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I'm trying to solve for $\sum_{0}^{\infty} \dfrac{1}{n^2}$ using the residue theorem. The integral in question is $\int_C f(z)\pi \cot(\pi z) dz$ where $f(z)=\dfrac{1}{z^2}$. I am bounding over the square with vertices $\pm (N+1/2)\pm(N+1/2)i$. Bounding the integrand, I get $\dfrac{1}{z^2} \pi$ * $\dfrac{1+e^{-2\pi y}}{1-e^{-2\pi y}}$. I'm not sure how to get the integral to reduce to 0. What further bounds/limits should I take?

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The cotangent function is given by

$$\cot (\pi z)=i\left(\frac{e^{i\pi z}+e^{-i\pi z}}{e^{i\pi z}-e^{-i\pi z}}\right)$$

So, on the contour $C_N$, which is the square with vertices $(N+1/2)(1+i)$, $(N+1/2)(-1+i)$, $(N+1/2)(-1-i)$, and $(N+1/2)(1-i)$, we have

$$\lim_{N\to \infty}\left(\left. \cot (\pi z)\right|_{z\in C_N}\right)=\pm i$$

for the top and bottom parts of $C_N$. For the left and right parts, we exploit the fact that the integrand is an odd function of $z$ and integration over $y$ is from $y=-(n+1/2)$ to $y=n+1/2$. Therefore integration over the left segment for which $z=-(n+1/2)+iy$ and the right swgment for which $z=(n+1/2)+iy$ cancel.

Therefore, one can easily show that

$$\lim_{N\to \infty}\oint_{C_N} \frac{\pi \cot (\pi z)}{z^2}\,dz=0$$

We also have by the Residue Theorem that

$$\lim_{N\to \infty}\oint_{C_N} \frac{\pi \cot (\pi z)}{z^2}\,dz=\lim_{N\to \infty}\sum_{n=-N}^N\text{Res}\left(\frac{\pi \cot (\pi z)}{z^2}, z=n\right)$$

The residues at $n\ne 0$ are easily seen to $1/n^2$. To evaluate the residue at zero, we can write

$$\begin{align} \text{Res}\left(\frac{\pi \cot (\pi z)}{z^2}, z=0\right)&=\frac12\lim_{z\to 0}\frac{d^2 }{dz^2}\left(\pi (\pi z)\cot(\pi z)\right)\\\\ &=\pi^2\lim_{z\to 0}\left(\pi z\cot(\pi z)\csc^2(\pi z)-\csc^2(\pi z)\right)\\\\ &=-\frac{\pi^2}{3} \end{align}$$

Putting it all together, gives the expected result

$$2\sum_{n=1}^\infty\frac{1}{n^2}-\frac{\pi^2}{3}=0\implies \bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}}$$

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  • $\begingroup$ Thank you. Can you expand on how to approach finding the 2 limits? $\endgroup$
    – user121955
    Nov 22, 2015 at 0:09
  • $\begingroup$ You're welcome. My pleasure. There are four limits in the answer. What specifically is the biggest obstacle? $\endgroup$
    – Mark Viola
    Nov 22, 2015 at 0:49
  • $\begingroup$ How do you get $\lim_{N\to \infty}\left(\left. \cot (\pi z)\right|_{z\in C_N}\right)=\pm1$ and $\lim_{N\to \infty}\oint_{C_N} \frac{\pi \cot (\pi z)}{z^2}\,dz=0$? $\endgroup$
    – user121955
    Nov 22, 2015 at 18:01
  • $\begingroup$ A typo ... forgot the factor of $i$. The first limit applies only to the top and bottom segments of $C_N$ where $z=x\pm i(N+1/2)$. On those segments, $\cot (\pi z)=i \left(\frac{1+e^{\mp i2x}e^{\pm(2n+1)\pi}}{1-e^{\mp i2x}e^{\pm (2n+1)\pi}}\right)$. The limit is obviously, $\mp i$. For the second, we use the previous result along with $\left|\frac{1}{z^2}\,dz\right|\sim O\left(\frac1N\right)$ on the top and bottom segments. For the left and right segments, we exploit the oddness of the integrand. $\endgroup$
    – Mark Viola
    Nov 22, 2015 at 19:28
  • $\begingroup$ @hermes A typo. Thank you for the catch. +1 for your comment. - Mark $\endgroup$
    – Mark Viola
    Nov 23, 2015 at 23:52

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