Chapter 4 the Riemann Zeta Function and L-Functions

Home , Analytic continuation, Laurent series

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 ∞ X L(s, χ) = χ(n)n−s. n=1

P∞ −s We view ζ(s) = n=1 n as the L-function of the principal character modulo 1, (1) (1) more precisely, ζ(s) = L(s, χ0 ), where χ0 (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 X Z b Z b f(n) = f(x)dx + f(a) + (x − [x])f 0(x)dx. n=a a a

105 Remark. This result often occurs in the more symmetric form

b Z b Z b X 1 1 0 f(n) = f(x)dx + 2 (f(a) + f(b)) + (x − [x] − 2 )f (x)dx. n=a a a

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

Z n+1 Z n+1 x − [x]f 0(x)dx = (x − n)f 0(x)dx n n h in+1 Z n+1 Z n+1 = (x − n)f(x) − f(x)dx = f(n + 1) − f(x)dx. n n n By summing over n we get

b Z b X Z b (x − [x])f 0(x)dx = f(n) − f(x)dx, a n=a+1 a 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 → R such that Z |f(x, z)| 6 MK (x) for x ∈ D, z ∈ K, MK (x)dx < ∞. D Then the function F given by Z F (z) := f(x, z)dx D is analytic on U, and for every k > 1, Z F (k)(z) = f (k)(x, z)dx, D

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 X Z N Z N n−s = x−sdx + 1 + (x − [x])(−sx−1−s)dx n=1 1 1 1 − N 1−s Z N = + 1 − s (x − [x])x−1−sdx. s − 1 1 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, 1 Z ∞ (4.1) ζ(s) = + 1 − s (x − [x])x−1−sdx. s − 1 1

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

−1−s −1−σ |(x − [x])x | 6 x

R ∞ −1−σ and 1 x dx < ∞. 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} 1 and since it is of the shape 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. Q −s −1 (i) L(s, χ) = p(1 − χ(p)p ) for s ∈ C, Re s > 1. (q) (ii) If χ 6= χ0 , then L(s, χ) converges, and is analytic on {s ∈ C : Re s > 0}. (q) (iii) L(s, χ0 ) can be continued to an analytic function on {s ∈ C : Re s > 0, s 6= 1}, and for s in this set we have

(q) Y −s L(s, χ0 ) = ζ(s) · (1 − p ). p|q

(q) Hence L(s, χ0 ) 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 Pq P2q Theorem 3.8), we have m=1 χ(m) = 0, m=q+1 χ(m) = 0, etc. Hence

N X χ(n) = χ(tq + 1) + ··· + χ(tq + r) 6 r < q. n=1 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,

(q) Y −s −1 Y −s L(s, χ0 ) = (1 − p ) = ζ(s) (1 − p ). p-q p|q The right-hand side is deﬁned and analytic on {s ∈ C : Re s > 0, s 6= 1}, and so it (q) can be taken as an analytic continuation of L(s, χ0 ) 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 ∈ C with 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) (1) as L(s, χ0 ). 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 2 (q) coprime with q and so, χ is real if and only if χ = χ0 .

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

2 (4.2) 3 + 4 cos θ + cos 2θ = 2(1 + cos θ) > 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,

(q) 3 4 2 F (s) := L(s, χ0 ) · L(s + it, χ) · L(s + 2it, χ ).

By the conditions imposed on χ and t, the function L(s + 2it, χ2) is analytic around (q) s = 1. Further, L(s, χ0 ) 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 P∞ a (s − s )n around s = s with a 6= 0. This ord -function is additive on n=n0 n 0 0 n0 s=s0 products. Thus,

(q) 2 ords=1(F ) = 3 · ords=1 L(s, χ0 ) + 4 · ords=1 L(s + it, χ) + ords=1 L(s + 2it, χ ) > 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   3 4 Y 1 1 1 log |F (σ)| = log  · ·  (q) −σ 1 − χ(p)p−σ−it 1 − χ(p)2p−σ−2it p 1 − χ0 (p)p X 1 1 1 = 3 log + 4 log + log . 1 − p−σ 1 − χ(p)p−σ−it 1 − χ(p)2p−σ−2it p-q

Recall that

∞ n 1 X z 1 1 log = , log = Re log for z ∈ with |z| < 1. 1 − z n 1 − z 1 − z C n=1

