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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:

$$\int_{0}^{\infty}e^{-t}t^{\frac{s}{2}-1}(\sum_{n=1}^{\infty}(\frac{1}{\pi^{-s/2}n^s}))dt$$

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:

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

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  • $\begingroup$ 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
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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.

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