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

Question

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.

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
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 08 '13 at 22:40
The book "the Theory of Riemann zeta function" by Titchmarsh contains several proofs of the functional equation. — zy_, Dec 09 '13 at 09:01
Related: Riemann's thinking on symmetrizing the zeta functional equation — Grigory M, Dec 13 '13 at 14:37
Related: How does one motivate the analytic continuation of the Riemann zeta function? — Grigory M, Dec 13 '13 at 14:38
@AlexB. What theory is necessary to understand Tate's proof? — Meow, Apr 17 '14 at 09:58
@Alyosha You need to be fairly comfortable with $p$-adic numbers, and to know some basic measure theory. That would be enough to give it a go, and fill in the gaps as you progress. It doesn't require that much high-flying theory, the main difficulty for novices is that it takes a completely new point of view, compared to the classical one. Note that Tate also proves the analytic class number formula in the same big sweep, so it would help if you knew the statement of that. — Alex B., Apr 17 '14 at 10:06
@AlexB. I've encountered a few things that need $p$-adic numbers, but never a particularly good introduction on them. I know this is not really to do with the original question, but do you know of a good introduction? Regardless, thanks for the speedy response. — Meow, Apr 17 '14 at 10:08
Related: Contour approach to Riemann zeta functional equation — Grigory M, May 25 '14 at 14:42

score 15 · Answer 1 · edited Apr 19 '14 at 11:36

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:

The functional equation takes simpler form for the $\xi$-function.
The $\xi$-function is (more or less) the Mellin transform of $\theta$-function.
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.

See http://mathoverflow.net/questions/58004/how-does-one-motivate-the-analytic-continuation-of-the-riemann-zeta-function/97401#97401 — Tom Copeland, Nov 29 '14 at 04:22
Related: https://math.stackexchange.com/questions/3882936/proving-property-of-xis-in-riemann-zeta-function/3883068#3883068 — pshmath0, Sep 25 '22 at 21:06
Thank you for your answer, @Grigory M
However, I note that your original definition of \xi (s) is only valid of Re(s) > 1. If the real value of s is 1 or less, the summation in the definition diverges.

In order for the final line to draw a valid conclusion, you must show that \xi(s) = \xi(1-s) over a finite interval (or at least enough points to constitute a limit point, which is a seed for a holomorphic function). You have not done this if the map s\mapsto 1-s does send the region Re(s)>1 to a sufficient overlap of itself. — Steven Xu, Feb 14 '23 at 06:53

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

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