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

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.

Answer 2

You could also consider the function $$g(x) = \begin{cases} 1-|x|, & \text{if } |x|\leq1 \\ 0, & \text{otherwise} \\ \end{cases}$$ whose Fourier transform is easily proven to be $\hat{g}(\xi)=\left(\dfrac{\sin(\pi\xi)}{\pi\xi}\right)^2$.

Using Poisson's summation formula, and the function $$f(x) = \begin{cases} (1-|x|)e^{2\pi iax}, & \text{if } |x|\leq1 \\ 0, & \text{otherwise} \\ \end{cases}$$ whose Fourier transform is $\hat{g}(\xi-a)$ the result follows naturally as shown next:

$$\sum\limits_{n\in\mathbb{Z}}\hat{f}(n)=\sum\limits_{n\in\mathbb{Z}}f(n)\Longrightarrow\sum\limits_{n\in\mathbb{Z}}\hat{g}(n-a)=\sum\limits_{n\in\mathbb{Z}}f(n)=f(0)=1,$$

since $f(n)=0 \quad \forall n\neq0$.

We have that

$$\sum\limits_{n\in\mathbb{Z}}\left(\dfrac{\sin(\pi(n-a))}{\pi(n-a)}\right)^2=1$$

Using the trigonometric property for sin of subtraction, it follows that

$$\pi^2=\sum\limits_{n\in\mathbb{Z}}\left(\dfrac{\sin(\pi a)}{n-a}\right)^2$$

and finally,

$$\dfrac{\pi^2}{\sin^2(\pi a)}=\sum\limits_{n\in\mathbb{Z}}\dfrac{1}{(n+a)^2}$$