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Mathstud28 · #2 December 27th, 2008, 12:50 AM

Quote:

Originally Posted by chiph588@

I have a couple question regarding the Riemann zeta function and its extension onto the complex plane. I know the function is given by the reflection formula $\zeta(s) = 2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s)$ .

My first question is how is this proved?

My second question is how does this extend $\zeta(s)$ to the complex plane?

The way I look at it is to plug in a value already known, which would be some $s>1$ and solve for $\zeta(1-s)$ . But this only solves for all $s \in (-\infty,0)$ leaving $\mathbb{C}$ \ $(-\infty,0) \cup [1,\infty)$ unaccounted for...

Thanks guys

I obviously know nothing about CA, but just so you can cross-check whatever anybody else here gives you, I will transcribe a proof from "Riemann's Zeta Function" H.M. Edwards (Dover)

" For negative real values of s, Riemann evaluated the integral $\zeta(s)=\frac{\Gamma(-s)}{2\pi i}\int_{+\infty}^{+\infty}\frac{(-x)^s}{e^x-1}\frac{dx}{x}{\color{red}(*)}$ as follows. Let $D$ be the domain in the $s$ -plane which consists of all other points other than those which lie within $\varepsilon$ of the positve real axis or within $\varepsilon$ of one of the singularities $x=\pm 2\pi i n$ of the integrand. Let $\partial D$ be the boundary of $D$ oriented in the usual way. Then, ignoring for the moment the fact that $D$ is not compact, Cauchy's theorem gives

$\frac{\Gamma(-s)}{2\pi i}\int_{\partial D}\frac{(-x)^s}{e^x-1}\frac{dx}{x}=0{\color{blue}(*)}$ .

Now one component of this integral is the integral $\color{red}(*)$ with the orientation reversed, whereas the others are integrals of the circles $|x\pm 2\pi i n|=\varepsilon$ oriented clockwise. Thus when the circles are oriented in the usual counterclockwise sense $\color{blue}(*)$ becomes

$-\zeta(s)-\sum\frac{\Gamma(-s)}{2\pi i}\int_{|x\pm2\pi i n|=\varepsilon}\frac{(-x)^s}{e^x-1}\frac{dx}{x}=0~{\color{green}(*)}$

The integrals over the circles can be evaluated by setting $x=2\pi i n+y$ for $|y|=\varepsilon$ to find

$\begin{aligned}\frac{\Gamma(-s)}{2\pi i}\int_{|y|=\varepsilon}\frac{(-2\pi i n-y)^s}{e^{2\pi i n+y}-1}\frac{dy}{2\pi i n+y}&=\frac{-\Gamma(-s)}{2\pi i}\int_{|y|=\varepsilon}(-2\pi i n-y)^{s-1}\frac{y}{e^y-1}\frac{dy}{y}\\ &=-\Gamma(-s)(-2\pi i n)^{s-1}\end{aligned}$

by the Cauchy integral formula. Summing over all integers $n$ other than $n=0$ and using $\color{green}(*)$ then gives

$\begin{aligned}\zeta(s)&=\sum_{n=1}^{\infty}\Gamma(-s)\left[(-2\pi i n)^{s-1}+(2\pi i n)^{x-1}\right]\\ &=\Gamma(-s)(2\pi)^{s-1}\left[i^{s-1}+(-i)^{s-1}\right]\sum_{n=1}^{\infty}\frac{1}{n^{s-1}}\end{aligned}$

Finally using the simplification

$\begin{aligned}i^{s-1}+(-i)^{s-1}&=\frac{1}{i}\left[e^{s\ln(i)}-e^{s\ln(-i)}\right]\\ &=\frac{1}{i}\left[e^{\frac{s\pi i}{2}}-e^{\frac{-s\pi i}{2}}\right]\\ &=2\sin\left(\frac{s\pi}{2}\right)\end{aligned}$

One obtains the desired formula

$\zeta(s)=\Gamma(-s)(2\pi)^{s-1}2\sin\left(\frac{s\pi}{2}\right)\zeta(1-s){\color{black}(*)}$

In order to prove rigorously that the above holds for negative s, it suffices to modify the above argument letting $D_n$ be the intersection of $D$ with the disk $|s|\leqslant (2n+1)\pi$ and letting $n\to\infty$ .; then the integral $\color{blue}(*)$ splits into two parts, one being an integral over the circle $|s|=(2n+1)\pi$ with the points within $\varepsilon$ of the positive real axis deleted, and the other being an integral whose limit as $n\to\infty$ is the left side of $\color{green}(*)$ . The first of these two parts approaches zero because the length of the part of integration is less than $2\pi(2n+1)\pi$ , because the modulus of $\frac{(-x)^s}{x}$ on the circle is $\left|x^{s-1}\right|\leqslant \left[(2n+1)\pi\right]^{s-1}$ for $s\leqslant -\delta<0$ . Thus the second part, which by Cauchy's theorem is the negative of the first part, also approaches zero, which implies $\color{green}(*)$ and hence $\color{black}(*)$ .

This completes the proof of the functional equation in the case $s<0$ . However, both sides of $\color{black}(*)$ are analytic functions of $s$ , so this suffices to prove $\color{black}(*)$ for all values of $s$ (except for $s=0,1,2,\cdots$ where one or more of terms in $\color{black}(*)$ have poles)

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Note two things

A) In the first integral the limits of integration are meant to represent a path of integration which begins at $+\infty$ , moves to the left down the positive real axis, circles the origin once in the counterclockwise direction, and returns up the positive real axis to $+\infty$

B) I only did this because I almost never see anyone on this site give an answer regarding the RZ function. If this does not make sense I am sorry. Take this proof with a grain of salt, for I know nothing of this and am just copying it from a book. I could easily have copied the wrong thing. If something confuses you or this does not seem like the pertinent proof just disgregard the thing in its entirety. And if need be there is an alternate proof.

The following users thank Mathstud28 for this useful post:
chiph588@
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