Step 1. Analytic Properties of the Riemann zeta function [2 lectures]

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1 Step. Analytic Properties of the Riemann zeta function [2 lectures] The Riemann zeta function is the infinite sum of terms /, n. For each n, the / is a continuous function of s, i.e. lim s s 0 n = s n, s 0 for all s 0 C, an is ifferentiable, i.e. sn = s s e slogn = logn)e slogn = logn, for all s C. We know from secon year analysis that we can a a finite number of continuous ifferentiable) functions to get a continuous ifferentiable) function, but this is not necessarily true for an infinite sum. We nee more than the sum converges. Definition 4.3 A sequence {x n } n in C is a Cauchy Sequence if, an only if, ε > 0, N : n,m N, x n x m < ε. Depening on how you have constructe C you may efine that a series converges if, an only if, it is a Cauchy Sequence. Unfortunately this was not how convergence of a sequence was efine in MATH200 say. Theorem 4.4 A sequence {x n } n converges in C if, an only if, {x n } n is a Cauchy Sequence. Proof ) Assume that {x n } n converges to x say. Let ε > 0 be given. Then by the ε N efinition of convergence, N : x n x < ε/2. Let m,n N then x n x m = x n x+x x m x n x + x x m < ε, an we have verifie the efinition of a Cauchy Sequence. ) This follows as a efining property of C, basically that C is complete. 7

2 Definition 4.5 Let {F n z)} n be a sequence of functions efine on a set D C. The sequence {F n } n converges to F on D iff ε > 0, z D, N = N ε,z) : n N, F n z) F z) < ε. The sequence {F n } n converges uniformly to F on D iff ε > 0, N = N ε) : z D, n N, F n z) F z) < ε. Be careful to unerstan the ifference in these efinitions, for convergence the value of N epens on the point z, whereas for uniform convergence the N will work simultaneously for all z in the omain D. Equivalently the efinition of uniform convergence can be given in terms of a Cauchy sequence as ε > 0, N = N ε) : z D, n,m N, F n z) F m z) < ε. All efinitions for sequences translate into efinitions for series, as in Definition 4.6 A series j= f jz) converges uniformly on D if, an only if, the sequence of partial sums F n z) = n j= f jz) converges uniformly on D. That is, n ε > 0, N = N ε) : z D, n,m N, f j z) < ε. This leas to j=m+ Weierstrass s M-Test for Uniform Convergence of a Series: If there exists a sequence {M i } i of non-negative numbers for which f j z) < M j for all z D an all i an j= M j converges then j= f jz) converges uniformly on D. Verification In fact, assuming k > l, the partial sums satisfy F k z) F l z) = f k z)+f k z)+...+f l+ z) f k z) + f k z) f l+ z) < M k +M k +...+M l+. On account of the convergence of the series the last sum can be mae smaller than a given ε > 0. Hence the sequence of partial sums F m z) converges uniformly. 8

3 The Question then arises what properties of a uniformly converging sequence are inherite by its limit? We have Weierstrass s Theorem for Series see Backgroun Notes 0.7) that states Theorem 4.7 Weierstrass s Theorem for Series. Assume f z),f 2 z),f 3 z),... are holomorphic in an opeet D, an i= f iz) converge uniformly on every close an boune subset of D. Then i) The F z) = i= f iz) is holomorphic on D, ii) For all k, the series i= fk) i z) converges on D, an converge uniformly on every close an boune subset of D with limit F k) z). So the series can be ifferentiate term-by-term.) Our first example of use of this theorem is the Riemann zeta function. The Riemann zeta function is a particular example of a Dirichlet Series, n= a nn s. For a give C such a series may converge or not an if it converges it may converge absolutely or not. Results on how such regions are connecte for general Dirichlet Series are not give in this course. We will, though, make use of the fact that if a series n= a n converges absolutely then we have an infinite triangle inequality n= a n n= a n. For Res + δ the Riemann zeta function converges absolutely an to see this use the Comparison Test on n= /ns, for each term satisfies / /n +δ an n= /n+δ converges. To go further we nee to show that the series efining ζs) converges uniformly oome omain. Theorem 4.8 Assume δ > 0. For Res +δ the Riemann zeta function ζs) = converges uniformly, is holomorphic in Rs >, with erivative ζ logn s) = for Res >. 9 n= n=

