On a very well known identity

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 the identity:

$$\pi \cot \pi a = \lim_{N \rightarrow +\infty} \sum_{n=-N}^{N} \frac{1}{n+a}$$

Solution



First of all we observe that:

$$\begin{aligned} \sum_{k=-n}^{n}\dfrac{1}{k+a}&=\sum_{k=-n}^{0}\dfrac{1}{k+a}+\sum_{k=1}^{n}\dfrac{1}{k+a}\\&=\sum_{k=0}^{n}\dfrac{1}{a-k}+\sum_{k=1}^{n}\dfrac{1}{a+k}\\&=\left[\dfrac{1}{a}+\dfrac{1}{a-1}+...+\dfrac{1}{a-n}\right]+\left[\dfrac{1}{a+1}+\dfrac{1}{a+2}+\cdots+\dfrac{1}{a+n}\right]\\&=\dfrac{1}{a}+\dfrac{2\,a}{a^2-1}+\dfrac{2\,a}{a^2-4}+ \cdots+\dfrac{2\,a}{a^2-n^2}\\&=\dfrac{1}{a}+\sum_{k=1}^{n}\dfrac{2\,a}{a^2-k^2}\end{aligned}$$

So it suffices to prove that:

$$ \frac{1}{a}+\sum_{n=1}^{\infty}\frac{2a}{a^2-n^2}=\pi \cot \pi a$$

And this is where the Fourier series comes in.  We evaluate the coefficients.

$\displaystyle {\color{gray} \blacksquare} \;\; a_0= \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\, {\rm d} x = \frac{2\sin a\pi }{a\pi}$

$\displaystyle {\color{gray} \blacksquare} \;\; a_n =\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos nx \, {\rm d}x = \frac{2 (-1)^n a \sin \pi a}{\pi (a^2-n^2)}$

Then $b_n$ coefficients are actually zero since $\cos ax$ is even.

Hence:

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

as wanted.