# Proving the relation: $\sum_{n\in\Bbb Z} \frac{2a}{a^2+4 \pi^2 n^2} = \sum_{n\in\Bbb Z} e^{-a \left\lvert n \right\rvert}$

I came across this problem in an exercise for Fourier analysis. I tried solving just $$e^{−a|n|}$$ to get the Fourier transform of a similar form as seen on the LHS because it looked familiar. But in that case in the denominator, I get $$f^2$$ (where $$f$$ is the frequency as given in the equation below) instead of $$n^2$$ (I'm integrating with respect to n). I used the following relation:

Fourier transform of $$x(n)$$: $$X(f)=\int_{-\infty}^\infty x(n)e^{-2\pi jfn}dn$$

Solving this gives: $$\frac{2a}{a^2+4 \pi^2 f^2}$$

I think my approach is wrong here.

• in what context are you seeing this problem? what have you tried? – gt6989b Feb 26 '19 at 16:03
• I tried solving just $e^{-a \left\lvert n \right\rvert}$ to get the Fourier transform of a similar form as seen on the LHS, but in the denominator, it comes out to be $f^2$ instead of $n^2$ (I'm integrating with respect to n). So I'm at a complete loss in this problem. Am I not even looking at it right? – Swapnil Feb 26 '19 at 16:07
• Welcome to MSE. What have you tried? Are you stuck on some concept? Is your calculation wrong? What do you need help with? – Andrei Feb 26 '19 at 16:07
• My guess is Poisson Summation. Related: math.stackexchange.com/questions/3123991/… – mrtaurho Feb 26 '19 at 16:39
• If you "got it" please fix the post as quid suggested in meta. Having non-fixed badly received question may hamper your use of the site later. We are somewhat strict about a few things. The CEO of StackExchange compared some of the rules that are necessary here to the rules around the Burning Man -event. Newcomers reportedly find those rather unintuitive :-) – Jyrki Lahtonen Feb 26 '19 at 18:14