# 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?

-
Any textbook in complex analysis has at least one proof written out or presented as an exercise. –  zyx Apr 12 '12 at 0:00
@zyx I only have "Complex Variables and Applications, Second Edition" by Ruel Churchill. Its about 40 years old now. It contains no proof. –  Argon Apr 12 '12 at 0:05
That's surprising, I was thinking of Churchill with its computational style of problems as exactly the kind of book that would have it. –  zyx Apr 12 '12 at 6:39
@Argon: I don't know about the second edition, but exercise 14 in section 57 of the fifth edition has an equivalent result (alternating sum = $\pi^2/12$). –  Hans Lundmark Apr 15 '12 at 9:52

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 $\color{green}{-\frac{\pi^2}{3}}+2\cdot\color{red}{\frac{\pi^2}{6}}=\color{blue}{0}$).

-

All the proofs I know of are in Kalman's article. There is one that use complex analysis (residues, to be concrete).

-