Show that \[ \sum_{k\in \ZZ} {1\over \qty{ k - {1\over 2}}^2 } = \pi^2 .\]
Solution
For \(N\ge1\), let \(Q_N\) be the positively oriented square \[ |\Re z|\le N+\tfrac12, \qquad |\Im z|\le N+\tfrac12, \] and consider \[ F(z)=\frac{\pi\cot(\pi z)}{(z-\tfrac12)^2}. \]
1. On \(\partial Q_N\), \(|\cot(\pi z)|\) is bounded by a constant independent of \(N\).
1.1. On the vertical sides, \(\Re z\) is a half-integer, so \(|\cot(\pi z)|=|\tanh(\pi\Im z)|\le1\).
1.2. On the horizontal sides, \(|\Im z|=N+1/2\). For real \(x\) and \(y\neq0\), \(\abs{\cot(x+iy)}^2=\frac{\cos^2x+\sinh^2y}{\sin^2x+\sinh^2y}\le\coth^2y\), so \(\abs{\cot(\pi z)}\le\coth(\pi(N+\tfrac12))\le\coth(\pi/2)\).
1.3. Also \(|z-1/2|\ge N\) on \(\partial Q_N\), while the perimeter is \(O(N)\). Hence\[ \oint_{\partial Q_N}F(z)\,dz\longrightarrow0. \]
2. The poles inside \(Q_N\) are the integers \(k=-N,\dots,N\) and \(z=1/2\).
2.1. Since \(\pi\cot(\pi z)\) has residue \(1\) at every integer,\[ \operatorname{Res}(F;k)=\frac1{(k-1/2)^2}. \]
2.2. Writing \(w=z-1/2\),\[ \pi\cot(\pi z)=-\pi\tan(\pi w)=-\pi^2w+O(w^3), \]so\[ F(z)=-\frac{\pi^2}{w}+O(w), \]and therefore\[ \operatorname{Res}(F;1/2)=-\pi^2. \]
3. The residue theorem gives\[ \frac1{2\pi i}\oint_{\partial Q_N}F(z)\,dz =\sum_{k=-N}^N\frac1{(k-1/2)^2}-\pi^2. \]Letting \(N\to\infty\) yields\[ \sum_{k\in\mathbb Z}\frac1{(k-1/2)^2}=\pi^2. \]