Hence for r, ϕ ∈ R with 0 < r < 1,

∞ iϕ n ! 1 1 X (re ) log = Re log = Re 1 − reiϕ 1 − reiϕ n n=1 ∞ ∞ X rn X rn = Re (einϕ) = · cos nϕ. n n n=1 n=1

This leads to

∞ ∞ ∞ ! X X p−nσ X p−nσ X p−nσ log |F (σ)| = 3 + 4 · cos nϕ + cos 2nϕ n n p n p p-q n=1 n=1 n=1 ∞ X X p−nσ = (3 + 4 cos nϕ + cos 2nϕ ) 0, n p p > p-q n=1 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 P∞ −s for all n. Suppose that Lf (s) = n=1 f(n)n 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 analytic 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 expansion of g(s) around σ1, we get

∞ (k) X g (σ1) g(σ) = · (σ − σ )k. k! 1 k=0

Since σ1 is larger than the abscissa of convergence σ0 of Lf (s), we have

111 ∞ (k) (k) X k −σ1 g (σ1) = Lf (σ1) = f(n)(− log n) n for k > 0. n=1 Hence

∞ ∞ ! X 1 X g(σ) = f(n)(− log n)kn−σ1 (σ − σ )k k! 1 k=0 n=1 ∞ ∞ ! X 1 X = f(n)(log n)kn−σ1 (σ − σ)k. k! 1 k=0 n=1 Now all terms are non-negative, hence it is allowed to interchange the summations. Thus,

∞ ∞ ! X X 1 g(σ) = f(n)n−σ1 (log n)k(σ − σ)k k! 1 n=1 k=0 ∞ ∞ X X = f(n)n−σ1 e(log n)(σ1−σ) using ez = zk/k! n=1 k=0 ∞ ∞ X X = f(n)n−σ1 nσ1−σ = f(n)n−σ. n=1 n=1

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. P∞ −s For instance, if χ is a non-principal character mod q, then L(s, χ) = n=1 χ(n)n 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 1 > 2 and derive a contradiction. P∞ −s P∞ −s The series ζ(s) = n=1 n and n=1 χ(n)n converge absolutely if Re s > 1. So by Theorem 2.12, ∞ X −s F (s) = Lf (s) = f(n)n for s ∈ C, Re s > 1, n=1 where f = E ∗ χ, i.e., X f(n) = χ(d) for n ∈ Z>0. d|n 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   1 if p|q, k  X  k + 1 if p q, χ(p) = 1, f(pk) = χ(p)j = -  1 if p - q, χ(p) = −1, k even, j=0   0 if p - q, χ(p) = −1, k odd. k k Therefore, f(p ) > 0 for all prime powers p . 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 2 if n = m is a square, then f(n) > 1. Hence ∞ ∞ X −σ X −2σ 1 Lf (σ) = f(n)n > m = ∞ if σ 6 2 . n=1 m=1 1 So σ0(f) > 2 . 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 Z ∞ −t z−1 Γ(z) := e t dt for z ∈ C with Re z > 0, 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; 1 √ (iii) Γ(n) = (n − 1)! for n = 1, 2,... and Γ( 2 ) = 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,...}; π (vi) Γ(z)Γ(1 − z) = for z ∈ \ ; sin πz C Z 22z−1 (vii) Γ(2z) = √ · Γ(z)Γ(z + 1 ) for z ∈ , z 6= 0, − 1 , −1, − 3 , −2,... (duplication π 2 C 2 2 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

1 −s/2 1 −s/2 1 (4.3) ξ(s) := 2 s(s − 1)π Γ( 2 s)ζ(s) = (s − 1)π Γ( 2 s + 1)ζ(s),

1 1 1 where we have used the identity 2 sΓ( 2 s) = Γ( 2 s + 1). In his memoir from 1859, Riemann proved the famous identity Z ∞ 1 1 (s/2)−1 −(s+1)/2 (4.4) ξ(s) = 2 + 2 s(s − 1) ω(t) t + t dt if Re s > 1, 1

P∞ −πm2t where ω(t) := m=1 e . 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) = 1 2 .

