[Math] $\sum_{n=-\infty}^{\infty}\frac{1}{(n+\alpha)^2}=\frac{\pi^2}{(\sin \pi \alpha)^2}$

Question

Now we want to prove:
$$\sum_{n=-\infty}^{\infty}\frac{1}{(n+\alpha)^2}=\frac{\pi^2}{(\sin \pi \alpha)^2}$$
$\alpha$ >0 and not an integer.
According to poisson summation formula
$$\sum_{n=-\infty}^{\infty}f(x+n)=\sum_{n=-\infty}^{\infty}\hat{f}(n)e^{2\pi inx}.$$
So, if we let $f(x)=\frac{1}{x^2}$, then
$$\sum_{n=-\infty}^{\infty}f(x+n)=\sum_{n=-\infty}^{\infty}\frac{1}{(n+x)^2}$$
Therefore, if we can prove that
$$\sum_{n=-\infty}^{\infty}\left(\int_{-\infty}^{\infty}\frac{1}{x^2} e^{-2\pi inx}dx e^{2\pi inx}\right)=\frac{\pi^2}{(\sin \pi x)^2}$$
Then it will be done.
But I don't know how to prove
$$\sum_{n=-\infty}^{\infty}\left(\int_{-\infty}^{\infty}\frac{1}{x^2} e^{-2\pi inx}dx e^{2\pi inx}\right)=\frac{\pi^2}{(\sin \pi x)^2}$$
Who could give me some hints? Thanks!

Answer 1

The proper technique is documented at MSE 112161 and at MSE 3056578.