Let $f(x) = e^{i\alpha x}, x\in(-\pi,\pi)$ where $\alpha$ not an integer and an orthogonal basis $\{e^{ikx}\}$, show that
$$\sum_{n\in\mathbb{Z}} \frac{1}{(n+\alpha)^2} = \frac{\pi^2}{\sin^2(\alpha\pi)}$$
using Parseval's identity i.e. $\sum_{n\in\mathbb{Z}} \vert\langle f, e_n\rangle\vert^2 = \Vert f\Vert^2$ where $\{e_n\}$ is an orthogonal basis.
Naturally, I first computed the Fourier coefficient of $f$
$$\hat{f}(n) = \frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-inx}dx = \frac{1}{\pi (\alpha – n)}\sin((\alpha – n)\pi),$$
which yields the Fourier series
$$f(x) = \frac{1}{\pi}\sum_{n\in\mathbb{Z}} \frac{\sin((\alpha – n)\pi)}{\alpha – n}e^{inx}.$$
Computing $\langle f, e^{ikx}\rangle$ we get
\begin{align}
\langle f, e^{ikx}\rangle = \langle e^{i\alpha x}, e^{ikx}\rangle &= \langle \frac{1}{\pi}\sum_{n\in\mathbb{Z}} \frac{\sin((\alpha – n)\pi)}{\alpha – n}e^{inx}, e^{ikx}\rangle\\
&= \frac{1}{\pi} \sum_{n\in\mathbb{Z}} \frac{1}{\alpha – n}\langle \sin((\alpha – n)\pi)e^{inx}, e^{ikx}\rangle\\
&= \frac{1}{\pi} \sum_{n\in\mathbb{Z}} \frac{\sin((\alpha – n)\pi)}{\alpha – n}\langle e^{inx}, e^{ikx}\rangle
\end{align}
where
$$\langle e^{inx}, e^{ikx}\rangle = \int_{-\pi}^\pi e^{inx}e^{-ikx} dx = \begin{cases}2\pi, \text{ if }n=k\\ 0, \text{ if } n\neq k\end{cases}$$
Thus if $n=k$,
\begin{align}
\langle e^{i\alpha x},e^{ikx}\rangle = \frac{2\sin((\alpha – n)\pi)}{\alpha -n}
\end{align}
On the other hand,
$$\Vert f\Vert = \int_{-\pi}^\pi e^{i\alpha x}e^{-i\alpha x}dx = 2\pi\implies \Vert f\Vert^2 = 4\pi^2$$.
Therefore by Parseval,
\begin{align}
\sum_{n\in\mathbb{Z}} \vert\langle e^{i\alpha x}, e^{ikx}\rangle\vert^2 &= \sum_{n\in\mathbb{Z}}\left\vert\frac{2\sin((\alpha – n)\pi)}{\alpha -n}\right\vert^2\\
&= \sum_{n\in\mathbb{Z}}\frac{4\sin^2((\alpha – n)\pi)}{(\alpha – n)^2}\\
&= \Vert f\Vert^2 = 4\pi^2.
\end{align}
Thus
$$\sum_{n\in\mathbb{Z}}\frac{\sin^2((\alpha – n)\pi)}{(\alpha – n)^2} = \pi^2,$$
which is not the desired result. How do I resolve this? please point out the mistakes I have made. Thanks!
Best Answer
Notice that: $$(\sin((a-n)π))^{2}=(\sin(aπ)\cos(-nπ)+\cos(aπ)\sin(-nπ))^{2}$$
Now $\sin(-nπ)=0$ and $\cos(-nπ)= (-1)^{n}$, for all $n\in\mathbb{Z}$.
So you are left with: $$(\sin((a-n)π))^{2}=((-1)^{n}\sin(aπ))^{2}=\sin^{2}(aπ)$$ Dividing both sides of the equality you've arrived at with $\sin^{2}(aπ)$ gives the desired result.