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

ξ(s)πs/2 · Γ( 1 s + 1)−1 (4.5) ζ(s) = 2 . s − 1

By Theorem 4.9, ξ(s) has an analytic continuation to C. Further, πs/2 is analytic 1 −1 on C, and moreover, Γ( 2 s + 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}. 3 1 1 1 √ Using Γ( 2 ) = 2 Γ( 2 ) = 2 π, we get lim(s − 1)ζ(s) = ξ(1)π1/2Γ( 3 )−1 = 1, s→1 2 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 s/2 1 know also that π 6= 0 for s ∈ C, that Γ( 2 s+1) has a simple pole at s = −2, −4,... 1 and Γ( 2 s + 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 s/2 1 Theorem 4.9. Together with what we observed above about π and Γ( 2 s + 1), using (4.5), we obtain that for s = s0 with Re s0 6 0, 1 1 if s0 ∈ {−2, −4, −6,...}, ords=s0 ζ(s) = −ords=s0 Γ( 2 s + 1) = 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.

−s/2 1 Proof. We have ξ(s0) = 0, and so ξ(1 − s0) = 0 by Theorem 4.9. But π Γ( 2 s) 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. 1 1 Inserting ξ(0) = 2 and Γ(1) = 1 in (4.5) one easily shows that ζ(0) = − 2 . 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

q−1 X τ(1, χ) = χ(a)e2πia/q. a=0

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 √ q s/2 q ξ(s, χ) := Γ( 1 s)L(s, χ), c(χ) := if χ is even, π 2 τ(1, χ) √ q (s+1)/2 i q ξ(s, χ) := Γ 1 (s + 1)L(s, χ), c(χ) := if χ is odd. π 2 τ(1, χ)

116 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

Z ∞ −t z−1 Γ(z) := e t dt (z ∈ C, Re z > 0). 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, −t z−1 the function F (t, z) := e t 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,

δ−1 −t z−1 t for 0 6 t 6 1, |e t | 6 M(t) := −t R−1 −t/2 e t 6 Ce for t > 1, where C is some constant. Now we have

Z ∞ Z 1 Z ∞ M(t)dt = tδ−1dt + C e−t/2dt = δ−1 + 2C · e−1/2 < ∞. 0 0 1 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, Z ∞ Γ(z) = z−1 e−tdtz 0 h it=∞ Z ∞ = z−1 e−ttz + e−ttzdt = z−1Γ(z + 1), t=0 0 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 1 (4.7) Γ(z) = · Γ(z + n) for Re z > 0, n = 1, 2,.... z(z + 1) ··· (z + n − 1)

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 1 lim (z + n)Γ(z) = lim (z + n) Γ(z + n + 1) z→−n z→−n z(z + 1) ··· (z + n) 1 (−1)n = Γ(1) = . (−n)(−n + 1) ··· (−1) 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. π Theorem 4.16. We have Γ(z)Γ(1 − z) = for z ∈ \ . sin πz C Z Proof. We prove that zΓ(z)Γ(1 − z) = πz/ sin πz, or equivalently, πz (4.8) Γ(1 + z)Γ(1 − z) = for z ∈ A := ( \ ) ∪ {0}, sin πz C Z 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 1 take S := {2n : 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 π/2n (4.9) Γ1 + 1 · Γ1 − 1 = (n = 1, 2,...). 2n 2n sin π/2n

Notice that Z ∞ Z ∞ 1 1 −s 1/2n −t −1/2n Γ 1 + 2n · Γ 1 − 2n = e s ds · e t dt 0 0 Z ∞ Z ∞ = e−s−t(s/t)1/2ndsdt. 0 0 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 v u ∂s ∂s ∂(s, t) v + 1 (v + 1)2 ∂u ∂v = = 1 u ∂(u, v) ∂t ∂t − ∂u ∂v v + 1 (v + 1)2 −uv − u −u = = . (v + 1)3 (v + 1)2 It follows that Z ∞ Z ∞ 1 1 −u 1/2n ∂(s, t) Γ 1 + 2n · Γ 1 − 2n = e v · dudv 0 0 ∂(u, v) Z ∞ Z ∞ Z ∞ Z ∞ 1/2n −u 1/2n u −u v = e v 2 · dudv = e udu · 2 dv. 0 0 (v + 1) 0 0 (v + 1)

