Proof of Functional Equation Zeta

Question

$$ \pi^{-s/2}\zeta(s)\Gamma(s/2)=\pi^{-(1-s)/2}\zeta(1-s)\Gamma((1-s)/2) $$ (That's the equation that want to prove)

Hello guys, so I'm trying to prove the functional equation of Riemann Zeta, through the function of Jacobi Theta, did the following.

Be the Theta function

$$ \theta(z,t)=\sum_{n=-\infty}^{\infty} e^{2\pi i nz-\pi n^2t} $$

$$ \theta(0,t) =\theta(t) $$

It is owned by the following Theta function (as shown here)

$$ \theta(t)=\frac{1}{\sqrt(t)}\theta\left(\frac1t\right) $$

We must also

$$ \theta(t)=\sum_{n=-\infty}^{\infty}e^{-\pi n^2 t}=1+2\sum_{n=1}^{\infty} e^{-\pi n^2 t} $$

Considering the $\Gamma(s/2)$ function have to

$$ \Gamma\left ( \frac s2 \right )=\int_{0}^{\infty} e^{-x}x^{s/2-1} dx $$

Making $x=\pi n^2t$

$$\pi^{-s/2}\zeta(s)\Gamma(s/2)=\frac12\int_0^\infty \left ( \theta(t)-1 \right )t^{s/2-1}\;dt$$

After many accounts, replacements and others, we have to

$$\pi^{-s/2}\zeta(s)\Gamma(s/2)=\frac12\int_1^\infty \left ( \theta(u)-1 \right )\left ( u^{(1-s)/2-1}+u^{s/2-1} \right )\;du+\frac{1}{s(s-1)}$$

In short, what I want is to know (please, step by step), making this full result in what I want, ie

SOLVE $$\frac12\int_1^\infty \left ( \theta(u)-1 \right )\left ( u^{(1-s)/2-1}+u^{s/2-1} \right )\;du+\frac{1}{s(s-1)}$$

or

$$\frac12\int_1^\infty \left ( \theta(u)-1 \right )\left ( u^{(1-s)/2-1}+u^{s/2-1} \right )\;du+\frac{1}{s(s-1)}=\pi^{-(1-s)/2}\zeta(1-s)\Gamma((1-s)/2)$$

Answer 1

HINT

Let $\xi(s)=\pi^{-s/2}\zeta(s)\Gamma(s/2)$ the completed zeta function. The function $\xi(s)$ extends to a meromorphic function on $\Bbb C$, regular except for simple poles at $s = 0; 1$, which satisfies the functional equation $\xi(s) = \xi(1 - s)$.

You found that

$$ \xi(s)=\frac12\int_1^\infty \left ( \theta(u)-1 \right )\left ( u^{(1-s)/2-1}+u^{s/2-1} \right )\;du+\frac{1}{s(s-1)} $$ so you don't have to evaluate more.

The RHS is is manifestly symmetrical under $s \leftrightarrow 1 - s$, and analytic since $\theta(u)$ decreases exponentially as $u\to \infty$. This concludes the proof of the functional equation $\xi(s) = \xi(1 - s)$ and analytic continuation of $\xi(s)$.

Answer 2

The proof was finished with the result you got, see alexjo's anwser. Moreover, if you want to see that the RHS is what you want, you have to "come back" (say) to the original expression. Doing the same process for $\Gamma((1-s)/2)$ we get $$\pi^{-(1-s)/2}\zeta(1-s)\Gamma\left(\frac{1-s}{2}\right)=\frac{1}{2}\int_0^\infty (\theta(t)-1)t^{(1-s)/2}\,\frac{dt}{t}.$$

Following the same computations you did for the right hand side you get $$\frac12\int_1^\infty \left ( \theta(u)-1 \right )\left ( u^{(1-s)/2-1}+u^{s/2-1} \right )\;du+\frac{1}{s(s-1)}=\pi^{-(1-s)/2}\zeta(1-s)\Gamma\left(\frac{1-s}{2}\right).$$ Precise computation can be find in Edward's book Riemman's zeta function (Section 1.7).

Answer 3

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 https://www.youtube.com/watch?v=K6L4Ez4ZVZc 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 the variable name to t on the first integral, so that we can substitute it back to x later. $$ \xi(s) = \int_0^1 \psi(t) t^{s/2-1}\;dt + \int_1^\infty \psi(x) x^{s/2-1}\;dx$$ Make the substitution for $\psi(t)=\psi(\frac1{t})t^{-\frac12}+\frac12 t^{-\frac12} - \frac12 $. Put the simple terms in their own integral, $$ \xi(s) = \int_0^1 \psi(\frac1{t})t^{s/2-3/2}\;dt + \int_0^1 \frac12 t^{s/2-3/2} - \frac12 t^{s/2-1}\;dt + \int_1^\infty \psi(x) x^{s/2-1}\;dx$$ Substitute $t = \frac1{x}$ so $dt = -x^{-2} dx$. As t goes from 0 to 1, x goes from infinity to 1. Also do the integration on the simple terms. $$ \xi(s) = - \int_\infty^1 \psi(x)x^{-s/2+3/2}x^{-2}\;dx + [ \frac12 \frac{t^{s/2-1/2}}{s/2-1/2} - \frac12 \frac{t^{s/2}}{s/2}]_0^1 + \int_1^\infty \psi(x) x^{s/2-1}\;dx$$ Switch the integral direction, to flip the sign. Put values in for the integral range and simplify. $$ \xi(s) = \int_1^\infty \psi(x)x^{-s/2-1/2}\;dx + \frac1{s-1} - \frac1{s} + \int_1^\infty \psi(x) x^{s/2-1}\;dx$$ Join the two integrals back together. Join the two fractions with a common denominator.Take out a power of x as a divisor, to get the powers of x in the right form. $$ \xi(s) = \int_1^\infty \frac{\psi(x)}{x}(x^{(1-s)/2}+x^{s/2})\;dx - \frac1{s(1-s)}$$ which remains the same if you substitute 1-s for s so, $$ \xi(s) = \xi(1-s)$$