A well known identity

[Tolaso J Kos]

Let \( f \) be a two \( 2\pi \) 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:

$$\pi \cot \pi a = \sum_{n=-\infty}^{\infty} \frac{1}{n+a}, \;\; a \notin \mathbb{Z} $$

[Papapetros Vaggelis]

Comment

Firstly, \(\displaystyle{\lim_{n\to +\infty}\dfrac{1}{n+a}=0}\) . Let \(\displaystyle{n\in\mathbb{N}}\) . Then,

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

and the sequence \(\displaystyle{\left(\sum_{k=1}^{n}\dfrac{2\,a}{a^2-k^2}\right)_{n\in\mathbb{N}}}\) converges.

Therefore, \(\displaystyle{\sum_{n=-\infty}^{\infty}\dfrac{1}{n+a}=\dfrac{1}{a}+\sum_{n=1}^{\infty}\dfrac{2\,a}{a^2-n^2}\,\,(\star)}\) .

Solution

The function \(\displaystyle{f}\) is \(\displaystyle{2\,\pi}\) periodical, continuous at \(\displaystyle{\left[-\pi,\pi\right]}\) and even in

\(\displaystyle{\left[-\pi,\pi\right]}\) .

Let's compute the \(\displaystyle{\rm{Fourier}}\) coefficients.

\(\displaystyle{\begin{aligned} a_0&=\dfrac{1}{\pi}\,\int_{-\pi}^{\pi}f(x)\,\mathrm{d}x\\&=\dfrac{1}{\pi}\,\int_{-\pi}^{\pi}\cos\,a\,x\,\mathrm{d}x\\&=\left[\dfrac{\sin\,a\,x}{a\,\pi}\right]_{-\pi}^{\pi}\\&=\dfrac{2\,\sin\,a\,\pi}{a\,\pi}\end{aligned}}\) .

\(\displaystyle{\begin{aligned} a_{n}&=\dfrac{1}{\pi}\,\int_{-\pi}^{\pi}f(x)\,\cos\,(n\,x)\,\mathrm{d}x\\&=\dfrac{2}{\pi}\,\int_{0}^{\pi}\cos\,(a\,x)\,\cos\,(n\,x)\,\mathrm{d}x\\&=\dfrac{1}{\pi}\,\int_{0}^{\pi}\left(\cos\,(a+n)\,x+\cos\,(a-n)\,x\right)\,\mathrm{d}x\\&=\dfrac{1}{\pi}\,\left[\dfrac{\sin\,(n+a)\,x}{n+a}+\dfrac{\sin\,(a-n)\,x}{a-n}\right]_{0}^{\pi}\\&=\dfrac{1}{\pi}\,\left(\dfrac{(-1)^{n}\,\sin\,a\,\pi}{n+a}+\dfrac{(-1)^{n}\,\sin\,a\,\pi}{a-n}\right)\end{aligned}}\)

\(\displaystyle{b_{n}=\dfrac{1}{\pi}\,\int_{-\pi}^{\pi}f(x)\,\sin\,n\,x\,\mathrm{d}x=0\,,\forall\,n\in\mathbb{N}}\) since :

\(\displaystyle{f(-x)\,\sin\,(n\,(-x))=f(x)\,(-\sin\,n\,x)=-f(x)\,\sin\,n\,x\,,\forall\,x\in\left[-\pi,\pi\right]\,,\forall\,n\in\mathbb{N}}\) .

Now, we can write :

\(\displaystyle{\begin{aligned} f(x)&=\cos\,a\,x\\&=\dfrac{a_0}{2}+\sum_{n=1}^{\infty}\left(a_{n}\,\cos\,n\,x+b_{n}\,\sin\,x\right)\\&=\dfrac{\sin\,a\,\pi}{a\,\pi}+\sum_{n=1}^{\infty}\left[\dfrac{1}{\pi}\,\left(\dfrac{(-1)^{n}\,\sin\,a\,\pi}{n+a}+\dfrac{(-1)^{n+1}\,\sin\,a\,\pi}{a-n}\right)\right]\,\cos\,n\,x\,\,,x\in\left[-\pi,\pi\right]\end{aligned}}\)

Setting \(\displaystyle{x=\pi}\) we get :

\(\displaystyle{\cos\,a\,\pi=\dfrac{\sin\,a\,\pi}{a\,\pi}+\sum_{n=1}^{\infty}\left[\dfrac{1}{\pi}\,\left(\dfrac{(-1)^{n}\,\sin\,a\,\pi}{n+a}+\dfrac{(-1)^{n}\,\sin\,a\,\pi}{a-n}\right)\right]\,\cos\,n\,\pi}\)

or

\(\displaystyle{\cos\,a\,\pi=\sin\,a\,\pi\,\left(\dfrac{1}{a\,\pi}+\sum_{n=1}^{\infty}\dfrac{1}{\pi}\,\left(\dfrac{(-1)^{n}}{n+a}+\dfrac{(-1)^n}{a-n}\right)\,(-1)^{n}\right)}\)

or

\(\displaystyle{\pi\,\dfrac{\cos\,a\,\pi}{\sin\,a\pi}=\dfrac{1}{a}+\sum_{n=1}^{\infty}\left(\dfrac{1}{n+a}+\dfrac{1}{a-n}\right)\implies}\)

\(\displaystyle{\implies \pi\,\cot\,a\,\pi=\dfrac{1}{a}+\sum_{n=1}^{\infty}\dfrac{2\,a}{a^2-n^2}}\)

and using the relation \(\displaystyle{(\star)}\) we have that :

\(\displaystyle{\pi\,\cot\,a\,\pi=\sum_{n=-\infty}^{\infty}\dfrac{1}{n+a}}\)

and the exercise comes to an end.

Note

\(\displaystyle{\sin\,a\,\pi=0\iff \exists\,k\in\mathbb{Z}: a\,\pi=k\,\pi\iff \exists\,k\in\mathbb{Z}: a=k\in\mathbb{Z}}\), a contradiction,

so : \(\displaystyle{\sin\,a\,\pi\neq 0}\) .