sequences and series - Complex Analysis Solution to the Basel Problem ($sum_{k=1}^infty frac{1}{k^2}$)

Most of us are aware of the famous "Basel Problem":

$$\sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}$$

I remember reading an elegant proof for this using complex numbers to help find the value of the sum. I tried finding it again to no avail. Does anyone know a complex number proof for the solution of the Basel Problem?

Answer

The most straightforward way I know is to consider the contour integral
$$
\frac{1}{2\pi i}\oint\pi\cot(\pi z)\frac{1}{z^2}\mathrm{d}z\tag{1}
$$
around circles whose radii are $\frac12$ off an integer.

The function $\pi\cot(\pi z)$ has residue $1$ at every integer. Thus the integral in $(1)$ equals the residue of $\pi\cot(\pi z)\dfrac{1}{z^2}$ at $z=0$ plus twice the sum in question (one for the positive integers and one for the negative integers).

The integral in $(1)$ tends to $\color{blue}{0}$ as the radius goes to $\infty$.

The Laurent expansion of $\pi\cot(\pi z)\dfrac{1}{z^2}$ at $z=0$ is
$$
\frac{1}{z^3}-\frac{\pi^2}{3z}-\frac{\pi^4z}{45}-\frac{2\pi^6z^3}{945}-\dots\tag{2}
$$
The only term that contributes to the residue at $z=0$ is the $\dfrac1z$ term. That is, the residue at $z=0$ of $(2)$ is $\color{green}{-\frac{\pi^2}{3}}$. Thus, the sum in question must be $\color{red}{\frac{\pi^2}{6}}$ (so that $\color{green}{-\frac{\pi^2}{3}}+2\cdot\color{red}{\frac{\pi^2}{6}}=\color{blue}{0}$).