119 In the last product, the ﬁrst integral is equal to 1, while, as we will show below, the π/2n second integral is sin π/2n. This proves (4.9). Indeed, we have

Z ∞ 1/2n Z ∞ 1/2n ∞ Z ∞ v 1/2n 1 v 1 1/2n 2 dv = − v d = − + · dv 0 (v + 1) 0 v + 1 v + 1 0 0 v + 1 Z ∞ Z ∞ dw 1 dw = 2n = 2 2n . 0 w + 1 −∞ w + 1 We apply Lemma 0.32 and Theorem 0.33 from the Prerequisites. The zeros of 2n (πi/2n)+(kπi/n) g(z) := z + 1 in the upper half plane are zk := e (k = 0, . . . , n − 1) 1 and these are all simple, and by Lemma 0.32, the function z2n+1 has at zk a simple pole with residue 1 = 1 z−(2n−1) = 1 e−(2n−1)((πi/2n)+(kπi/n)) = − 1 e(πi/2n)+(kπi/n). 0 2n k 2n 2n g (zk) So by Theorem 0.33, on applying the sum formula for geometric sequences,

n−1 Z ∞ v1/2n Z ∞ dw πi · eπi/2n X dv = 1 = − ekπi/n (v + 1)2 2 w2n + 1 2n 0 −∞ k=0 πi · eπi/2n eπi − 1 π 2i π/2n = − = · = . 2n eπi/n − 1 2n eπi/2n − e−πi/2n sin π/2n This proves Theorem 4.16.

1 √ Corollary 4.17. Γ( 2 ) = π.

1 1 Proof. Substitute z = 2 in Theorem 4.16, and use Γ( 2 ) > 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) sin πz = 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 n! · nz Γ(z) = lim . n→∞ z(z + 1) ··· (z + n)

Proof. Deﬁne n! · nz F (z) := . n 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

Z 1 n! (1 − u)nuz−1du = , 0 z(z + 1) ··· (z + n) so Z 1 Z n Z ∞ z n z−1 t n z−1 Fn(z) = n (1 − u) u du = 1 − n t dt = gn(t)dt, 0 0 0 t n z−1 where gn(t) = 1 − n t if 0 < t 6 n and gn(t) = 0 if t > n. Now we have −1 x −1 −1 x −1 (1 − x ) 6 e for all x > 1, and limx→∞(1 − x ) = e . This implies

−t z−1 −t Re z−1 lim gn(t) = e t , |gn(t)| e t for t > 0, n > 0, n→∞ 6

R ∞ −t Re z−1 while moreover, 0 e t dt = Γ(Re z) < ∞. So by the Dominated Convergence Theorem, Z ∞ Z ∞ −t z−1 lim Fn(z) = lim gn(t)dt = e t dt = Γ(z). n→∞ n→∞ 0 0

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

z+1 Fn(z + 1) n! · n (z + 1) ··· (z + n + 1) lim = lim · z n→∞ zFn(z) n→∞ (z + 1) ··· (z + n + 1) (n + 1)!(n + 1) n z+1 = lim = 1. n→∞ n + 1

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

Fn(z + 1) Γ(z + 1) lim Fn(z) = lim = = Γ(z). n→∞ n→∞ z 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 N ! X 1 γ := lim − log N. N→∞ n n=1 Corollary 4.20. We have ∞ Y ez/n Γ(z) = e−γzz−1 for z ∈ \{0, −1, −2,...}. 1 + z/n C n=1

Proof. Let z ∈ C \{0, −1, −2,...}. Then for N ∈ Z>0 we have N z · N! ez log N Γ(z) = lim = z−1 lim N→∞ z(z + 1) ··· (z + N) N→∞ (1 + z)(1 + z/2) ··· (1 + z/N) N z/n −1 (log N−1− 1 −···− 1 )z Y e = z lim e 2 N N→∞ 1 + z/n n=1 ∞ Y ez/n = e−γzz−1 . 1 + z/n n=1

Corollary 4.21. For the logarithmic derivative of Γ we have ∞ Γ0(z) 1 X z = −γ − + for z ∈ \{0, −1, −2,...}. Γ(z) z n(z + n) C n=1

Proof. Apply Corollary 0.47 from the Prerequisites.

