Find all residues of \[ \frac{\cot z}{z^2} \] and use them to evaluate \[ \frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\cdots. \]
Solution
Set \[ F(z)\coloneqq\frac{\cot z}{z^2}. \]
1. For every nonzero integer \(n\),\[ \operatorname{Res}_{z=n\pi}F(z) = \frac{1}{n^2\pi^2}. \]
At \(z=n\pi\), the function \(\sin z\) has a simple zero and \[ (\sin z)'\big|_{z=n\pi} = \cos(n\pi) \neq 0. \] Therefore \(\cot z=\cos z/\sin z\) has a simple pole there with residue \[ \frac{\cos(n\pi)}{\cos(n\pi)}=1. \] Since \(z^{-2}\) is holomorphic at \(n\pi\neq0\), multiplication gives the stated residue.
2. At the origin,\[ \operatorname{Res}_{z=0}F(z) = -\frac13. \]
The Taylor expansions at \(0\) give \[ \sin z = z-\frac{z^3}{6}+O(z^5), \qquad \cos z = 1-\frac{z^2}{2}+O(z^4). \] Hence \[ \begin{aligned} \cot z &= \frac{1-z^2/2+O(z^4)} {z(1-z^2/6+O(z^4))}\\ &= \frac1z \left( 1-\frac{z^2}{3}+O(z^4) \right)\\ &= \frac1z-\frac z3+O(z^3). \end{aligned} \] Therefore \[ F(z) = \frac1{z^3} -\frac1{3z} +O(z), \] whose residue is \(-1/3\).
3. Let\[ R_N\coloneqq\left(N+\frac12\right)\pi \]and let \(\Gamma_N\) be the positively oriented square with vertices\[ \pm R_N\pm iR_N. \]There is a constant \(C\) independent of \(N\) such that\[ \abs{\cot z}\leq C \]for every \(z\in\Gamma_N\).
On a vertical side, \[ z=\left(N+\frac12\right)\pi+iy \] or its negative counterpart. Periodicity and the identity \[ \cot\left(\frac\pi2+iy\right) = -i\tanh y \] give \[ \abs{\cot z}\leq1. \]
For \(z=x+iy\), \[ \abs{\sin z}^2 = \sin^2x+\sinh^2y \] and \[ \abs{\cos z}^2 = \cos^2x+\sinh^2y. \] Thus on either horizontal side, where \(\abs{y}=R_N\geq\pi/2\), \[ \abs{\cot z}^2 \leq \frac{1+\sinh^2R_N}{\sinh^2R_N} = \coth^2R_N \leq \coth^2(\pi/2). \] Hence one may take \[ C\coloneqq\coth(\pi/2). \]
4. The contour integrals satisfy\[ \lim_{N\to\infty} \int_{\Gamma_N}F(z)\,dz = 0. \]
Every point of \(\Gamma_N\) satisfies \[ \abs{z}\geq R_N, \] and the perimeter of \(\Gamma_N\) is \(8R_N\). By step 3, \[ \abs{F(z)} \leq \frac{C}{R_N^2} \] on the contour. Therefore \[ \abs{ \int_{\Gamma_N}F(z)\,dz } \leq 8R_N\frac{C}{R_N^2} = \frac{8C}{R_N} \longrightarrow0. \]
5. For every \(N\geq1\),\[ \frac{1}{2\pi i} \int_{\Gamma_N}F(z)\,dz = -\frac13 + \frac{2}{\pi^2} \sum_{n=1}^N\frac1{n^2}. \]
The poles inside \(\Gamma_N\) are exactly \[ n\pi, \qquad -N\leq n\leq N. \] The residue theorem, together with steps 1 and 2, gives \[ \begin{aligned} \frac{1}{2\pi i} \int_{\Gamma_N}F(z)\,dz &= -\frac13 + \sum_{n=1}^N \left( \frac{1}{n^2\pi^2} + \frac{1}{n^2\pi^2} \right)\\ &= -\frac13 + \frac{2}{\pi^2} \sum_{n=1}^N\frac1{n^2}. \end{aligned} \]
6. The Basel sum is\[ \boxed{ \sum_{n=1}^{\infty}\frac1{n^2} = \frac{\pi^2}{6} }. \]
5. By step 4, the left-hand side tendsto \(0\). Hence\[ 0 = -\frac13 + \frac{2}{\pi^2} \sum_{n=1}^{\infty}\frac1{n^2}, \]which rearranges to the displayed value.
7. Q.E.D.
Steps 1 and 2 give all residues of \(F\), and step 6 gives the requested series evaluation.