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The exercise

Let $f$ be a $2\pi$ periodical function defined as $f(x)=\cos ax, \; |x|\leq \pi, \; a \notin \mathbb{Z}$. Expand $f$ in a Fourier series and prove that: $$\pi \cot \pi a = \sum_{n=-\infty}^{\infty} \frac{1}{n+a}, \; a \notin \mathbb{Z}$$

The Fourier series is pretty straight forward. By evaluating the coefficients one gets that:

$$\cos ax = \frac{\sin \pi a}{\pi a}+ \frac{2a \sin \pi a}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^n \cos nx}{(a-n)(a+n)} \overset{x=\pi}{\implies }\\ \overset{x=\pi}{\implies}\pi \cot \pi a =\frac{1}{a}+ 2a\sum_{n=1}^{\infty}\frac{1}{(a-n)(a+n)}$$

How do I derive the series from $-\infty$ to $\infty$? I cannot see how to manipulate the last series in order to get what I want.

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HINT:

Note that for the substitution $n \to -n$ the series is unchanged. Thus, the sum over non-zero integers is twice the sum over the positive integers only.

Can you finish now?

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  • $\begingroup$ Sure, thanks for you answer. $\endgroup$ – Tolaso Aug 4 '15 at 23:28
  • $\begingroup$ Pleased to hear! And you're most welcome. It was my pleasure. $\endgroup$ – Mark Viola Aug 4 '15 at 23:33
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Hint: Evaluate $$\frac1{a+n}+\frac1{a-n}\,.$$

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  • $\begingroup$ We can't split a convergent series into a sum of two divergent (harmonic like) series $\endgroup$ – Mark Viola Aug 4 '15 at 23:20
  • $\begingroup$ Well, yes, that's a good observation. Nevertheless, it rather says that interpreting the desired formula $\sum_{n=-\infty}^{\infty}\frac1{n+a}$ is not straightforward (if possible in any way). $\endgroup$ – Berci Aug 4 '15 at 23:24
  • $\begingroup$ I did not get your point. $\endgroup$ – Tolaso Aug 4 '15 at 23:31
  • $\begingroup$ @berci The series you wrote diverges. $\endgroup$ – Mark Viola Aug 4 '15 at 23:34
  • $\begingroup$ Well, how is this sum over $n=-\infty..\infty$ defined exactly? $\endgroup$ – Berci Aug 4 '15 at 23:45

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