= Solution
Evaluate the absolutely convergent <complex Fourier series> at $x=\pi$. The factors $(-1)^ne^{in\pi}$ equal one, and the $n=0$ term is $1/\alpha^2$. Pairing positive and negative integers gives
$$
\cos(\pi\alpha)=\frac{\alpha\sin(\pi\alpha)}\pi
\left[\frac1{\alpha^2}-2\sum_{n=1}^\infty\frac1{n^2-\alpha^2}\right].
$$
Since $\sin(\pi\alpha)\ne0$ for nonintegral complex $\alpha$, rearranging proves
$$
\boxed{\sum_{n=1}^\infty\frac1{n^2-\alpha^2}
=\frac1{2\alpha^2}-\frac{\pi}{2\alpha\tan(\pi\alpha)}.}
$$
The last term means $\pi\cot(\pi\alpha)/(2\alpha)$, so half-integer arguments are included by the finite cotangent value zero.
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