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

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1 RIEMANN S FIRST PROOF OF THE ANALYTIC CONTINUATION OF ζ(s AND L(s, χ FELIX RUBIN SEMINAR ON MODULAR FORMS, WINTER TERM 6 Abstrat. In this hapter, we will see a proof of the analyti ontinuation of the Riemann eta funtion ζ(s and the Dirihlet L funtion L(s, χ via the Hurwit eta funtion. This then gives rise to a funtional equation for ζ(s and a diret omputation for the value of this funtion at negative integer points. 1. The Hurwit eta funtion We have already seen the definition of the Riemann eta funtion ζ(s and the Dirihlet L funtion L(s, χ as series: 1 χ(n ζ(s = n s and L(s, χ = n s where s = σ + it C, σ > 1, t R and χ is a Dirihlet harater modulo k, k N. These notations will be used throughout this hapter. We an unify the treatment of the above two funtions by introduing the Hurwit eta funtion: Definition 1.1. For σ > 1, we define ζ(s, a = 1 n= (n+a, < a 1, the s Hurwit eta funtion. Remark 1.. (1 For a = 1, ζ(s, 1 = ζ(s. ( L(s, χ = k s k χ(rζ(s, r k, if χ is a Dirihlet harater modulo k. The seond statement in the remark follows from the following alulation: Write n = qk + r, 1 r k, q =, 1,,... We then have: χ(n χ(qk + r L(s, χ = n s = (qk + r s = 1 1 k s χ(r (q + r k s q= = k s χ(rζ(s, r k. Theorem 1.3. The series ζ(s, a onverges absolutely for σ > 1 with uniform onvergene in every half-plane σ > 1 + δ, δ >. Hene, ζ(s, a is an analyti funtion of s in the half-plane σ > 1. n= n= Proof. We have (n + a s = (n + a σ (n + a (1+δ where the first equality follows from x s = e s ln x = e σ ln x e it ln x = e σ ln x = x σ, for x C, and the theorem follows. n= q= Date: Deember 14, 6. 1

2 FELIX RUBIN SEMINAR ON MODULAR FORMS, WINTER TERM 6 We will be able to analytially ontinue the Hurwit eta funtion beyond the line σ = 1 by an integral representation of ζ(s, a obtained from the integral formula for the Gamma funtion. We need the following properties of the Gamma funtion: For σ >, we an define the Gamma funtion by Γ(s = x s 1 e x dx. This funtion an be analytially ontinued to the whole omplex plane with simple poles at, 1,, 3,... with residues ( 1n n! at s = n, n N. Moreover, the gamma funtion satisfies the funtional equations Γ(s + 1 = sγ(s and Γ(sΓ(1 s = π sin(πs. Finally, we note that Γ(n + 1 = n!, if n N. For more detail on the Gamma funtion, see for example []. The next theorem gives an integral representation for the Hurwit eta funtion. Theorem 1.4. For σ > 1 and < a 1, we have (1.1 Γ(sζ(s, a = x s 1 e ax dx 1 e x Proof. First, keep s real, s > 1 and later extend it to omplex s by analyti ontinuation. By a hange of variable x = (n + at, n, we have for the Gamma funtion: Γ(s = (n + a s Γ(s = e x x s 1 dx = (n + a s e (n+at t s 1 dt e nt e at t s 1 dt Summing over all n, this gives: ζ(s, aγ(s = n= e nt e at t s 1 dt, where the left hand side onverges for σ > 1. By applying Beppo-Levi s theorem (1.5 below, we an interhange the sum and the integral to obtain: ζ(s, aγ(s = n= e nt e at t s 1 dt. Sine t >, < e t < 1 and thus n= e nt = 1. This gives: t n= e nt e at t s 1 = e at t s 1, for almost all t [,. Hene: t ζ(s, aγ(s = e at t s 1 dt, where s > 1, s R. 1 e t In order to extend this formula to the half plane σ > 1, note that the left hand side of the equation is analyti for σ > 1. To show that the right hand side is analyti, we assume that for some δ >, > 1, we have 1 + δ σ. Beause t σ 1 t δ for t [, 1], t σ 1 t 1, for t 1 and e t 1 t, for t, this gives: 1 1 e at t s 1 dt 1 e t e at t σ 1 dt 1 e t e at t σ 1 dt 1 e t 1 1 e at t σ 1 dt = ( 1 e t 1 e (1 at t δ e t dt e1 a 1 e at t 1 dt 1 e t e at t σ 1 dt 1 e t t δ 1 dt = e1 a, and δ e at t 1 dt = Γ(ζ(, a. 1 e t Hene, the integral in (1.1 onverges uniformly in every strip 1 + δ σ and therefore represents an analyti funtion in the half plane σ > 1. From this, the theorem follows. Beppo-Levi s theorem an be found in any ourse on measure and integration theory or probability theory and says the following:

