n+ 1)2z

}= lim

n→∞

{22n+2(n!)2

√n

(2n+ 1)!

}since

limn→∞

22z · n2z

(2n+ 1)2z= lim

n→∞e2z log(2n/(2n+1)) = 1.

This shows that indeed F (z) is constant.

Remark. By an argument similar to the proof of Corollary 4.23 (exercise), one can

derive the multiplication formula of Legendre-Gauss,

Γ(nz) = (2π)−(n−1)/2nnz−1/2Γ(z)Γ(z + 1

n

)· · ·Γ

(z + n−1

n

)for n ∈ Z>2.

4.5 Proof of the functional equation for the Rie-

mann zeta function

There are various methods to prove Theorem 4.9, see E.C. Titchmarsh, The theory

of the Riemann zeta function. We give Riemann’s proof based on a functional

equation for the Jacobi theta function θ(z) =∑∞

m=−∞ e−πm2z. We start with some

preparations.

4.5.1 The Jacobi theta function

We need a version of the Poisson summation formula for infinite sums, which will

be used in the deduction of the functional equation for the theta function.

Theorem 4.24. Let f be a complex function such that:

(i) there is δ > 0 such that f(z) is analytic on U(δ) := {z ∈ C : |Im z| < δ};

124

(ii) there are ε > 0, r > 0 such that

|f(z)| 6 |z|−1−ε for z ∈ U(δ), |z| > r.

Then∞∑

m=−∞

f(m) = limN→∞

N∑n=−N

∫ ∞−∞

f(t)e−2πintdt.

The idea is to apply Theorem 3.25 to the function F (z) :=∑∞

m=−∞ f(z + m).

We first prove some properties of this function.

Lemma 4.25. (i) The function F (z) is analytic on an open subset of C containing

[0, 1].

(ii) For every n ∈ Z we have∫ 1

0F (t)e−2πintdt =

∫∞−∞ f(t)e−2πintdt.

(iii) F (0) = F (1) =∑∞

m=−∞ f(m).

Proof. (i) Let U := {z ∈ C : −δ < Re z < 1 + δ, |Im z| < 12δ}. By taking δ

sufficiently small we may assume that |z| 6 2 for z ∈ U . We prove that there are

positive reals Am such that∑∞

m=−∞Am converges and |f(z +m)| 6 Am for z ∈ U ,

m ∈ Z. Since all summands f(z+m) are analytic, by Corollary 0.46 it then follows

that F (z) is analytic on U . Note that for z ∈ U , m ∈ Z with |m| > r + 3 we have

|z + m| > |m| − |z| > |m| − 2 > r and |f(z + m)| 6 (|m| − 2)−1−ε =: Am. To treat

the remaining m, let R := sup{|f(z)| : |Im z| 6 12δ, |z| 6 r + 5}; this quantity is

finite since |f(z)| is continuous on the region indicated. Then for z ∈ U , m ∈ Zwith |m| < r + 3 we have |f(z +m)| 6 R =: Am. Clearly,

∑∞m=−∞Am converges.

(ii) We apply the Fubini-Tonelli Theorem. We have∫ 1

0

( ∞∑m=−∞

|f(t+m)e−2πint|)dt 6

∫ 1

0

( ∞∑m=−∞

Am

))dt <∞,

so∫ 1

0

F (t)e−2πintdt =

∫ 1

0

( ∞∑m=−∞

f(t+m))e−2πintdt =

∞∑m=−∞

∫ 1

0

f(t+m)e−2πintdt

=∞∑

m=−∞

∫ 1

0

f(t+m)e−2πin(t+m)dt =∞∑

m=−∞

∫ m+1

m

f(t)e−2πintdt

=

∫ ∞−∞

f(t)e−2πintdt.

125

In the last step we have used that∫∞−∞ f(t)e−2πintdt converges, due to our assumption

|f(z)| 6 |z|−1−ε for z ∈ U(δ), |z| > r.

(iii) Obvious, since∑∞

m=−∞ f(m) converges.

Proof of Theorem 4.24. By combining Theorem 3.25 with Lemma 4.25 we obtain

∞∑m=−∞

f(m) = 12

(F (0) + F (1)

)= lim

N→∞

N∑n=−N

∫ 1

0

F (t)e−2πintdt

= limN→∞

N∑n=−N

∫ ∞−∞

f(t)e−2πintdt.

The Jacobi theta function is given by

θ(z) :=∞∑

m=−∞

e−πm2z (z ∈ C, Re z > 0).

