Compute sum (possibly using Parseval's formula)

  • #1
psie
108
10
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?
 
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  • #2
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 [itex](0, \frac \pi 2)[/itex] 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 [itex]1/(n(n^2 - \alpha^2))[/itex] taken over odd positive [itex]n[/itex]. This is similar to the dependence of the coeffcients in the fourier series, except you need an additional factor of [itex]n[/itex] in the denominator. You can get that by integrating the fourier series - which if done over [itex](0, \frac \pi 2)[/itex] will also kill off the terms in even [itex]n[/itex].
 
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