3 RIEMANN S FIRST PROOF OF THE ANALYTIC CONTINUATION OF ζ(s AND L(s, χ 3 Theorem 1.5. (Beppo-Levi Let {f n } n N be an inreasing sequene of nonnegative, measurable funtions from any spae Ω into R (i.e. f 1 f f 3..., almost everywhere. Then, lim f n (ωdµ(ω = lim n(ωdµ(ω, n Ω Ω n where µ denotes the measure on the spae Ω. We need another representation of ζ(s, a, in order to give its analyti ontinuation beyond σ = 1. Consider the ontour C, whih is a loop around the negative real axis as shown in figure 1. We deompose C into the parts C 1, C and C 3. C is a irle around with radius < π (pos. oriented. C 1 and C 3 are the lower and upper edges of a ut in the omplex plane along the negative real axis. We use the parameteriations = re i on C 1, = re πi on C 3 with r [, and = e iθ with θ [, π] on C. Figure 1. The ontour C around the negative real axis The next result allows us to represent the Hurwit eta funtion as an integral along C ombined with the Gamma funtion, whih will lead us to the analyti ontinuation in the next setion. Theorem 1.6. If < a 1, σ > 1, the following holds (1. ζ(s, a = Γ(1 si(s, a, where I(s, a = 1 s 1 e a πi d is an entire funtion of s. C Note here, that for the definition of s 1 (s 1 ln( in the integrand, we write e and use the prinipal branh of the logarithm whih is defined on C\{x R; x }. This branh does make a jump of πi when rossing the negative real axis. Therefore if we ontinue the logarithm analytially from above the negative real axis and from below the negative real axis to this axis and take the integral along the ontour C, this integral will not anel out in general and moreover, the integral does not depend on the radius, as long as π. Proof. The integral along C is an entire funtion in s sine the integrand is an entire funtion (1 e on C.We thus only need to prove that the integrals along C 1 and C 3 onverge uniformly on any disk s M, M >. Along C 1, we have for r 1: s 1 = r σ 1 e i(σ 1+it = r σ 1 e πt r M 1 e πm sine s M. Similarly on C 3 : s 1 = r σ 1 e πi(σ 1+it = r σ 1 e t r M 1 e πm sine s M. We thus have for r 1 on C 1 and C 3 : s 1 e a rm 1 e πm e ar = rm 1 e πm e (1 ar r e r 1. But e r 1 > er, when r > ln, hene the integrand is bounded by Ar M 1 e ar, where the onstant A only depends on M. But r M 1 e ar dr onverges for >, and hene the integral does onverge uniformly on every disk s M. Therefore, I(s, a is an entire funtion of s.

