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?
 A: All the proofs I know of are in Kalman's article. There is one that uses complex analysis (residues, to be concrete).
A: 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.  
A: 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 $\color{green}{-\frac{\pi^2}{3}}+2\cdot\color{red}{\frac{\pi^2}{6}}=\color{blue}{0}$).