As another consequence, we derive an inﬁnite product expansion for sin πz. Corollary 4.22. We have ∞ Y z2 sin πz = πz · 1 − for z ∈ . n2 C n=1

122 Proof. For z ∈ C we have by Theorem 4.16, Corollary 4.18 and Corollary 4.20, π π sin πz = = Γ(z)Γ(1 − z) Γ(z)(−z)Γ(−z) ∞ ∞ −1 γz Y −z/n z −γz Y −z/n z = π(−z) e z e 1 + n · (−z)e e 1 − n n=1 n=1 ∞ ∞ Y z z Y z2 = πz 1 − n 1 + n = πz · 1 − n2 . n=1 n=1

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

Corollary 4.23. We have

22z−1 Γ(2z) = √ · Γ(z)Γ(z + 1 ) for z ∈ , z 6= 0, − 1 , −1, − 3 , −2,.... π 2 C 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 + 1 )/Γ(2z) is constant on A. Substituting z = 1 gives that the √ 2 2 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

n! · nz 2n+1 · n! · nz Γ(z) = lim = lim , n→∞ z(z + 1) ··· (z + n) n→∞ 2z(2z + 2) ··· (2z + 2n)

n! · nz+1/2 Γ(z + 1 ) = lim 2 n→∞ (z + 1/2)(z + 3/2) ··· (z + n + 1/2) 2n+1 · n! · nz+1/2 = lim , n→∞ (2z + 1)(2z + 3) ··· (2z + 2n + 1) (2n + 1)! · (2n + 1)2z Γ(2z) = lim n→∞ 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,

22zΓ(z)Γ(z + 1 ) F (z) = 2 Γ(2z) 22n+2(n!)2n2z+1/2 2z(2z + 1) ··· (2z + 2n + 1) = 22z lim · n→∞ 2z(2z + 1) ··· (2z + 2n + 1) (2n + 1)! · (2n + 1)2z √ 22n+2(n!)2 n = lim n→∞ (2n + 1)! since 22z · n2z lim = lim e2z log(2n/(2n+1)) = 1. n→∞ (2n + 1)2z n→∞ 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,

−(n−1)/2 nz−1/2 1 n−1 Γ(nz) = (2π) n Γ(z)Γ z + n ··· Γ z + 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 P∞ −πm2z equation for the Jacobi theta function θ(z) = m=−∞ e . 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

−1−ε |f(z)| 6 |z| for z ∈ U(δ), |z| > r. Then ∞ N X X Z ∞ f(m) = lim f(t)e−2πintdt. N→∞ m=−∞ n=−N −∞

P∞ 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]. R 1 −2πint R ∞ −2πint (ii) For every n ∈ Z we have 0 F (t)e dt = −∞ f(t)e dt. P∞ (iii) F (0) = F (1) = m=−∞ f(m).

1 Proof. (i) Let U := {z ∈ C : −δ < Re z < 1 + δ, |Im z| < 2 δ}. By taking δ suﬃciently small we may assume that |z| 6 2 for z ∈ U. We prove that there are P∞ 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 −1−ε |z + m| > |m| − |z| > |m| − 2 > r and |f(z + m)| 6 (|m| − 2) =: Am. To treat 1 the remaining m, let R := sup{|f(z)| : |Im z| 6 2 δ, |z| 6 r + 5}; this quantity is ﬁnite since |f(z)| is continuous on the region indicated. Then for z ∈ U, m ∈ Z P∞ with |m| < r + 3 we have |f(z + m)| 6 R =: Am. Clearly, m=−∞ Am converges. (ii) We apply the Fubini-Tonelli Theorem. We have Z 1 ∞ Z 1 ∞ X −2πint X |f(t + m)e | dt 6 Am )dt < ∞, 0 m=−∞ 0 m=−∞ so ∞ ∞ Z 1 Z 1 X X Z 1 F (t)e−2πintdt = f(t + m) e−2πintdt = f(t + m)e−2πintdt 0 0 m=−∞ m=−∞ 0

∞ ∞ X Z 1 X Z m+1 = f(t + m)e−2πin(t+m)dt = f(t)e−2πintdt m=−∞ 0 m=−∞ m Z ∞ = f(t)e−2πintdt. −∞

125 R ∞ −2πint In the last step we have used that −∞ f(t)e dt converges, due to our assumption −1−ε |f(z)| 6 |z| for z ∈ U(δ), |z| > r. P∞ (iii) Obvious, since m=−∞ f(m) converges.