4 Proof Wehavetoshowonlythattheseriesefiningζs)convergesuniformly Res +δ. We coul simply apply Weierstrass s M-test with M n = /n +δ or o it irectly as N ζs) = n= n=n+ n=n+ by infinite triangle inequality n=n+ N n +δ since Res +δ u u = +δ δn δ. Given any ε > 0, then /δn δ is less than ε when N is sufficiently large, inepenent of s, showing uniform convergence. We can apply Weierstrass s Theorem since any close an boune subset of Res > is containe in Res + δ for some δ > 0. Each term in the series is holomorphic on C with erivative s ) = s e slogn = logn. The state result then follows from Weierstrass s Theorem. In Theorem.8 we saw ζs) = p p s ), for Res >. This with Theorem 4.8 shows that the Euler prouct is continuous an ifferentiable. This was not obvious. For each prime p, the /p s ) is a continuous function of s, i.e. s lim ) = ), s s 0 p s p s 0 for all s 0 C, an is ifferentiable, i.e. ) = ) 2 p s p s = logp p s p s ) 2, 0 p ) = s s ) 2 p s s p s

5 for all s C. We know from secon year analysis that we can multiply a finite number of continuousifferentiable) functions to get a continuousifferentiable) function, but this is not necessarily true for an infinite prouct. There is a Weierstrass s Theorem for Infinite Proucts but it is not given here. Earlier in the course we looke at replacing sums by integrals. In the same vein we can use Partial Summation to replace the Dirichlet Series efining the Riemann zeta-function by an integral. Theorem 4.9 For s we have n N n = + s s + N s s s N where is the fractional part of u so = u [u]). n N n N u u s+ 5) Proof By Partial Summation an this argument, for s =, has beeeen before in the previous Chapter) = N s N )) s Let N to get = N N s n N = N N s +s N N n n u = N N N +s u [u] s u s+ = N N s +s N s) u u s+ u u s+ u u N s u [u]) u us+ u s+ = N N + s N s ) N s s s u us+. 6)

6 Theorem 4.0 For Res >, ζs) = + s s u+su. 7) Further, the right han sie of 7) efines a function holomorphic in Res > 0, except for a simple pole, resiue at s =. This is an example of an analytic continuation of ζs). The right han sie of 7) efines a function holomorphic on a region containing Res >, where it agrees with the Dirichlet series efinition of ζs). It is in fact possible to continue analytically ζs) to all of C, though we o not o so in this course. For the remainer of the course when we talk of the Riemann zeta function we will be thinking of the function given by the right han sie of 7) for Res > 0. When Res > we will know that it has an alternative efinition as a Dirichlet Series. Note On Problem Sheet 3 you are aske to show that ) n = n= ) ζs), 2 s for Res >. It can be shown that the Dirichlet Series on the left is, in fact, holomorphic in Res > 0 an so this equality gives another analytic continuation of ζs) to Res > 0. It can be shown though that if a function has an analytic continuation, then that continuation is unique. Proof Remember that Res = σ >. Let N, then N s = N σ 0 while an so the integral in 6) converges. Hence u +s 8) u +σ ζs) = + s s u+su 9) as require. In fact, from 8), we see that the integral in 9) converges absolutely for Res > 0. There is a theory for uniform convergence of integrals; see Weierstrass s Theorem for Infinite Integrals in Backgroun Notes 0.7. But this is not irectly 2

7 applicable here since the integran u s is not a continuous function of u for fixe s. Instea, consier the infinite integral in 9) as an infinite sum of integrals over intervals of length : u +su = n+ u = f u +s n s). n= We cahow but will only o so for level 4) that for any δ > 0, n n= i) the series n= f ns) converges uniformly on Res > δ, ii) each f n s) is holomorphic on Res > δ, with erivative f n s) s = n+ n logu)u. u+s This is sufficient, by Weierstrass s Theorem for Series, part i of Theorem 4.7 above, to imply that the integral in 9) is holomorphic on Res > 0. Diffferentiate 5) w.r.t s, allowable since the sum an integral are finite. This is harer to justify is we ifferentiate 7). In this way we get Theorem 4. for integral N an s we have N n= Let N to erive Corollary 4.2 For Res > 0, logn = s ) 2 + s) s logn +N s s) 2 ζ s) = s ) 2 N N u +su+s logu u. u +s logu u s+u+s u. 0) u s+ As in the proof of Theorem 4.0 this only follows for Res >, when N s logn s 0 an N s s) 2 0 as N. But once erive, the right han sie of 0) efines a holomorphic function for Res > 0,s. 3