4 4 FELIX RUBIN SEMINAR ON MODULAR FORMS, WINTER TERM 6 To prove (1., we write πii(s, a = ( C 1 + C + C 3 s 1 g(d, where g( = e a. On C 1 and C 3, g( = g( r and on C, we write = e iθ, with θ π. πii(s, a = + = i sin(πs π r s 1 e is g( rdr + i r s 1 e πis g( rdr s 1 e (s 1iθ e iθ g(e iθ dθ π r s 1 g( rdr + i s e isθ g(e iθ dθ. Now, let : We have lim r s 1 g( rdr = Γ(sζ(s, a for σ > 1, aording to equation (1.1. It remains to show that the seond integral from to π goes to ero when in order to prove (1.. Note that g( is analyti in < π exept for a simple pole at =. Hene, g( is analyti everywhere in < π and therefore bounded by some onstant A. This gives for σ > 1: π π s e isθ g(e iθ dθ σ e tθ A dθ πaeπ t σ 1 when πi(s, a = sin(πsγ(sζ(s, a and the result follows from Γ(sΓ(1 s = π sin(πs.. The analyti ontinuation of the Hurwit eta funtion We an use equality (1. to extend the definition of the Hurwit eta funtion beyond the line σ = 1, sine the right hand side of this equation is meaningful for any s C. We therefore define ζ(s, a := Γ(1 si(s, a, s C, < a 1. Theorem.1. The funtion ζ(s, a so defined is analyti on C \ {1} with a simple pole at s = 1 with residue 1. Proof. We know that I(s, a is entire, hene any possible singularities ome from Γ(1 s. Γ(1 s has poles at s = 1,, 3,... However, we have seen that ζ(s, a is analyti at s =, 3,..., so s = 1 is the only possible pole of ζ(s, a. For s = n N, I(n, a = 1 n 1 e a πi C 1 e d = 1 πi C n 1 e a 1 e d = Res = sine the integrals along C 1 and C 3 anel out. Hene, for n = 1, Now, we ompute I(1, a = Res = ( lim e a 1 e = lim e a 1 e = lim (s 1ζ(s, a = lim(1 sγ(1 si(s, a ( n 1 e a 1 e 1 e = 1 = I(1, a lim Γ( s = Γ(1 = 1. Therefore, ζ(s, a has a simple pole at s = 1 with residue 1. By the equations given in remark 1., we an now give analyti ontinuations of the Riemann eta and the Dirihlet L funtions by using these equations as definitions of the respetive funtions.

5 RIEMANN S FIRST PROOF OF THE ANALYTIC CONTINUATION OF ζ(s AND L(s, χ 5 Theorem.. residue 1. (1 ζ(s is analyti on C with a simple pole at s = 1 with ( For the prinipal harater χ 1 modulo k, L(s, χ 1 is analyti on C with a simple pole at s = 1 with residue φ(k k. (3 If χ χ 1, L(s, χ is an entire funtion of s. φ(k denotes Euler s totient whih gives the number of elements n k, n N, suh that (k, n = 1. Proof. The first statement is trivial. To prove the other two, we note that { if χ χ1 χ(r = φ(k if χ = χ 1 This formula is a onsequene of the definition of Dirihlet haraters of redued residue lasses (see for example [3]. Now, ζ(s, r k has a simple pole at s = 1 with residue 1. Hene χ(rζ(s, r k has a simple pole at s = 1 with residue χ(r. Res s=1 (L(s, χ = lim (s 1L(s, χ = lim (s 1k s = 1 k χ(rζ(s, r k { χ(r = φ(k k if if χ χ1 χ = χ 1 3. Evaluation of ζ( n, a for n N We an now expliitly alulate ζ( n, a for n N: ζ( n, a = Γ(1+nI( n, a = n!i( n, a. In the proof of theorem.1 we have seen that for n N, I( n, a = Res = ( n 1 e a. The alulation of this residue leads to the Beroulli polynomials. Definition 3.1. x C, we define the Bernoulli polynomials B n (x by the equation ex e 1 = B n(x n= n! n, where < π. This leads to the expliit result Theorem 3.. n N and < a 1, we have ζ( n, a = Bn+1(a n+1. Proof. We already know that ζ( n, a = n!i( n, a. But ( n 1 e a I( n, a = Res = 1 e ( = Res = n Hene, the theorem follows. n ea = Res = ( e 1 m= B m (a m = B n+1(a m! (n + 1! 4. A funtional equation for the Riemann eta funtion Here, we give e funtional equation for ζ(s, whih leads us to the so alled trivial eros of the Riemann eta funtion.

