Show that \(\displaystyle f(z) = \frac{\pi^2}{\sin^2 \pi z}\) and \(\displaystyle g(z) = \sum_{n = - \infty}^{ \infty} \frac{1}{(z-n)^2}\) have the same principal part at each integer point \(k \in \mathbb{Z}\).
Show that \(\displaystyle h(z) = \frac{\pi^2}{\sin^2 \pi z} - g(z)\) is bounded on \(\mathbb{C}\) and conclude that \(\displaystyle \frac{\pi^2}{\sin^2 \pi z} = \sum_{n = - \infty}^{ \infty} \frac{1}{(z-n)^2}\).
Solution
Write \[ g(z)=\sum_{n\in\mathbb Z}\frac1{(z-n)^2}. \] The series converges locally uniformly on \(\mathbb C\setminus\mathbb Z\), since on a compact set away from the integers its tails are dominated by a constant multiple of \(\sum n^{-2}\). Hence \(g\) is meromorphic there and is \(1\)-periodic.
Near an integer \(k\), \[ g(z)=\frac1{(z-k)^2}+\text{holomorphic terms}. \] Also \[ \sin(\pi z)=(-1)^k\pi(z-k)+O((z-k)^3), \] so \[ \frac{\pi^2}{\sin^2(\pi z)} =\frac1{(z-k)^2}+O(1). \] Thus the two meromorphic functions have the same principal part at every integer.
Therefore \[ h(z)=\frac{\pi^2}{\sin^2(\pi z)}-g(z) \] extends to an entire \(1\)-periodic function.
It remains to bound \(h\) on one vertical fundamental strip. For \(z=x+iy\) with \(0\le x\le1\), \[ |\sin(\pi z)|^2=\sin^2(\pi x)+\sinh^2(\pi y), \] so the first term tends uniformly to \(0\) as \(|y|\to\infty\). Moreover \[ |g(x+iy)| \le \sum_{n\in\mathbb Z}\frac1{(x-n)^2+y^2} \longrightarrow0 \] uniformly in \(x\in[0,1]\) as \(|y|\to\infty\); for example, compare the sum with \(|y|^{-2}+\int_{-\infty}^{\infty}(t^2+y^2)^{-1}dt\). Hence \(h\) is bounded for large \(|y|\). On the remaining compact part of the strip it is bounded by continuity. Periodicity then makes \(h\) bounded on all of \(\mathbb C\).
By Liouville’s theorem \(h\) is constant, and the limit along \(iy\) as \(y\to\infty\) is \(0\). Thus \(h\equiv0\), i.e. \[ \frac{\pi^2}{\sin^2(\pi z)} =\sum_{n\in\mathbb Z}\frac1{(z-n)^2}. \]