Show that \[ \sum_{k\in \mathbb{Z}} { (-1)^k \over (k+a)^2} = \pi^2 \cos(\pi a)\csc^2(\pi a) \quad \text{for } a\in \mathbb{R}\setminus\mathbb{Z} .\]
Solution
Fix \(a\in\mathbb R\setminus\mathbb Z\) and set \[ F(z)=\frac{\pi\csc(\pi z)}{(z+a)^2}. \] Integrate over the squares \(\Gamma_N\) with vertices \((N+\tfrac12)(\pm1\pm i)\). On \(\Gamma_N\), \(|\csc(\pi z)|\) is uniformly bounded, while \(|z+a|\gg N\) and the contour length is \(O(N)\), so \[ \int_{\Gamma_N}F(z)\,dz\to0. \]
1. At an integer \(k\),\[ \operatorname{Res}(F,k)=\frac{(-1)^k}{(k+a)^2}. \]At the double pole \(z=-a\),\[ \operatorname{Res}(F,-a) =\left.\frac{d}{dz}(\pi\csc(\pi z))\right|_{z=-a} =-\pi^2\csc(\pi a)\cot(\pi a) =-\pi^2\cos(\pi a)\csc^2(\pi a). \]
2. For large \(N\), the residue theorem gives\[ \frac1{2\pi i}\int_{\Gamma_N}F(z)\,dz =\sum_{k=-N}^N\frac{(-1)^k}{(k+a)^2} -\pi^2\cos(\pi a)\csc^2(\pi a). \]Letting \(N\to\infty\) yields\[ \sum_{k\in\mathbb Z}\frac{(-1)^k}{(k+a)^2} =\pi^2\cos(\pi a)\csc^2(\pi a). \]