Exercise 4.6 in Gasquet and Witomski (1999) [1]: Find the Fourier expression of the function $f$ with period 2 defined on $[-1,1)$ for $z\in\mathbb{C}\setminus\mathbb{Z}$ by
$$
f(t) = \exp\left\{i\pi zt\right\}\,.
$$
Deduce from Parseval's equality the relation:
$$
\color{blue}{
\frac{\pi^2}{\sin^2(\pi x)} = \sum_{n=-\infty}^\infty\frac{1}{(x-n)^2}\,,\qquad\hbox{for all }x\in\mathbb{R}\setminus\mathbb{Z}\,. \qquad\qquad\qquad\qquad (*)}
$$
This is what I have done so far:
Compute Fourier coefficients:
$$
\begin{align}
c_n =& \frac{1}{2}\int_{-1}^1 f(t)\exp\left\{-i\pi nt\right\}\,dt =
\frac{1}{2}\int_{-1}^1 \exp\left\{i\pi(z-n)t\right\}\,dt \\
c_n =& \left.\frac{i}{2\pi(z-n)}\exp\left\{i\pi(z-n)t\right\}\right|_{t=-1}^{t=1}\\
c_n =& \frac{1}{\pi(z-n)}\sin\left(\pi(z-n)\right)\,.
\end{align}
$$
Parseval's equality:
$$
\frac{1}{2}\left\lVert f\right\rVert_2^2 = \frac{1}{2}\int_{-1}^1|\exp\{i\pi zt\}|^2dt = 1 = \sum_{n=-\infty}^\infty |c_n|^2 = \sum_{n=-\infty}^\infty \frac{1}{\pi^2(z-n)^2}\sin^2\left(\pi(z-n)\right)\,.
$$
So this is the closest I got to expression $(*)$ above:
$$
\color{red}{\pi^2 = \sum_{n=-\infty}^\infty \frac{1}{(z-n)^2}\sin^2\left(\pi(z-n)\right)\,.}
$$
I tried defining explicity $z$ as a complex number, $z=x+iy$, $x,y\in\mathbb{R}$ and using relations such as $\sin(ix)=i\sinh(x)$, but could't arrive to the relation stated in the heading.
[1] Gasquet, C. and Witomski, P. (1999). Fourier Analysis and Applications: Filtering, Numerical Computation, Wavelets. Springer. https://link.springer.com/book/10.1007/978-1-4612-1598-1.
Best Answer
$\sin (y-n\pi)=\pm \sin y$ for any real number $y$. So $\sin ^{2}(\pi (x-n))=\sin^{2}(\pi x)$. Now divide by $\sin^{2}(\pi x)$ which is permissible if $x$ is not an integer.