Verify yourself that θ(z) converges and is analytic on {z ∈ C : Re z > 0}.

Theorem 4.26. θ(z−1) =√z · θ(z) for z ∈ C, Re z > 0, where

√z is chosen such

that | arg√z| < π/4.

Remark. Let A := {z ∈ C : Re z > 0}. We may choose the argument of z ∈ Asuch that | arg z| < π/2. Then indeed, we may choose

√z such that | arg

√z| < π/4.

Proof. Both θ(z−1) and√zθ(z) are analytic on A. In view of Corollary 0.37, it

suffices to prove the identity in Theorem 4.26 on any infinite subset of A of our

choice having a limit point in A. For this subset we take R>0. Thus, it suffices to

prove that∞∑

m=−∞

e−πm2/x =

√x ·

∞∑m=−∞

e−πm2x for x > 0.

We apply Theorem 4.24 to f(z) := e−πz2/x with x > 0 fixed. Verify that f

satisfies all conditions of that Theorem. Thus, for any x > 0,

∞∑m=−∞

e−πm2/x = lim

N→∞

N∑n=−N

∫ ∞−∞

e−(πt2/x)−2πintdt.

126

We compute the integrals by substituting u = t√x. Thus,∫ ∞

−∞e−(πt

2/x)−2πintdt =√x ·∫ ∞−∞

e−πu2−2πin

√x·udu

=√x ·∫ ∞−∞

e−π(u+in√x)2−πn2xdu

=√xe−πn

2x

∫ ∞−∞

e−π(u+in√x)2du.

In the lemma below we prove that the last integral is equal to 1. Then it follows

that∞∑

m=−∞

e−πm2/x = lim

N→∞

N∑n=−N

√xe−πn

2x =√x

∞∑n=−∞

e−πn2x,

since the last series converges. This proves our Theorem.

Lemma 4.27. Let z ∈ C. Then∫∞−∞ e

−π(u+z)2du = 1.

Proof. The following proof was suggested to me by Michiel Kosters. Let

F (z) :=

∫ ∞−∞

e−π(u+z)2

du.

We show that this defines an analytic function on C. We apply Theorem 0.42.

First, (u, z) 7→ e−π(u+z)2

is continuous, hence measurable, on R×D(0, R). Second,

for every fixed u ∈ R, z 7→ e−π(u+z)2

is analytic on C. Third, let K be a compact

subset of C, and choose R > 0 such that |z| 6 R for z ∈ K. Then for z ∈ K we

have

|e−π(u+z)2 | = e−Reπ(u+z)2 = e−(πu2+2πuRe z+πRe z2)

6 e−πu2+2πRu+πR2

= e−π(u−R)2+2πR2

,

and∫∞−∞ e

−π(u−R)2+2πR2du converges. So by Theorem 0.42, F is analytic on C.

Knowing that F is analytic on C, in order to prove that F (z) = 1 for z ∈ C it is

sufficient to prove, for any infinite set S ⊂ C with a limit point in C, that F (z) = 1

for z ∈ S. For the set S we take R. For z ∈ R we obtain, by substituting v = u+ z,

F (z) =

∫ ∞−∞

e−π(u+z)2

du =

∫ ∞−∞

e−πv2

dv = 2

∫ ∞0

e−πv2

dv.

127

Now a second substitution t = πv2 yields

F (z) = π−1/2∫ ∞0

e−tt−1/2dt = π−1/2Γ(12) = 1.

4.5.2 Proof of the functional equation for the zeta function

Define

ξ(s) := 12s(s− 1)π−s/2Γ(1

2s)ζ(s).

Theorem 4.9. The function ξ has an analytic continuation to C. For this contin-

uation, we have ξ(1− s) = ξ(s) for s ∈ C. and moreover, ξ(0) = ξ(1) = 12.

Proof (Riemann). Let for the moment s ∈ C, Re s > 1. Recall that

Γ(12s) =

∫ ∞0

e−tt(s/2)−1dt.

Substituting t = πn2u gives

Γ(12s) =

∫ ∞0

e−πn2u(πn2u)(s/2)−1d(πn2u) = πs/2ns

∫ ∞0

e−πn2uu(s/2)−1du.

Hence

π−s/2Γ(12s)n−s =

∫ ∞0

e−πn2uu(s/2)−1du,

and so, by summing over n,

π−s/2Γ(12s)ζ(s) =

∞∑n=1

∫ ∞0

e−πn2u · u(s/2)−1du.

