For$\alpha \in \mathbb{R} \backslash \mathbb{Z}$, consider the fnunction $[0,2\pi] \to \mathbb{C} : x \mapsto \frac{\pi}{\sin \pi \alpha} e^{i(\pi - x)\alpha}$, and prove that $\sum_{n=-\infty}^\infty \frac{1}{(n+\alpha)^2} = \frac{\pi^2}{(\sin \pi \alpha)^2}$. Recall that $\{x \mapsto (2\pi)^{-1/2} e^{inx} : n \in \mathbb{Z}\}$ is an orthonormal basis for $L^2_\mathbb{C}[0,2\pi]$
My Attempt:
We use Parseval's Theorem, the fact that $\|f\|^2 = \sum_{n \in \mathbb{Z}} |\langle f,e_n \rangle|^2$, with the inner product norm on $L^2_\mathbb{C}[0,2\pi]$. Firstly, we compute the left-hand side. We have \begin{align*} \|f\|^2 & = \int_0^{2\pi} f(x)^2 dx = \frac{\pi^2}{(\sin \pi \alpha)^2} \int_0^{2\pi} e^{2i(\pi - x)\alpha} dx \\ & = \frac{\pi^2}{(\sin \pi \alpha)^2} e^{2i\pi\alpha} \int_0^{2\pi} e^{-2ix \alpha} dx = \frac{\pi^2}{(\sin \pi \alpha)^2} e^{2i\pi\alpha} \left[ -\frac{1}{2i\alpha}e^{-2ix \alpha} \right]_0^{2\pi} \\ & = \frac{\pi^2}{(\sin \pi \alpha)^2} e^{2i\pi\alpha} \left( -\frac{1}{2i\alpha}e^{-4i\pi \alpha} + \frac{1}{2i\alpha} \right) = \frac{\pi^2}{(\sin \pi \alpha)^2} e^{2i\pi\alpha} \left( \frac{1 - e^{-4i\pi \alpha}}{2i\alpha} \right) \\ & = \frac{\pi^2}{(\sin \pi \alpha)^2} \left( \frac{e^{2i\pi\alpha} - e^{-2i\pi \alpha}}{2i\alpha} \right) = \frac{\pi^2}{(\sin \pi \alpha)^2} \left( \frac{\sin (2\pi \alpha)}{\alpha} \right) = \frac{2 \pi^2 \cos (\pi \alpha)}{\sin (\pi \alpha) \alpha}. \end{align*} Here we use the trigonometric identities $$ \sin \beta = \frac{e^{i\beta} - e^{-i\beta}}{2i}, \ \textrm{ and } \ \sin(2\beta) = 2\sin(\beta)\cos(\beta). $$ On the left-hand side we have \begin{align*} \langle f,e_n \rangle & = \int_0^{2\pi} \frac{1}{\sqrt{2\pi}} e^{inx} \frac{\pi}{\sin \pi \alpha} e^{i(\pi - x)\alpha} dx = \frac{\pi}{\sin (\pi \alpha) \sqrt{2\pi}} \int_0^{2\pi} e^{inx + i(\pi - x)\alpha} dx \\ & = \frac{\pi}{\sin (\pi \alpha) \sqrt{2\pi}} e^{i \pi \alpha} \int_0^{2\pi} e^{(in- i\alpha)x} dx = \frac{\pi}{\sin (\pi \alpha) \sqrt{2\pi}} e^{i \pi \alpha} \left[ \frac{1}{in - i\alpha} e^{(in- i\alpha)x} \right]_0^{2\pi} \\ & = \frac{\pi}{\sin (\pi \alpha) \sqrt{2\pi}} e^{i \pi \alpha} \left( \frac{1}{in - i\alpha} e^{(in- i\alpha)2\pi} - \frac{1}{in - i\alpha} \right) \\ & = \frac{\pi}{\sin (\pi \alpha) \sqrt{2\pi}} e^{i \pi \alpha} \left( \frac{e^{i(n-\alpha)2\pi} - 1}{i(n - \alpha)} \right) = \frac{\pi}{\sin (\pi \alpha) \sqrt{2\pi}} \left( \frac{-(e^{i \pi \alpha} - e^{-i\pi\alpha})}{i(n - \alpha)} \right) \\ & = \frac{\pi}{\sin (\pi \alpha) \sqrt{2\pi}} \frac{-2 \sin(\pi \alpha)}{n - \alpha} = \frac{-2\pi}{(n-\alpha)\sqrt{2\pi}} \end{align*} Hence, we have that $$ |\langle f,e_n \rangle|^2 = \frac{2\pi}{(n-\alpha)^2}. $$
This is what I tried, things don't seem to add up to the desired outcome and I can't find any mistakes. Anyone has suggestions? I find the computations I made rather laborious, perhaps there are some tricks to do this simpler? I'm not so familiar with trigonometry and complex integration. Thanks in advance.