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.