Finding $\sum_{n } \frac{1}{(n+a)^2}$ using Poisson Summation formula

Asked Feb 22 '23 at 14:09

Active Feb 23 '23 at 05:25

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This is an exercise from the book Fourier analysis: An Introduction, by E.Stein and R.Shakarchi.

Let $ g(x) = 1-|x|$ if $|x| \le 1$ and $g(x)=0$ otherwise. The exercise asks us to apply Poisson summation to the function $g$ to obtain : $$ \sum_{n=-\infty}^{\infty}\frac{1}{(n+a)^2}= \frac{\pi^2}{\sin^2(\pi a)}.$$

The Poisson summation formula given in text book is, $$\sum_{n \in \mathbb{Z}} f(x+n) = \sum_{n \in \mathbb{Z}}\hat{f}(n)\exp(2i\pi nx), $$ under appropriate conditions for $f$.

I was able to show that $\hat g(s) = \frac{\sin^2(\pi s)}{\pi^2 s^2}.$ If I put $f=g$ I get $\hat g (n) =0$ for non zero $n$ and $\hat g(0) = 1$. So RHS in above formula will be $1$ while the LHS will be of the form $ 1-|x+n_0| $ where $ n_0 $ depends on $x$.

I do not understand how to proceed here.

edited Feb 23 '23 at 05:25

asked Feb 22 '23 at 14:09

Eloon_Mask_P

1

Does this answer your question? Prove $\sum_{n= -\infty}^{\infty} \frac{1}{(t+n)^2} = \frac{\pi^2 }{\sin^2(\pi t)}$ using $f(x)=\max (1-|x|,0)$ and Poisson summation formula – Anne Bauval Feb 22 '23 at 14:16
The answer to the question in the referred post is unclear. Specifically the author claims that g is $\hat g$s own Fourier transform which is not elaborated. Some one had asked same in the comment section and it was not answered. – Eloon_Mask_P Feb 22 '23 at 14:22

Finding $\sum_{n } \frac{1}{(n+a)^2}$ using Poisson Summation formula

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