6 6 FELIX RUBIN SEMINAR ON MODULAR FORMS, WINTER TERM 6 Definition 4.1. For x R and σ > 1, define the Dirihlet series e πinx F (x, s = F (x, s is alled the periodi eta funtion. Note that F (x, s is periodi in x with period 1 and F (1, s = ζ(s. Moreover, the series onverges absolutely for σ > 1. Proposition 4.. If < a 1 and σ > 1, the Hurwit formula holds: ζ(1 s, a = Γ(s ( (π s e πis πis F (a, s + e F ( a, s In the proof of this proposition, we need the following lemma: Lemma 4.3. For < r < π, denote with S(r the region ( S(r = C\ B(nπi, r, n Z where B(, r denotes the open disk with enter and radius r. Then, if < a 1, the funtion g( = ea 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 Hurwit s formula. Proof. Consider the funtion I N (s, a = 1 s 1 e a πi C(N d, where C(N is the ontour shown in figure, and N N. Note that we use again the prinipal branh of the logarithm to define s 1 in the integrand, as we did in the proof of formula (1.. n s Figure We now show that if σ <, the integral along the outer irle tends to as N. Hene, I N (s, a I(s, a, for N and σ <. Parameteriing = Re iθ, θ π, we have s 1 = R s 1 e iθ(s 1 = R σ 1 e tθ R σ 1 e π t. Sine the outer irle lies in the set S(r of the preeding lemma, the absolute value of the integrand is bounded by Ae π t R σ 1, A being the bound for g( implied by the lemma. Hene, the absolute value of the integral along the outer irle is bounded by πae π t R σ, whih tends to as R, for σ <. We thus have: lim I N(1 s, a = I(1 s, a if σ > 1. N

7 RIEMANN S FIRST PROOF OF THE ANALYTIC CONTINUATION OF ζ(s AND L(s, χ 7 Now On the other hand, we have by Cauhy s residue theorem: ( s e a I N (1 s, a = Res =nπi 1 e = n= N,n ( (Res s e a ( s e a =nπi + Res = nπi 1 e 1 e ( s e a Res =nπi 1 e = lim ( e a nπi nπi s 1 e = enπia (nπi s, hene I N (1 s, a = N enπia (nπi s ( i s = e πis so I N (1 s, a = e is (π s + N e nπia ( nπi s. But i s = e is and e nπia n s + e πis (π s e nπia n s. Letting N and multiplying both sides with Γ(s, we obtain the desired result. We an now state the funtional equation for the Riemann eta funtion. Theorem 4.4. For all s C, we have ζ(1 s = (π s Γ(s os( πs ζ(s or equivalently ζ(s = (π s 1 Γ(1 s sin( πs ζ(1 s. Proof. Take a = 1 in Hurwit s formula to obtain for σ > 1: ζ(1 s = ζ(1 s, 1 = Γ(s ( (π s e πis πis F (1, s + e F (1, s = Γ(s (π s ( e πis ζ(s + e πis ζ(s = Γ(s (π s os(πs ζ(s Hene, the theorem is proved for σ > 1. The result follows by analyti ontinuation. Taking s = n + 1, n N, we find that the fator os( πs vanishes and we get the trivial eros of ζ(s: ζ( n =, n N. Referenes [1] Tom M. Apostol. Introdution to Analyti Number Theory. Undergraduate Texts in Mathematis. Springer, first edition, [] E. Freitag, R. Busam. Funktionentheorie. Springer Lehrbuh. Springer, first edition, [3] Melvyn B. Nathanson. Elementary Methods Number Theory. Graduate Texts in Mathematis. Springer, first edition,.