How did Euler prove this identity? While studying Fourier analysis last semester, I saw an interesting identity:
$$\sum_{n=1}^{\infty}\frac{1}{n^2-\alpha^2}=\frac{1}{2\alpha^2}-\frac{\pi}{2\alpha\tan\pi\alpha}$$
whenever $\alpha \in \mathbb{C}\setminus \mathbb{Z}$, which I learned two proofs using Fourier series and residue calculus.
More explicitly, we can deduce the theorem using Fourier series of $f(\theta)=e^{i(\pi - \theta)\alpha}$ on $[0,2\pi]$ or contour integral of the function $g(z)=\frac{\pi}{(z^2-\alpha^2)\tan\pi z}$ along large circles.
But these techniques, as long as I know, wasn't fully developed at Euler's time. 
So what was Euler's method to prove this identity? Is there any proof at elementary level?
 A: For the partial fraction decomposition of the cotangent
$$\pi \cot \pi z = \frac{1}{z} + \sum_{n\ge 1}\left( \frac{1}{z-n} + \frac{1}{z+n}\right) = \lim_{k\to\infty} \sum_{n=-k}^k \frac{1}{z-n}$$
hence
\begin{align*}
\sum_{n=1}^\infty \frac{1}{n^2-\alpha ^2}
&= \lim_{k\to\infty}\sum_{n=1}^k \frac{1}{n^2-\alpha^2}\\
&= -\frac{1}{2\alpha}\lim_{k\to\infty} \sum_{n=1}^k \frac{1}{\alpha-n} + \frac{1}{\alpha+n}\\
&= -\frac{1}{2\alpha}\lim_{k\to\infty} \left(-\frac{1}{\alpha} +\sum_{n=-k}^k  \frac{1}{\alpha-n}\right)\\
&= -\frac{1}{2\alpha}\left(\pi \cot \pi \alpha - \frac{1}{\alpha}\right)\\
&= \frac{1-\pi \alpha\cot \pi \alpha}{2\alpha^2}\\
&=\frac{1}{2\alpha^2}-\frac{\pi}{2\alpha\tan\pi\alpha}
\end{align*}
A: According to Wikipedia (https://en.wikipedia.org/wiki/Basel_problem),
 Euler was the first to give a  representation of the sine function as an infinite product:
$$(*) \hspace{2cm}\sin (\pi \alpha)=\pi \alpha \prod\limits_{n=1}^{\infty}(\frac{n^2-\alpha^2}{n^2}),$$
which was formally proved by Weierstrass about 100 years later.
Now taking "$\ln$" on by sides of (*) gives
$$\ln(\sin (\pi \alpha))=\ln(\pi \alpha)+ \sum \limits_{n=1}^{\infty} \ln (\frac{n^2-\alpha^2}{n^2}),$$
and after taking derivatives on both sides we arrive at 
$$\sum_{n=1}^{\infty}\frac{1}{n^2-\alpha^2}=\frac{1}{2\alpha^2}-\frac{\pi}{2\alpha\tan\pi\alpha}.$$
