RIEMANN's FIRST PROOF of the ANALYTIC CONTINUATION of Ζ(S) and L(S, Χ) 1. the Hurwitz Zeta Function We Have Already Seen

RIEMANN’S FIRST PROOF OF THE ANALYTIC CONTINUATION OF ζ(s) AND L(s, χ)

FELIX RUBIN SEMINAR ON MODULAR FORMS, WINTER TERM 2006

Abstract. In this chapter, we will see a proof of the analytic continuation of the Riemann zeta function ζ(s) and the Dirichlet L function L(s, χ) via the Hurwitz zeta function. This then gives rise to a functional equation for ζ(s) and a direct computation for the value of this function at negative integer points.

1. The Hurwitz zeta function We have already seen the definition of the Riemann zeta function ζ(s) and the Dirichlet L function L(s, χ) as series: ∞ ∞ X 1 X χ(n) ζ(s) = and L(s, χ) = ns ns n=1 n=1 where s = σ + it ∈ C, σ > 1, t ∈ R and χ is a Dirichlet character modulo k, k ∈ N. These notations will be used throughout this chapter. We can unify the treatment of the above two functions by introducing the Hur- witz zeta function: P∞ 1 Definition 1.1. For σ > 1, we define ζ(s, a) = n=0 (n+a)s , 0 < a ≤ 1, the Hurwitz zeta function. Remark 1.2. (1) For a = 1, ζ(s, 1) = ζ(s). −s Pk r (2) L(s, χ) = k r=1 χ(r)ζ(s, k ), if χ is a Dirichlet character modulo k. The second statement in the remark follows from the following calculation: Write n = qk + r, 1 ≤ r ≤ k, q = 0, 1, 2, .... We then have: ∞ k ∞ k ∞ X χ(n) X X χ(qk + r) 1 X X 1 L(s, χ) = = = χ(r) ns (qk + r)s ks (q + r )s n=1 r=1 q=0 r=1 q=0 k k X r = k−s χ(r)ζ(s, ). k r=1 Theorem 1.3. The series ζ(s, a) converges absolutely for σ > 1 with uniform convergence in every half-plane σ > 1 + δ, δ > 0. Hence, ζ(s, a) is an analytic function of s in the half-plane σ > 1. Proof. We have ∞ ∞ ∞ X X X |(n + a)−s| = (n + a)−σ ≤ (n + a)−(1+δ) n=0 n=0 n=0 where the first equality follows from |xs| = |es ln x| = |eσ ln xeit ln x| = eσ ln x = xσ, for x ∈ C, and the theorem follows.

Date: December 14, 2006. 1 2 FELIX RUBIN SEMINAR ON MODULAR FORMS, WINTER TERM 2006

We will be able to analytically continue the Hurwitz zeta function beyond the line σ = 1 by an integral representation of ζ(s, a) obtained from the integral formula for the Gamma function. We need the following properties of the Gamma function: R ∞ s−1 −x For σ > 0, we can deﬁne the Gamma function by Γ(s) = 0 x e dx. This function can be analytically continued to the whole complex plane with simple poles (−1)n at 0, −1, −2, −3, ... with residues n! at s = −n, n ∈ N. Moreover, the gamma function satisﬁes the functional equations Γ(s + 1) = sΓ(s) and Γ(s)Γ(1 − s) = π sin(πs) . Finally, we note that Γ(n + 1) = n!, if n ∈ N. For more detail on the Gamma function, see for example [2]. The next theorem gives an integral representation for the Hurwitz zeta function. Theorem 1.4. For σ > 1 and 0 < a ≤ 1, we have Z ∞ xs−1e−ax (1.1) Γ(s)ζ(s, a) = −x dx 0 1 − e Proof. First, keep s real, s > 1 and later extend it to complex s by analytic continuation. By a change of variable x = (n + a)t, n ≥ 0, we have for the Gamma function: Z ∞ Z ∞ Γ(s) = e−xxs−1dx = (n + a)s e−(n+a)tts−1dt 0 0 Z ∞ ⇒ (n + a)−sΓ(s) = e−nte−atts−1dt 0 P∞ R ∞ −nt −at s−1 Summing over all n, this gives: ζ(s, a)Γ(s) = n=0 0 e e t dt, where the left hand side converges for σ > 1. By applying Beppo-Levi’s theorem (1.5 below), we can interchange the sum and the integral to obtain: ζ(s, a)Γ(s) = R ∞ P∞ −nt −at s−1 0 n=0 e e t dt. −t P∞ −nt 1 Since t > 0, 0 < e < 1 and thus n=0 e = 1−e−t . This gives: P∞ −nt −at s−1 e−atts−1 n=0 e e t = 1−e−t , for almost all t ∈ [0, ∞). Hence:

