The identity I want to prove is the following (from Stein's book, an introduction to Fourier analysis):

$$\pi^{-s/2} \Gamma(s/2) \zeta (s)=\frac{1}{2} \int_{0}^{\infty}t^{\frac{s}{2}-1}(v(t)-1)dt$$

for $s>1$, where $v(t)$ is the theta function and $\Gamma(s)$ is the gamma function.

So when I manipulate the LHS of the equation I get the following:


The thing is What can I do next, and How do I am going to get the negative part of the theta function?.

Can someone help me to prove this identity please?, Thanks a lot in advance :)

Theta function:

$$\nu(s)=\sum_{n=-\infty}^{\infty}e^{-\pi n^{2}s}$$

  • $\begingroup$ Is the one that goes with the sum of the exponential to the -$\pi (n^{2})s$ from -infinity to infinity $\endgroup$ – user162343 Aug 25 '15 at 13:58

To start with let's write (following the wikipedia convention) $$\nu(t):=\vartheta(0,it)=\sum_{n=-\infty}^{\infty}e^{-\pi n^2 t }=\sum_{n=-\infty}^{-1}e^{-\pi n^2 t }+1+\sum_{n=1}^{\infty}e^{-\pi n^2 t }=2\sum_{n=1}^{\infty}e^{-\pi n^2 t }+1$$

Therefore: $$ I=\frac{1}{2}\int_{0}^{\infty}dt \left(\nu(t)-1\right)t^{s/2-1}=\sum_{n=1}^\infty\int_{0}^{\infty}dt e^{-\pi n^2 t }t^{s/2-1} $$

The integral is easily be calculated in terms of $\Gamma$-functions (substituting $\pi n^2 t \rightarrow x$ and use the defintion of $\Gamma$ )

and we end up with

$$ I=\pi^{-s/2}\Gamma(s/2)\sum_{n=1}^\infty\frac{1}{n^s}=\pi^{-s/2}\Gamma(s/2)\zeta(s) $$ Q.E.D.

| cite | improve this answer | |

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.