# $\sum_{n=-\infty}^{\infty}\frac{1}{(n+\alpha)^2}=\frac{\pi^2}{(\sin \pi \alpha)^2}$?

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!

## 1 Answer

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

• This one is better done with a rectangular contour aligned with half-integers (real and imaginary). – Marko Riedel Dec 30 '13 at 21:15