Z ∞ e−atts−1 ζ(s, a)Γ(s) = −t dt, where s > 1, s ∈ R. 0 1 − e In order to extend this formula to the half plane σ > 1, note that the left hand side of the equation is analytic for σ > 1. To show that the right hand side is analytic, we assume that for some δ > 0, c > 1, we have 1 + δ ≤ σ ≤ c. Because tσ−1 ≤ tδ for t ∈ [0, 1], tσ−1 ≤ tc−1, for t ≥ 1 and et − 1 ≥ t, for t ≥ 0, this gives:

Z ∞ e−atts−1 Z ∞ e−attσ−1 Z 1 Z ∞ e−attσ−1 | −t |dt ≤ −t dt = ( + ) −t dt 0 1 − e 0 1 − e 0 1 1 − e Z 1 −at σ−1 Z 1 (1−a)t δ Z 1 1−a e t e t 1−a δ−1 e −t dt ≤ t dt ≤ e t dt = , and 0 1 − e 0 e − 1 0 δ Z ∞ e−attσ−1 Z ∞ e−attc−1 Z ∞ e−attc−1 −t dt ≤ −t dt ≤ −t dt = Γ(c)ζ(c, a). 1 1 − e 1 1 − e 0 1 − e Hence, the integral in (1.1) converges uniformly in every strip 1 + δ ≤ σ ≤ c and therefore represents an analytic function in the half plane σ > 1. From this, the theorem follows. Beppo-Levi’s theorem can be found in any course on measure and integration theory or probability theory and says the following: RIEMANN’S FIRST PROOF OF THE ANALYTIC CONTINUATION OF ζ(s) AND L(s, χ) 3

Theorem 1.5. (Beppo-Levi) Let {fn}n∈N be an increasing sequence of non- negative, measurable functions from any space Ω into R (i.e. 0 ≤ f1 ≤ f2 ≤ f3 ≤ ..., almost everywhere). Then, Z Z lim fn(ω)dµ(ω) = lim fn(ω)dµ(ω), n→∞ Ω Ω n→∞ where µ denotes the measure on the space Ω. We need another representation of ζ(s, a), in order to give its analytic continuation beyond σ = 1. Consider the contour C, which is a loop around the negative real axis as shown in ﬁgure 1. We decompose C into the parts C1, C2 and C3. C2 is a circle around 0 with radius c < 2π (pos. oriented). C1 and C3 are the lower and upper edges of a ”cut” in the complex plane along the negative real axis. We −πi πi use the parameterizations z = re on C1, z = re on C3 with r ∈ [c, ∞) and iθ z = ce with θ ∈ [−π, π] on C2.

Figure 1. The contour C around the negative real axis

The next result allows us to represent the Hurwitz zeta function as an integral along C combined with the Gamma function, which will lead us to the analytic continuation in the next section. Theorem 1.6. If 0 < a ≤ 1, σ > 1, the following holds (1.2) ζ(s, a) = Γ(1 − s)I(s, a), 1 R zs−1eaz where I(s, a) = 2πi C 1−ez dz is an entire function of s. Note here, that for the deﬁnition of zs−1 in the integrand, we write e(s−1) ln(z) and use the principal branch of the logarithm which is deﬁned on C\{x ∈ R; x ≤ 0}. This branch does make a jump of 2πi when crossing the negative real axis. Therefore if we continue the logarithm analytically from above the negative real axis and from below the negative real axis to this axis and take the integral along the contour C, this integral will not cancel out in general and moreover, the integral does not depend on the radius c, as long as c ≤ 2π.

