5
$\begingroup$

$$ \pi^{-s/2}\zeta(s)\Gamma(s/2)=\pi^{-(1-s)/2}\zeta(1-s)\Gamma((1-s)/2) $$ (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

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

$\endgroup$
0
5
$\begingroup$

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 $\xi(s)$.

$\endgroup$
2
$\begingroup$

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).

$\endgroup$
1
$\begingroup$

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, $$ \xi(s) = \xi(1-s)$$

$\endgroup$
8
  • $\begingroup$ All those details make it impossible to read. You only need to show the main steps.. $\endgroup$ – reuns Aug 20 '17 at 22:02
  • $\begingroup$ Sorry, I appreciate that this may be true for you, but not for me. In general I would agree that less steps is better, but I was confused without the other steps. The sequence of splitting, substitution, usub, integration and merging is fairly precises. With your experience you can probably see and understand this with some insight, but for me, I need to see it all layed out, to start to understand what is going on. What I am looking for is some understanding of why the result holds which still eludes me, $\endgroup$ – Peter Driscoll Aug 20 '17 at 22:13
  • $\begingroup$ I check all the steps but when explaining to someone I only show the main steps and give some hints on how to obtain them. See any question and answer on MSE. $\endgroup$ – reuns Aug 20 '17 at 22:15
  • $\begingroup$ This functional equation is a direct consequence of $e^{-\pi x^2}$ being its own Fourier transform as well as $\sum_n \delta(x-n)$ the Dirac comb distribution $\endgroup$ – reuns Aug 20 '17 at 22:18
  • $\begingroup$ Sure. In general, that is the right way to do it. But in this case I was confused by the derivation. It is so weird the way it all works out. $\endgroup$ – Peter Driscoll Aug 20 '17 at 22:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.