Compute sum (possibly using Parseval's formula)

[psie]

Homework Statement

Prove the formula
$$\sum _{k=0}^{\infty }\frac{\left(-1\right)^k}{\left(2k+1\right)\left(\left(2k+1\right)^2-\alpha ^2\right)}=\frac{\pi }{4\alpha ^2}\left(\frac{1}{\cos \left(\frac{\alpha \pi \ }{2}\right)}-1\right),$$ where $\alpha\notin \mathbb Z$. (Hint: study the series established in the previous exercise on the interval $(0,\pi/2)$.)

Relevant Equations

See the previous exercise below.

Previously I worked the following exercise:

Determine the the Fourier series of $\cos{(\alpha t)}$ ($|t|\leq\pi$), where $\alpha$ is a complex number but not an integer. Use this to verify $$\pi\cot{( \alpha \pi)}=\sum_{k=-\infty}^\infty \frac{1}{\alpha-k}.$$

The Fourier coefficients of $\cos{(\alpha t)}$ ($|t|\leq\pi$) are \begin{align} a_0&=\frac{2\sin(\alpha\pi)}{\alpha\pi}, \nonumber \\ a_n&=\frac{2\alpha\sin(\alpha\pi)(-1)^{n+1}}{\pi(k^2-\alpha^2)} \quad n\geq 1. \nonumber \end{align} So
$$\cos{(\alpha t)}=\frac{\sin(\alpha \pi)}{\alpha \pi }-\sum_{k=1}^\infty \frac{2\alpha \sin(\alpha \pi)(-1)^k}{\pi(k^2-\alpha^2)}\cos(kt).$$
If you plug in $t=\pi$ in the previous equation and rearrange a bit, one can arrive at
$$\pi\cot{( \alpha \pi)}=\sum_{k=-\infty}^\infty \frac{1}{\alpha-k}.$$
But I'm stuck at verifying the formula in the homework statement. This problems appears in a section that introduces the Parseval formula, $$\frac1{\pi}\int_{\mathbb T}|f(t)|^2dt=\frac12|a_0|^2+\sum_{n=1}^\infty |a_n|^2,$$ so I'm pretty sure it has something to do with this formula, but the fact that $\alpha$ is complex confuses me. I'm not sure my approach is right either. Does anyone see any pattern here?

 

[pasmith]

If you have complex quantities in the result, then Parseval's Theorem will not assist you.

The question also suggests that you consider this series on $(0, \frac \pi 2)$ which is only a quarter period, so Parseval's Theorem cannot be applied.

Note that the sum you want to find is of terms of magnitude $1/(n(n^2 - \alpha^2))$ taken over odd positive $n$. This is similar to the dependence of the coeffcients in the fourier series, except you need an additional factor of $n$ in the denominator. You can get that by integrating the fourier series - which if done over $(0, \frac \pi 2)$ will also kill off the terms in even $n$.