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

Question

The function in question is $f(x)= \frac{\pi}{\sin\pi\alpha}e^{i(\pi-x)\alpha}$ where $\alpha$ is not an integer.

With a bit of elbow grease, one can find that the Fourier series for $f$ is $\sum_{n=-\infty}^{\infty} \frac{e^{inx}}{n+\alpha}$, so we apply Parseval's and try to work out $||f||^2$ to get the identity.

So $$||f||^2 = \frac{1}{2\pi}\int_{0}^{2\pi}\frac{\pi^2}{\sin^2\pi\alpha}e^{2i(\pi-x)\alpha} dx$$ And some more elbow grease, I got to $$||f||^2=\frac{\pi^2 \sin2\pi\alpha}{2\pi\alpha\sin^2\pi\alpha}$$ Which seems really close as there seems to be something close to a squeeze theorem going on, but alas there are no limits to make the excess junk go away. How can I simplify further to get the desired $||f||^2=\frac{\pi^2}{sin^2\pi\alpha}$? Or perhaps I missed something along the way?

Answer 1

You just need to take the square modulus of $f$, not the square. Indeed,

\begin{equation} \vert f(x)\vert^2=\frac{\pi^2}{(\sin \pi\alpha)^2} \vert e^{i(\pi-x)\alpha}\vert^2=\frac{\pi^2}{(\sin \pi\alpha)^2}, \end{equation}

so integrating with the correct normalization gives the desired formula for $\|f\|^2$.

Answer 2

You could do it without integration.

Let $$S_p=\sum_{n=-p}^{p} \frac{1}{(n+\alpha)^2}$$ $$S_p =-\frac{1}{\alpha ^2}+\psi ^{(1)}(-\alpha )+\psi ^{(1)}(\alpha )-\psi ^{(1)}(p-\alpha +1)-\psi ^{(1)}(p+\alpha +1)$$ Using the reflection formula of the polygamma function $$\psi ^{(1)}(-\alpha )+\psi ^{(1)}(\alpha )=\frac{1}{\alpha ^2}+\pi ^2 \csc ^2(\pi \alpha )$$ which makes $$S_p=\pi ^2 \csc ^2(\pi \alpha )-\psi ^{(1)}(p-\alpha +1)-\psi ^{(1)}(p+\alpha +1)$$ Now, using the asymptotics $$\psi ^{(1)}(q)=\frac{1}{q}+\frac{1}{2 q^2}+\frac{1}{6 q^3}+O\left(\frac{1}{q^5}\right)$$ apply it twice an continue with Taylor series to get $$S_p=\pi ^2 \csc ^2(\pi \alpha )-\left(\frac 2{p}+\frac 1{p^2}+\frac{6 \alpha ^2+1}{3 p^3}+O\left(\frac{1}{p^4}\right) \right)$$