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

∞ N X X Z 1 f(m) = 1 F (0) + F (1) = lim F (t)e−2πintdt 2 N→∞ m=−∞ n=−N 0

N X Z ∞ = lim f(t)e−2πintdt. N→∞ n=−N −∞

The Jacobi theta function is given by ∞ X −πm2z θ(z) := e (z ∈ C, Re z > 0). m=−∞ 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 ∈ , Re z > 0, where z is chosen such √ C that | arg z| < π/4.

Remark. Let A := {z ∈ : Re z > 0}. We may choose the argument of z ∈ A C √ √ such 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 ∞ ∞ X 2 √ X 2 e−πm /x = x · e−πm x for x > 0. m=−∞ m=−∞

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,

∞ N Z ∞ X 2 X 2 e−πm /x = lim e−(πt /x)−2πintdt. N→∞ m=−∞ n=−N −∞

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

Z ∞ √ Z ∞ √ e−(πt2/x)−2πintdt = x · e−πu2−2πin x·udu −∞ −∞ √ Z ∞ √ = x · e−π(u+in x)2−πn2xdu −∞ √ Z ∞ √ = xe−πn2x e−π(u+in x)2 du. −∞ In the lemma below we prove that the last integral is equal to 1. Then it follows that ∞ N ∞ X 2 X √ 2 √ X 2 e−πm /x = lim xe−πn x = x e−πn x, N→∞ m=−∞ n=−N n=−∞ since the last series converges. This proves our Theorem.

R ∞ −π(u+z)2 Lemma 4.27. Let z ∈ C. Then −∞ e du = 1.

Proof. The following proof was suggested to me by Michiel Kosters. Let Z ∞ F (z) := e−π(u+z)2 du. −∞

We show that this deﬁnes an analytic function on C. We apply Theorem 0.42. 2 First, (u, z) 7→ e−π(u+z) is continuous, hence measurable, on R × D(0,R). Second, 2 for every ﬁxed u ∈ R, z 7→ e−π(u+z) 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−Re π(u+z)2 = e−(πu2+2πuRe z+πRe z2) −πu2+2πRu+πR2 −π(u−R)2+2πR2 6 e = e ,

R ∞ −π(u−R)2+2πR2 and −∞ e du 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, Z ∞ Z ∞ Z ∞ F (z) = e−π(u+z)2 du = e−πv2 dv = 2 e−πv2 dv. −∞ −∞ 0

127 Now a second substitution t = πv2 yields Z ∞ −1/2 −t −1/2 −1/2 1 F (z) = π e t dt = π Γ( 2 ) = 1. 0

4.5.2 Proof of the functional equation for the zeta function

Deﬁne 1 −s/2 1 ξ(s) := 2 s(s − 1)π Γ( 2 s)ζ(s).

Theorem 4.9. The function ξ has an analytic continuation to C. For this contin- 1 uation, we have ξ(1 − s) = ξ(s) for s ∈ C. and moreover, ξ(0) = ξ(1) = 2 .

Proof (Riemann). Let for the moment s ∈ C, Re s > 1. Recall that Z ∞ 1 −t (s/2)−1 Γ( 2 s) = e t dt. 0 Substituting t = πn2u gives Z ∞ Z ∞ 1 −πn2u 2 (s/2)−1 2 s/2 s −πn2u (s/2)−1 Γ( 2 s) = e (πn u) d(πn u) = π n e u du. 0 0 Hence Z ∞ −s/2 1 −s −πn2u (s/2)−1 π Γ( 2 s)n = e u du, 0 and so, by summing over n,

∞ Z ∞ −s/2 1 X −πn2u (s/2)−1 π Γ( 2 s)ζ(s) = e · u du. n=1 0 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 P∞ R ∞ is a sequence of measurable functions such that n=1 0 |fn(u)|du converges, then R ∞ P∞ all integrals 0 fn(u)du (n > 1) converge, the series n=1 fn(u) converges almost everywhere on (0, ∞) and moreover,