Proof. The integral along C2 is an entire function in s since the integrand is an z entire function (1 − e 6= 0 on C2).We thus only need to prove that the integrals along C1 and C3 converge uniformly on any disk |s| ≤ M, M > 0. Along C1, we have for r ≥ 1: |zs−1| = rσ−1|e−πi(σ−1+it)| = rσ−1eπt ≤ rM−1eπM since |s| ≤ M.

Similarly on C3: |zs−1| = rσ−1|eπi(σ−1+it)| = rσ−1e−πt ≤ rM−1eπM since |s| ≤ M.

We thus have for r ≥ 1 on C1 and C3: zs−1eaz rM−1eπM e−ar rM−1eπM e(1−a)r r er | 1−ez | ≤ 1−e−r = er −1 . But e −1 > 2 , when r > ln 2, hence the integrand is bounded by ArM−1e−ar, where the constant A only depends on R ∞ M−1 −ar M. But c r e dr converges for c > 0, and hence the integral does converge uniformly on every disk |s| ≤ M. Therefore, I(s, a) is an entire function of s. 4 FELIX RUBIN SEMINAR ON MODULAR FORMS, WINTER TERM 2006

To prove (1.2), we write 2πiI(s, a) = (R + R + R )zs−1g(z)dz, where g(z) = C1 C2 C3 eaz iθ 1−ez . On C1 and C3, g(z) = g(−r) and on C2, we write z = ce , with −π ≤ θ ≤ π.

Z c Z π ⇒ 2πiI(s, a) = rs−1e−πisg(−r)dr + i cs−1e(s−1)iθceiθg(ceiθ)dθ ∞ −π Z ∞ + rs−1eπisg(−r)dr c Z ∞ Z π = 2i sin(πs) rs−1g(−r)dr + ics eisθg(ceiθ)dθ. c −π R ∞ s−1 Now, let c → 0: We have limc→0 c r g(−r)dr = Γ(s)ζ(s, a) for σ > 1, according to equation (1.1). It remains to show that the second integral from −π to π goes to zero when c → 0 in order to prove (1.2). Note that g(z) is analytic in |z| < 2π except for a simple pole at z = 0. Hence, zg(z) is analytic everywhere in |z| < 2π and therefore bounded by some constant A. This gives for σ > 1: Z π Z π A |cs eisθg(ceiθ)dθ| ≤ cσ e−tθ dθ ≤ 2πAeπ|t|cσ−1 → 0 when c → 0 −π −π c ⇒ πI(s, a) = sin(πs)Γ(s)ζ(s, a) π and the result follows from Γ(s)Γ(1 − s) = sin(πs) .

2. The analytic continuation of the Hurwitz zeta function We can use equality (1.2) to extend the deﬁnition of the Hurwitz zeta function beyond the line σ = 1, since the right hand side of this equation is meaningful for any s ∈ C. We therefore deﬁne

ζ(s, a) := Γ(1 − s)I(s, a), ∀s ∈ C, 0 < a ≤ 1. Theorem 2.1. The function ζ(s, a) so deﬁned is analytic on C \{1} with a simple pole at s = 1 with residue 1. Proof. We know that I(s, a) is entire, hence any possible singularities come from Γ(1 − s). Γ(1 − s) has poles at s = 1, 2, 3, .... However, we have seen that ζ(s, a) is analytic at s = 2, 3, ..., so s = 1 is the only possible pole of ζ(s, a). For s = n ∈ N,

