# Different methods to prove $\zeta(s)=2^s\pi^{s-1}\sin\left(\frac{s\pi}{2}\right) \Gamma (1-s) \zeta (1-s)$.

I've recently encountered this strangely attractive equation (Riemann's functional equation), along with Riemann's original proof. $$\displaystyle\zeta(s)=2^s\pi^{s-1}\sin\left(\frac{s\pi}{2}\right) \Gamma (1-s) \zeta (1-s)$$ There are probably countless proofs of Riemann's functional equation, but as of yet there's not a single place where they're all concentrated.

Anyone care to share some particularly pretty proofs?

$\zeta(s)$: Riemann zeta function.

$\Gamma(s)$: Gamma function.

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The "modern" way of proving it was developed in Tate's Thesis. It's pretty advanced reading, but absolutely gorgeous, and hugely influential. – Alex B. Nov 30 '13 at 14:55
See Tate's Thesis for a modern proof. – Ahaan S. Rungta Nov 30 '13 at 15:06
Ideally this page would actually contain proofs. – Alyosha Dec 4 '13 at 7:30
Multiply both sides with $\Gamma(s)$ , and use the fact that $\displaystyle\Gamma(s)\zeta(s)=\int_0^\infty\frac{x^{s-1}}{e^x-1}dx$, coupled with the reflection formula $\displaystyle\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin{(\pi z)}}$ – Lucian Dec 8 '13 at 22:40
The book "the Theory of Riemann zeta function" by Titchmarsh contains several proofs of the functional equation. – y zh Dec 9 '13 at 9:01

AFAIK none of proofs is very short and easy. I'll post just a rough sketch (in particular all analytical issues are silently ignored) of a Riemann's original proof based on the Poisson summation formula.

Let's define $$\xi(s)=\pi^{-s/2}\Gamma(s/2)\zeta(s).$$ Riemann's functional equation takes the form $$\xi(s)=\xi(1-s).$$

By definition $$\xi(s)= \sum_{n=1}^\infty\pi^{-s/2}n^{-s}\int_0^\infty e^{-t}t^{s/2}\frac{dt}t$$ and after substitution $t=\pi n^2x$ we get $$\xi(s)= \int_0^\infty x^{s/2}\sum_{n=1}^\infty e^{-\pi n^2x}\frac{dx}x= \frac12\int_0^\infty x^{s/2}(\theta(x)-1)\frac{dx}x,$$ where $\theta(x)=\sum_{n\in\mathbb Z}e^{-\pi n^2x}$.

By Poisson summation formula this theta function satisfies $$\theta(x^{-1})=x^{1/2}\theta(x),$$ which allows us to rewrite last integral in the form \begin{multline} \xi(s)= \frac12\int_1^\infty x^{s/2}(\theta(x)-1)\frac{dx}x+ \frac12\int_1^\infty x^{-s/2}(x^{1/2}\theta(x)-1)\frac{dx}x=\\ =\frac12\int_1^\infty(x^{s/2}+x^{(1-s)/2})(\theta(x)-1)\frac{dx}x-\frac1{s(1-s)}, \end{multline} which is manifestly symmetric under change $s\mapsto1-s$.

Summary:

1. The functional equation takes simpler form for the $\xi$-function.
2. The $\xi$-function is (more or less) the Mellin transform of $\theta$-function.
3. The $\theta$-function satisfies some modularity functional equation coming from the Poisson summation formula.

...or to make a long story short,

Riemann's functional equation is the Mellin transform of the Poisson summation formula.

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– Tom Copeland Nov 29 '14 at 4:22