∞ ∞ ! X Z ∞ Z ∞ X fn(u)du, fn(u) du n=1 0 0 n=1

128 converge and are equal. In our situation we have that indeed (putting σ := Re s) ∞ Z ∞ ∞ Z ∞ X 2 X 2 |e−πn u · u(s/2)−1|du = e−πn uu(σ/2)−1du n=1 0 n=1 0 ∞ X −σ/2 1 −σ = π Γ( 2 σ)n (reversing the above argument) n=1

−σ/2 1 = π Γ( 2 σ)ζ(σ) converges. Thus, we conclude that for s ∈ C with Re s > 1, Z ∞ ∞ −s/2 1 (s/2)−1 X −πn2u (4.10) π Γ( 2 s)ζ(s) = ω(u) · u du, where ω(u) = e . 0 n=1 P∞ −πn2u Recall that θ(u) = n=−∞ e = 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. R ∞ R ∞ R 1 R 1 To overcome these, we split the integral 0 into 1 + 0 and then transform 0 R ∞ −1 into an integral 1 by means of a substitution v = u . After this substitution, the integral contains a term ω(v−1). By Theorem 4.26, we have

−1 1 −1 1 1/2 1 ω(v ) = 2 (θ(v ) − 1) = 2 v θ(v) − 2 1 1/2 1 1/2 1 1/2 1 = 2 v 2ω(v) + 1) − 2 = v ω(v) + 2 v − 2 . We work out in detail the approach sketched above. We keep for the moment our assumption Re s > 1. Thus, Z ∞ Z ∞ −s/2 1 (s/2)−1 −1 1−s/2 −1 π Γ( 2 s)ζ(s) = ω(u)u du − ω(v )v dv 1 1 Z ∞ Z ∞ (s/2)−1 1/2 1 1/2 1 1−s/2 −2 = ω(u)u du + v ω(v) + 2 v − 2 v v dv 1 1 Z ∞ Z ∞ 1 −(s+1)/2 −(s/2)−1 (s/2)−1 −(s+1)/2 = 2 v − v dv + ω(v) v + v dv 1 1 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 ∞ 1 2 −(s−1)/2 2 −s/2 1 1 1 2 − v + v = − = . s − 1 s 1 s − 1 s s(s − 1)

129 Hence Z ∞ −s/2 1 1 (s/2)−1 −(s+1)/2 π Γ( 2 s)ζ(s) = + ω(v) v + v dv. s(s − 1) 1

1 −s/2 1 For our function ξ(s) = 2 s(s − 1)π Γ( 2 s)ζ(s) this gives Z ∞ 1 1 (s/2)−1 −(s+1)/2 (4.4) ξ(s) = 2 + 2 s(s − 1) ω(v) v + v dv if Re s > 1. 1

R ∞ (s/2)−1 −(s+1)/2 Assume for the moment that F (s) := 1 ω(v) v + v 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, 1 we see that ξ(1 − s) = ξ(s). Further, we observe that ξ(0) = ξ(1) = 2 . 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. P∞ −πn2v a) f(v, s) is measurable on (1, ∞) × C. For ω(v) = n=1 e 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.

(s/2)−1 −(s+1)/2 b) s 7→ ω(v) v + v 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) R ∞ on (1, ∞) such that |f(v, s)| 6 MK (v) for s ∈ K and 1 MK (v)dv < ∞. Indeed, choose A > 0 such that |Re s| 6 A for s ∈ K. we ﬁrst have for v ∈ (1, ∞)

−πv −3πv −8πv 0 6 ω(v) 6 e 1 + e + e + ··· ∞ X e−πv e−πv · e−3kπv = 2e−πv 6 1 − e−3πv 6 k=0 and second, for v ∈ (1, ∞), s ∈ K,

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

Hence −πv (A/2)−1 |f(v, s)| 6 4e v =: MK (v).

130 Further, Z ∞ Z ∞ −v (A/2)−1) 1 MK (v)dv 6 4 e v dv 6 4 · Γ( 2 A) < ∞. 1 0 So f(v, s) satisﬁes all conditions of Theorem 0.42, and it follows that the function R ∞ F (s) = 1 f(v, s)dv is indeed analytic on C.

4.6 Exercises

P∞ 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.