We justify that the infinite integral and infinite sum can be interchanged. We use

the following special case of the Fubini-Tonelli theorem: if {fn : (0,∞) → C}∞n=1

is a sequence of measurable functions such that∑∞

n=1

∫∞0|fn(u)|du converges, then

all integrals∫∞0fn(u)du (n > 1) converge, the series

∑∞n=1 fn(u) converges almost

everywhere on (0,∞) and moreover,

∞∑n=1

∫ ∞0

fn(u)du,

∫ ∞0

(∞∑n=1

fn(u)

)du

128

converge and are equal. In our situation we have that indeed (putting σ := Re s)

∞∑n=1

∫ ∞0

|e−πn2u · u(s/2)−1|du =∞∑n=1

∫ ∞0

e−πn2uu(σ/2)−1du

=∞∑n=1

π−σ/2Γ(12σ)n−σ (reversing the above argument)

= π−σ/2Γ(12σ)ζ(σ)

converges. Thus, we conclude that for s ∈ C with Re s > 1,

(4.10) π−s/2Γ(12s)ζ(s) =

∫ ∞0

ω(u) · u(s/2)−1du, where ω(u) =∞∑n=1

e−πn2u.

Recall that θ(u) =∑∞

n=−∞ e−πn2u = 1 + 2ω(u).

We want to replace the right-hand side of (4.10) by something that converges

for every s ∈ C. Obviously, for s ∈ C with Re s < 0 there are problems if u ↓ 0.

To overcome these, we split the integral∫∞0

into∫∞1

+∫ 1

0and then transform

∫ 1

0

into an integral∫∞1

by means of a substitution v = u−1. After this substitution, the

integral contains a term ω(v−1). By Theorem 4.26, we have

ω(v−1) = 12(θ(v−1)− 1) = 1

2v1/2θ(v)− 1

2

= 12v1/2

(2ω(v) + 1)− 1

2= v1/2ω(v) + 1

2v1/2 − 1

2.

We work out in detail the approach sketched above. We keep for the moment our

assumption Re s > 1. Thus,

π−s/2Γ(12s)ζ(s) =

∫ ∞1

ω(u)u(s/2)−1du−∫ ∞1

ω(v−1)v1−s/2dv−1

=

∫ ∞1

ω(u)u(s/2)−1du+

∫ ∞1

(v1/2ω(v) + 1

2v1/2 − 1

2

)v1−s/2v−2dv

=

∫ ∞1

12

(v−(s+1)/2 − v−(s/2)−1

)dv +

∫ ∞1

ω(v)(v(s/2)−1 + v−(s+1)/2

)dv

where we have combined the terms without ω into one integral, and the terms

involving ω into another integral. Since we are still assuming Re s > 1, the first

integral is equal to

12

[− 2

s− 1v−(s−1)/2 +

2

sv−s/2

]∞1

=1

s− 1− 1

s=

1

s(s− 1).

129

Hence

π−s/2Γ(12s)ζ(s) =

1

s(s− 1)+

∫ ∞1

ω(v)(v(s/2)−1 + v−(s+1)/2

)dv.

For our function ξ(s) = 12s(s− 1)π−s/2Γ(1

2s)ζ(s) this gives

(4.4) ξ(s) = 12

+ 12s(s− 1)

∫ ∞1

ω(v)(v(s/2)−1 + v−(s+1)/2

)dv if Re s > 1.

Assume for the moment that F (s) :=∫∞1ω(v)

(v(s/2)−1 + v−(s+1)/2

)dv defines an

analytic function on C. Then we can use the right-hand side of (4.4) to define the

analytic continuation of ξ to C. By substituting 1 − s for s in the right-hand side,

we see that ξ(1− s) = ξ(s). Further, we observe that ξ(0) = ξ(1) = 12.

It remains to prove that F defines an analytic function on C. We apply as

usual Theorem 0.42. We check that f(v, s) := ω(v)(v(s/2)−1 + v−(s+1)/2

)satisfies the

conditions of that theorem.

a) f(v, s) is measurable on (1,∞) × C. For ω(v) =∑∞

n=1 e−πn2v is measurable,

being a pointwise convergent series of continuous, hence measurable function, and

also v(s/2)−1 + v−(s+1)/2 is measurable, since it is continuous.

b) s 7→ ω(v)(v(s/2)−1 + v−(s+1)/2

)is analytic on C for every fixed v. This is

obvious.

c) Let K be a compact subset of C. Then there is a measurable function MK(v)

on (1,∞) such that |f(v, s)| 6 MK(v) for s ∈ K and∫∞1MK(v)dv < ∞. Indeed,

choose A > 0 such that |Re s| 6 A for s ∈ K. we first have for v ∈ (1,∞)

0 6 ω(v) 6 e−πv(1 + e−3πv + e−8πv + · · ·

)6 e−πv ·

∞∑k=0

e−3kπv =e−πv

1− e−3πv6 2e−πv

and second, for v ∈ (1,∞), s ∈ K,

|v(s/2)−1 + v−(s+1)/2| 6 v(A/2)−1 + v(−(A+1)/2 6 2v(A/2)−1.