1 Z zn−1eaz 1 Z zn−1eaz zn−1eaz I(n, a) = z dz = z dz = Resz=0 z 2πi C 1 − e 2πi C2 1 − e 1 − e since the integrals along C1 and C3 cancel out. Hence, for n = 1,

eaz zeaz z I(1, a) = Resz=0 = lim = lim = −1 1 − ez z→0 1 − ez z→0 1 − ez Now, we compute lim(s − 1)ζ(s, a) = − lim(1 − s)Γ(1 − s)I(s, a) s→1 s→1 = −I(1, a) lim Γ(2 − s) = Γ(1) = 1. s→1 Therefore, ζ(s, a) has a simple pole at s = 1 with residue 1. By the equations given in remark 1.2, we can now give analytic continuations of the Riemann zeta and the Dirichlet L functions by using these equations as deﬁnitions of the respective functions. RIEMANN’S FIRST PROOF OF THE ANALYTIC CONTINUATION OF ζ(s) AND L(s, χ) 5

Theorem 2.2. (1) ζ(s) is analytic on C with a simple pole at s = 1 with residue 1.

(2) For the principal character χ1 modulo k, L(s, χ1) is analytic on C with a φ(k) simple pole at s = 1 with residue k .

(3) If χ 6= χ1, L(s, χ) is an entire function of s.

φ(k) denotes Euler’s totient which gives the number of elements n ≤ k, n ∈ N, such that (k, n) = 1.

Proof. The ﬁrst statement is trivial. To prove the other two, we note that

k X 0 if χ 6= χ χ(r) = 1 φ(k) if χ = χ1 r=1 This formula is a consequence of the deﬁnition of Dirichlet characters of reduced r residue classes (see for example [3]). Now, ζ(s, k ) has a simple pole at s = 1 with r residue 1. Hence χ(r)ζ(s, k ) has a simple pole at s = 1 with residue χ(r).

k −s X r ⇒ Ress=1(L(s, χ)) = lim(s − 1)L(s, χ) = lim(s − 1)k χ(r)ζ(s, ) s→1 s→1 k r=1 k 1 X 0 if χ 6= χ1 = χ(r) = k φ(k) if χ = χ r=1 k 1

3. Evaluation of ζ(−n, a) for n ∈ N We can now explicitly calculate ζ(−n, a) for n ∈ N: ζ(−n, a) = Γ(1+n)I(−n, a) = n!I(−n, a). In the proof of theorem 2.1 we have seen that for n ∈ N, I(−n, a) = z−n−1eaz Resz=0( 1−ez ). The calculation of this residue leads to the Beroulli polynomials.

Deﬁnition 3.1. ∀x ∈ C, we deﬁne the Bernoulli polynomials Bn(x) by the xz ze P∞ Bn(x) n equation ez −1 = n=0 n! z , where |z| < 2π. This leads to the explicit result

Bn+1(a) Theorem 3.2. ∀n ∈ N and 0 < a ≤ 1, we have ζ(−n, a) = − n+1 . Proof. We already know that ζ(−n, a) = n!I(−n, a). But z−n−1eaz zeaz I(−n, a) = Res = −Res z−n−2 z=0 1 − ez z=0 ez − 1 ∞ ! X Bm(a) Bn+1(a) = −Res z−n−2 zm = − z=0 m! (n + 1)! m=0 Hence, the theorem follows.

4. A functional equation for the Riemann zeta function Here, we give e functional equation for ζ(s), which leads us to the so called trivial zeros of the Riemann zeta function. 6 FELIX RUBIN SEMINAR ON MODULAR FORMS, WINTER TERM 2006

Deﬁnition 4.1. For x ∈ R and σ > 1, deﬁne the Dirichlet series ∞ X e2πinx F (x, s) = ns n=1 F (x, s) is called the periodic zeta function. Note that F (x, s) is periodic in x with period 1 and F (1, s) = ζ(s). Moreover, the series converges absolutely for σ > 1. Proposition 4.2. If 0 < a ≤ 1 and σ > 1, the Hurwitz formula holds:

Γ(s) − πis πis ζ(1 − s, a) = e 2 F (a, s) + e 2 F (−a, s) (2π)s In the proof of this proposition, we need the following lemma: Lemma 4.3. For 0 < r < π, denote with S(r) the region ! [ S(r) = C\ B(2nπi, r) , n∈Z where B(c, r) denotes the open disk with center c and radius r. Then, if 0 < a ≤ 1, eaz the function g(z) = 1−ez is bounded in S(r). (The bound depends on r.) For a proof of this lemma see [1]. We now turn to the proof of Hurwitz’s formula.

