# A proof of a known identity using Fourier series

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.

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?

• Sure, thanks for you answer. – Tolaso Aug 4 '15 at 23:28
• Pleased to hear! And you're most welcome. It was my pleasure. – Mark Viola Aug 4 '15 at 23:33

Hint: Evaluate $$\frac1{a+n}+\frac1{a-n}\,.$$

• We can't split a convergent series into a sum of two divergent (harmonic like) series – Mark Viola Aug 4 '15 at 23:20
• 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). – Berci Aug 4 '15 at 23:24
• I did not get your point. – Tolaso Aug 4 '15 at 23:31
• @berci The series you wrote diverges. – Mark Viola Aug 4 '15 at 23:34
• Well, how is this sum over $n=-\infty..\infty$ defined exactly? – Berci Aug 4 '15 at 23:45