Hence

|f(v, s)| 6 4e−πvv(A/2)−1 =: MK(v).

130

Further, ∫ ∞1

MK(v)dv 6 4

∫ ∞0

e−vv(A/2)−1)dv 6 4 · Γ(12A) <∞.

So f(v, s) satisfies all conditions of Theorem 0.42, and it follows that the function

F (s) =∫∞1f(v, s)dv is indeed analytic on C.

4.6 Exercises

We describe a general method to compute series∑∞

n=1 f(n), where f is an even

meromorphic function on C, i.e., f(z) = f(−z) for z ∈ C minus the poles of f . This

will be used in the first exercise.

(N+l)f'r

*t)(*i)

Ilú'o

-1-"

a

Let N be an integer > 1 and let SN be the square through the four points

(N + 12)(−1− i), (N + 1

2)(1− i), (N + 1

2)(1 + i), (N + 1

2)(1− i),

traversed counterclockwise. Assume that f has only finitely many poles, and that

none are lying at the non-zero integers.

1) Compute

∮SN

2πif(z)

e2πiz − 1· dz, using the Residue Theorem from complex analysis.

2) Prove that limN→∞

∮SN

2πif(z)

e2πiz − 1· dz = 0. Here, you have to use the general inequal-

ity ∣∣∣∣∫γ

g(z)dz

∣∣∣∣ 6 L(γ) · supz∈γ|g(z)|,

131

where γ is a path in C, g : γ → C is a continuous function, and L(γ) denotes the

length of γ. Applying this estimate with γ = SN , one has to show that the upper

bound converges to 0 as N →∞.

The following lemma, of which we have included a proof here, is crucial in 2).

Lemma 4.28. There is a constant c > 0, independent of N , such that |e2πiz−1| > c

holds for all integers N > 1 and all z ∈ SN .

Proof. We consider the four edges of the square separately. First consider the edge

from (N + 12)(−1− i) to (N + 1

2)(1− i). This can be parametrized by (N + 1

2)(t− i)

with −1 6 t 6 1. So for the points z on this edge we have

|e2πiz − 1| = |e2πi(N+ 12)(t−i) − 1| = |e2πi(N+ 1

2)te2π(N+ 1

2) − 1|

> e2π(N+ 12) − 1 > e3π − 1.

Next, consider the edge from (N+ 12)(1−i) to (N+ 1

2)(1+i). This can be parametrized

by (N + 12)(1 + it) with −1 6 t 6 1. So for the points z on this edge we have

|e2πiz − 1| = |e2πi(N+ 12)(1+it) − 1| = | − e−2π(N+ 1

2)t − 1| > 1.

Here we have used that e2πi(N+ 12) = −1. The other two edges can be treated in the

same manner.

Exercise 4.1. a) Let k be an integer, and let f be a meromorphic function on

C that has no pole or zero at z = k. Prove that the function2πif(z)

e2πiz − 1has a

simple pole at z = k with residue f(k).

b) The Bernouilli numbers Bn are given byz

ez − 1=∞∑n=0

Bn

n!zn (z ∈ C, |z| < 2π).

Using the method sketched above, prove that

ζ(2k) = (−1)k−122k−1 B2k

(2k)!· π2k for k = 1, 2, . . ..

c) Show that 12z + z

ez−1 is an even function, and conclude that B1 = −12, B3 =

B5 = · · · = 0.

d) Prove that Bn = − 1n+1

n−1∑l=0

(n+1l

)Bl for n > 1.

132

Remark. About “odd zeta values” ζ(3), ζ(5), . . . not so much is known. Apery

proved in 1978 that ζ(3) is irrational. Zudilin proved in 2001 that at least one

among the numbers ζ(5), ζ(7), ζ(9), ζ(11) is irrational, and Rivoal proved in 2000

that infinitely many among ζ(5), ζ(7), ζ(9), . . . are irrational. At the moment there

is not a single odd integer k > 5 for which irrationality of ζ(k) could be proved. It

is of course believed that ζ(k) is irrational for every odd integer k > 3.