1 R zs−1eaz Proof. Consider the function IN (s, a) = 2πi C(N) 1−ez dz, where C(N) is the contour shown in ﬁgure 2, and N ∈ N. Note that we use again the principal branch of the logarithm to deﬁne zs−1 in the integrand, as we did in the proof of formula (1.2).

Figure 2

We now show that if σ < 0, the integral along the outer circle tends to 0 as N → ∞. Hence, IN (s, a) → I(s, a), for N → ∞ and σ < 0. Parameterizing z = Reiθ, −π ≤ θ ≤ π, we have

|zs−1| = |Rs−1eiθ(s−1)| = Rσ−1e−tθ ≤ Rσ−1eπ|t|. Since the outer circle lies in the set S(r) of the preceding lemma, the absolute value of the integrand is bounded by Aeπ|t|Rσ−1, A being the bound for |g(z)| implied by the lemma. Hence, the absolute value of the integral along the outer circle is bounded by 2πAeπ|t|Rσ, which tends to 0 as R → ∞, for σ < 0. We thus have:

lim IN (1 − s, a) = I(1 − s, a) if σ > 1. N→∞ RIEMANN’S FIRST PROOF OF THE ANALYTIC CONTINUATION OF ζ(s) AND L(s, χ) 7

On the other hand, we have by Cauchy’s residue theorem: N X z−seaz I (1 − s, a) = − Res N z=2nπi 1 − ez n=−N,n6=0 N X z−seaz z−seaz = − Res + Res z=2nπi 1 − ez z=−2nπi 1 − ez n=1 Now

z−seaz z−seaz e2nπia Resz=2nπi = lim (z − 2nπi) = − , 1 − ez z→2nπi 1 − ez (2nπi)s N 2nπia N −2nπia −πis P e P e −s 2 hence IN (1 − s, a) = n=1 (2nπi)s + n=1 (−2nπi)s . But i = e and −s πis (−i) = e 2 so

−πis N 2nπia πis N −2nπia e 2 X e e 2 X e I (1 − s, a) = + . N (2π)s ns (2π)s ns n=1 n=1 Letting N → ∞ and multiplying both sides with Γ(s), we obtain the desired result. We can now state the functional equation for the Riemann zeta function.

Theorem 4.4. For all s ∈ C, we have πs ζ(1 − s) = 2(2π)−sΓ(s) cos( )ζ(s) 2 πs or equivalently ζ(s) = 2(2π)s−1Γ(1 − s) sin( )ζ(1 − s). 2 Proof. Take a = 1 in Hurwitz’s formula to obtain for σ > 1:

Γ(s) − πis πis ζ(1 − s) = ζ(1 − s, 1) = e 2 F (1, s) + e 2 F (1, s) (2π)s

Γ(s) − πis πis Γ(s) πs = e 2 ζ(s) + e 2 ζ(s) = 2 cos( )ζ(s) (2π)s (2π)s 2 Hence, the theorem is proved for σ > 1. The result follows by analytic continuation. πs Taking s = 2n + 1, n ∈ N, we find that the factor cos( 2 ) vanishes and we get the trivial zeros of ζ(s): ζ(−2n) = 0, ∀n ∈ N. References [1] Tom M. Apostol. Introduction to Analytic Number Theory. Undergraduate Texts in Mathe- matics. Springer, first edition, 1976. [2] E. Freitag, R. Busam. Funktionentheorie. Springer Lehrbuch. Springer, first edition, 1993. [3] Melvyn B. Nathanson. Elementary Methods Number Theory. Graduate Texts in Mathematics. Springer, first edition, 2000.