Riemann Zeta Function

chiph588@ · #1 $Old$ December 26th, 2008, 11:59 PM

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

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

Quote:

Originally Posted by chiph588@ $View Post$

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.

mr fantastic · #3 $Old$ December 27th, 2008, 05:19 AM

In the interests of referencing accuracy, the exact transcription (whch starts first line of p13) is as follows:

Quote:

Originally Posted by Mathstud28 $View Post$

[snip]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 $D$ denote the domain in the $s$ -plane which consists of all [snip] 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 of (*). 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 over 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

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

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

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

This relationship between ${\color{red}\zeta(s)}$ and ${\color{red}\zeta(1 - s)}$ is known as the functional equation of the zeta function.

In order to prove rigorously that [the above] holds for s < 0, it suffices to modify the above argument by 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 path of integration is less than $2\pi(2n+1)\pi$ , because the factor ${\color{red}(e^{x}-1)^{-1}}$ is bounded on the circle ${\color{red} |s| = (2n + 1) \pi}$ , and because the modulus of $\frac{(-x)^s}{x}$ on this circle is $\left|x^{s-1}\right|\leqslant \left[(2n+1)\pi\right]^{{\color{red}\delta} -1}$ 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 [reference to footnote] one or more of terms of $\color{black}(*)$ 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 $+\infty$ , Mr F says: A path cannot begin at + infinity.

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 $+\infty$ Mr F says: This is not the boundary of D.

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 substitution of $\Gamma$ for $\Pi$ as the symbol for the gamma function has also been made.

The substitution of various colours of * rather than numbers to label equations has also been made.

A reference to a footnote has been omitted.

Opalg · #4 $Old$ December 27th, 2008, 06:39 AM

Quote:

Originally Posted by chiph588@ $View Post$

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 zeta function is initially defined as $\zeta(s) = \sum_{n=1}^\infty n^{-s}$ . This series converges for all complex numbers s whose real part is greater than 1. To extend the domain of definition, the usual procedure (as far as I understand it—I'm not an expert in this area) is to use analytic continuation to define it for all s with real part greater than 0 (except for the pole at s=1). An intricate argument using contour integration (as in the previous comments) then establishes the functional equation for s in the strip 0<re(s)<1. The functional equation is then used to define $\zeta(s)$ in the remainder of the complex plane.