Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|cite|improve this question
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

3 Answers 3

up vote 5 down vote accepted

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

share|cite|improve this answer

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

share|cite|improve this answer

Although there are so many different contour integrals to use as evidenced by the different posts to this question and other similar questions, here is a neat one I used. Consider the integral: $$\int_C \frac{\frac{1}{z^2}}{e^{i \pi z}+1} dz $$ Where C traverses through a square whose diagonals intersect at the origin in the complex plane and whose side lengths are 2N, where N is a very large even number. It can be shown through parametrizing the sides of the square and using the estimation lemma that this integral along C vanishes for very large 2N. Now, we need to calculate the residues of every pole. Let $$ f(z)=\frac{\frac{1}{z^2}}{e^{i \pi z}+1} $$ We see there are first order poles at $z=\pm (2n-1) $ for all positive integers n. There is also a second order pole at z=0. We will calculate the residues for the first order poles. Let a be an odd integer $$ Res (f(z),a)=\lim \limits_{z \to a} {(z-a)f(z)}= \frac{i}{\pi a^2} $$ For the second order pole at z=0: $$ Res (f(z),0)=\lim \limits_{z \to 0} \frac{d}{dz} (z^2f(z))=\frac{-\pi i}{4} $$ We now utilize Cauchy's residue theorem, which states: $$ \int_C g(z) dz= 2\pi i\sum{Res(g(z))} $$ if the singularities of $ g(z) $ are located in the interior of C. In our case, $$ \int_C \frac{\frac{1}{z^2}}{e^{i \pi z}+1} dz= 2 \pi i (\frac{-\pi i}{4}) + 2 \pi i \sum_{n=0}^ \infty\frac{2i}{\pi (2n+1)^2} $$ The $"2"$ inside the summation accounts for the fact that the residue of a particular negative odd integer for $ f(z)$ is equivalent to the residue of its positive counterpart. We also know the left hand side, from what was established earlier, equals $0$. Thus, $$ 0= 2 \pi i (\frac{-\pi i}{4}) + 2 \pi i \sum_{n=0}^ \infty\frac{2i}{\pi (2n+1)^2} $$ Dividing both sides by $2 \pi i$ and adding $\frac{\pi i}{4}$ to both sides, we see: $$ \frac{\pi i}{4}=\sum_{n=0}^ \infty\frac{2i}{\pi (2n+1)^2} $$ Now, multiplying both sides by $\frac{\pi}{2i}$, we get the result: $$ \frac {\pi^2}{8}= \sum_{n=0}^ \infty\frac{1}{(2n+1)^2} $$

Rearranging the terms from the $ \sum_{n=1}^ \infty\frac{1}{n^2} $, we can extrapolate that $$ \frac{3}{4}\sum_{n=1}^ \infty\frac{1}{n^2}= \sum_{n=0}^ \infty\frac{1}{(2n+1)^2} $$ Thus, we see: $$\sum_{n=1}^ \infty\frac{1}{n^2}= \frac{\pi^2}{6} $$ As a fun fact, you can get more zeta values considering the contour integral: $$\int_C \frac{\frac{1}{z^{2n}}}{e^{i \pi z}+1} dz $$ where n is a positive integer and C is the same contour as used before.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.