Exercise 4.2. Let q be an integer > 2, and a an integer with 1 6 a 6 q/2.

a) Show that∑∞

m=−∞(qm+ a)−2 converges and prove that

∞∑m=−∞

(qm+ a)−2 =(π/q)2

(sin πa/q)2.

Use the method sketched above with f(z) = (qz + a)−2 (this will work also for

this f , although it is not even).

b) Let χ be a non-principal character modulo q with χ(−1) = 1. Express L(2, χ)

as a finite sum.

Exercise 4.3. In this exercise you have to use the identity

ζ(s) =1

s− 1+ 1− s

∫ ∞1

(x− [x])x−1−sdx (Re s > 0).

a) Prove that ζ(s) < 0 for s ∈ R with 0 < s < 1.

b) Around s = 1, ζ(s) has a Laurent series expansion

1

s− 1+∞∑n=0

an(s− 1)n.

Prove that a0 = γ, where γ is the Euler-Mascheroni constant,

γ = limN→∞

( N∑n=1

1

n− logN

).

Hint. Explain that a0 = 1−∫∞1

(x− [x])dx/x2 and compute the integral.

133

Exercise 4.4. For α ∈ C with Reα > 0 one defines the Hurwitz zeta function

ζ(s;α) :=∞∑n=0

(n+ α)−s.

So ζ(s; 1) = ζ(s).

a) Prove that ζ(s;α) converges for all s ∈ C with Re s > 1.

b) Prove that ζ(s;α) has an analytic continuation to {s ∈ C : Re s > 0, s 6= 1},with a simple pole with residue 1 at s = 1.

c) Let q be an integer with q > 2. Then for a ∈ Z with 1 6 a < q and s ∈ C with

Re s > 0, s 6= 1 we have

ζ(s; a/q) =qs

ϕ(q)

∑χ∈G(q)

χ(a)L(s, χ).

Exercise 4.5. In this exercise you are asked to prove the following theorem: ζ(s) has

an analytic continuation to C\{1}, with a pole with residue 1 at s = 1. Equivalently,

if G(s) := (s − 1)(ζ(s) − 1), then G(s) has an analytic continuation to C and

G(1) = 1.

a) For s ∈ C with Re s > 1 and k ∈ Z>0, define Fk(s) :=∫∞1

(x − [x])kx−s−kdx.

Prove that F0(s) = 1s−1 and

Fk(s) =1

k + 1

(ζ(s+ k)− 1 + (s+ k)Fk+1(s)

).

Hint. First express Fn,k(s) :=∫ n+1

n(x−n)kx−s−kdx in terms of Fn,k+1(s) and

then sum over n.

b) Let r ∈ Z>0. Prove that for s ∈ C with Re s > 1 we have

G(s) = 1 −r∑

k=1

(s− 1)s(s+ 1) · · · (s+ k − 2)G(s+ k)

(k + 1)!

− (s− 1)s(s+ 1) · · · (s+ r)

(r + 1)!Fr+1(s).

134

c) Prove by induction on r that for every r ∈ Z>0, the function G(s) has an

analytic continuation to Ur := {s ∈ C : Re s > −r}, and then deduce that G(s)

has an analytic continuation to C. Show also that G(1) = 1 and G(0) = 32,

hence ζ(0) = −12.

Exercise 4.6. a) Using Theorem 4.9 and Lemma 4.8, prove that

ζ(1− s) = 21−sπ−s cos(12πs)Γ(s) · ζ(s) for s ∈ C \ {0, 1}.

b) Prove that ζ(1− 2k) = −B2k

2k for k = 1, 2, . . ..

c) Determine the sign of ζ(s) for s ∈ R.

d) Compute ζ ′(0). Use the above functional equation in combination with Exercise

4.3 b) and Corollary 4.21.

Exercise 4.7. Let q be an integer > 2 and χ a primitive character mod q. Using

Theorem 4.13 and Lemma 4.8, prove the following:

a) L(s, χ) has an analytic continuation to C.

b) if χ is even, then L(s, χ) has simple zeros at s = 0,−2,−4, . . . and L(s, χ) 6= 0

if Re s < 0, s 6∈ {−2,−4, . . .};if χ is odd, then L(s, χ) has simple zeros at s = −1,−3,−5, . . . and L(s, χ) 6= 0

if Re s < 0, s 6∈ {−1,−3,−5, . . .}.

c) Prove a) and b) in the case that χ is non-principal, but not necessarily primi-

tive.

135