I'd like to prove
$$\sum_{n= -\infty}^{\infty} \frac{1}{(t+n)^2} = \frac{\pi^2 }{\sin^2(\pi t)}$$
by using the Poisson summation formula. There is a way to do it by firstly taking the Fourier transform of $f(x) = 1 – |x|$ for $|x|\le1$ and $f(x) = 0$ for $|x|>1$. There's a next step which I can't work out, and then you apply the Poisson summation formula.
I get the Fourier transform of $f(x)$ as $\frac{2(1-\cos y)}{y^2}$ but can't work out the next step so I can apply Poisson summation formula.
There's also a comment in a previous post back in April by user Lost about having done it. But I can't find any tips. I know it's possible.
Thanks in advance.
Best Answer
Here is an almost identical post:
$\sum_{n=-\infty}^{\infty}\frac{1}{(n+\alpha)^2}=\frac{\pi^2}{(\sin \pi \alpha)^2}$?
I think there is a proof for this formula in Ahlfors independent of Possion summation formula(page 189 or 188) using elementary methods, which you can look up.
Here is a computational proof. We let $$ f(x)=1-|x|, |x|\le 1; f(x)=0, |x|>1 $$ as you defined. It is easy to calculate the Fourier transform is given by $$ \tilde{f}(\xi)=(\frac{\sin[\pi \xi]}{\pi \xi})^{2} $$
I will leave the details to you. Then by summation formula we have $$ \sum^{\infty}_{-\infty}f(x+n)=\sum^{\infty}_{-\infty}\tilde{f}(n)e^{2\pi inx} $$ We want to apply the formula using the fact that $f$ is $\tilde{f}$'s own Fourier transform. This holds since $f$ is in Schwarz class. Then we have $$ \sum^{\infty}_{-\infty}(\frac{\sin[\pi(x+n)]}{\pi (x+n)})^{2}=\sum^{\infty}_{-\infty}f(n)e^{2\pi in x} $$ We know that $\sin[\pi x+n\pi]^{2}=\sin[\pi x]^{2}$, therefore the left hand side can be factorized as $$ \frac{\sin[\pi x]^{2}}{\pi^{2}}*\sum^{\infty}_{-\infty}\frac{1}{(x+n)^{2}} $$ While the right hand side is $f(0)=1$, because all other values simply vanish. Therefore we have $$ \sum^{\infty}_{-\infty}\frac{1}{(x+n)^{2}}=\frac{\pi^{2}}{\sin[\pi x]^{2}} $$ as desired.