Zeta Function - Part 11 - Riemann Functional Equation I

What people said

Recommends Agno · 2017-03 · Mathematics Stack Exchange · on “Riemann zeta-function regularization in string theory”

One of the easiest explanations I found on how the analytic continuation of $\zeta(s)$ can be done (without Bernouill-numbers) : [this video] This yields the famous functional equation that relates $\zeta(s)$ to $\zeta(1-s)$ (hence also $\zeta(2)$ to $\zeta(-1)$).

Recommends Peter Driscoll · 2017-08 · Mathematics Stack Exchange · on “Proof of Functional Equation Zeta”

This is to clarify the working, even though all the steps have already been given. Makes it simpler for me to understand. Based on the video [this video] but elaborated to my taste. $$ \xi(s) = \pi^{-s/2}\zeta(s)\Gamma(s/2)= \int_0^\infty \psi(x) x^{s/2-1}\;dx $$ where $\psi$ is given in terms of theta, $ \theta(x)= 2 \psi(x)+1 $ and $\theta$ has the functional equation, $ \theta(x)=\frac{1}{\sqrt{x}}\theta(\frac1{x}) $. The integral will be left in $\psi$, because the maths works out more easily, so rearrange the functional equation in terms of $\psi$. $$ \psi(x)=\psi(\frac1{x})x^{-\frac12}+\frac12 x^{-\frac12} - \frac12 $$ Split the integral into two ranges, 0 to 1, and 1 to infinity. Change th

Cites Antoni Parellada · 2017-06 · Mathematics Stack Exchange · on “Elements needed to derive the Riemann-Siegel Z function”

On this post [LINK] I got a comment to motivate the present question: To prove a zero $\zeta(s_0)=0, s_0 \approx 1/2+iy$ is exactly on the critical line, it is enough to prove the real function [LINK] $Z(t) = \zeta(1/2+it) e^{i \,\text{arg}(\Gamma(1/4+it/2)}\, \pi^{-it/2}$ changes of sign around $t=y$. We know $Z(t)$ is real for $t$ real thanks to the functional equation [LINK] . So I would like to ask for some help understanding where the expression $$\large Z(t) = \zeta\left(\frac{1}{2}+it\right)\quad e^{i \;\text{arg}\left(\Gamma(\frac{1}{4}+\frac{it}{2})\right)}\quad \pi^{\frac{-it}{2}}\tag 1$$ comes from. The terser version [LINK] $Z(t)=\zeta\left(\frac{1}{2}+it\right)\; e^{i\theta(t)}$

Cites Shree · 2019-03 · Mathematics Stack Exchange · on “Need reference for proof of functional equation for generalized L-functions”

When proving functional equation for Riemann zeta function one starts at the definition of gamma function $$\Gamma(s) = \int_0^{\infty} x^{s-1} e^x\mathrm dx\tag1$$ After a few steps we arrive at $$ \pi^{-\frac{s}{2}} \Gamma \left(\frac{s}{2} \right) \zeta(s) = \frac{1}{s(1-s)} - \int_1^{\infty} \left( x^{\frac{s}{2}} + x^{\frac{1-s}{2}} \right) \frac{\psi(x)}{x}\mathrm dx\tag2$$ And observing the RHS of the above equation doesn’t change when we replace $s$ with $(1-s)$ in equation $(2)$ to get functional equation below: $$ \pi^{-\frac{s}{2}} \Gamma \left(\frac{s}{2} \right) \zeta(s) = \pi^{-\frac{1-s}{2}} \Gamma \left(\frac{1-s}{2} \right) \zeta(1-s) \tag3 $$ You can watch detail proof here