*t)(*i)

(N+l)f'r a Ilú'o

Let N be an integer > 1 and let SN be the square through the four points -1-" 1 1 1 1 (N + 2 )(−1 − i), (N + 2 )(1 − i), (N + 2 )(1 + i), (N + 2 )(1 − i), traversed counterclockwise. Assume that f has only ﬁnitely many poles, and that none are lying at the non-zero integers. I 2πif(z) 1) Compute 2πiz · dz, using the Residue Theorem from complex analysis. SN e − 1 I 2πif(z) 2) Prove that lim · dz = 0. Here, you have to use the general inequal- N→∞ 2πiz SN e − 1 ity Z

g(z)dz 6 L(γ) · sup |g(z)|, γ z∈γ

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).

2πiz Lemma 4.28. There is a constant c > 0, independent of N, such that |e −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 1 1 1 from (N + 2 )(−1 − i) to (N + 2 )(1 − i). This can be parametrized by (N + 2 )(t − i) with −1 6 t 6 1. So for the points z on this edge we have

2πiz 2πi(N+ 1 )(t−i) 2πi(N+ 1 )t 2π(N+ 1 ) |e − 1| = |e 2 − 1| = |e 2 e 2 − 1| 2π(N+ 1 ) 3π > e 2 − 1 > e − 1.

1 1 Next, consider the edge from (N+ 2 )(1−i) to (N+ 2 )(1+i). This can be parametrized 1 by (N + 2 )(1 + it) with −1 6 t 6 1. So for the points z on this edge we have

2πiz 2πi(N+ 1 )(1+it) −2π(N+ 1 )t |e − 1| = |e 2 − 1| = | − e 2 − 1| > 1.

2πi(N+ 1 ) Here we have used that e 2 = −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 2πif(z) that has no pole or zero at z = k. Prove that the function has a C e2πiz − 1 simple pole at z = k with residue f(k).

∞ z X Bn b) The Bernouilli numbers B are given by = zn (z ∈ , |z| < 2π). n ez − 1 n! C n=0 Using the method sketched above, prove that B ζ(2k) = (−1)k−122k−1 2k · π2k for k = 1, 2,.... (2k)!

1 z 1 c) Show that 2 z + ez−1 is an even function, and conclude that B1 = − 2 , B3 = B5 = ··· = 0.

n−1 1 X n+1 d) Prove that Bn = −n+1 l Bl for n > 1. l=0

132 Remark. About “odd zeta values” ζ(3), ζ(5),... not so much is known. Ap´ery 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.

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

∞ X (π/q)2 (qm + a)−2 = . (sin πa/q)2 m=−∞

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

1 Z ∞ ζ(s) = + 1 − s (x − [x])x−1−sdx (Re s > 0). s − 1 1

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 X + a (s − 1)n. s − 1 n n=0

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

N X 1 γ = lim − log N . N→∞ n n=1

R ∞ 2 Hint. Explain that a0 = 1 − 1 (x − [x])dx/x and compute the integral.

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

∞ X ζ(s; α) := (n + α)−s. n=0

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

qs X ζ(s; a/q) = χ(a)L(s, χ). ϕ(q) χ∈G(q)

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.

R ∞ k −s−k a) For s ∈ C with Re s > 1 and k ∈ Z>0, deﬁne Fk(s) := 1 (x − [x]) x dx. 1 Prove that F0(s) = s−1 and 1 F (s) = ζ(s + k) − 1 + (s + k)F (s) . k k + 1 k+1

R n+1 k −s−k Hint. First express Fn,k(s) := n (x − n) x dx 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

r X (s − 1)s(s + 1) ··· (s + k − 2)G(s + k) G(s) = 1 − (k + 1)! k=1 (s − 1)s(s + 1) ··· (s + r) − F (s). (r + 1)! r+1

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) 3 has an analytic continuation to C. Show also that G(1) = 1 and G(0) = 2 , 1 hence ζ(0) = − 2 . Exercise 4.6. a) Using Theorem 4.9 and Lemma 4.8, prove that

1−s −s 1 ζ(1 − s) = 2 π cos( 2 πs)Γ(s) · ζ(s) for s ∈ C \{0, 1}.

B2k b) Prove that ζ(1 − 2k) = − 2k for k = 1, 2,....

c) Determine the sign of ζ(s) for s ∈ R. d) Compute ζ0(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 primitive.

135