- Suppose u is not an integer. Prove that
\[ \sum_ {n = - \infty} ^ {\infty} \frac {1}{(u + n) ^ {2}} = \frac {\pi^ {2}}{(\sin \pi u) ^ {2}} \]
by integrating
\[ f (z) = \frac {\pi \cot \pi z}{(u + z) ^ {2}} \]
over the circle \(| z | = R _ { N } = N + 1 / 2\) (N integral, \(N \geq | u | )\) , adding the residues of f inside the circle, and letting N tend to infinity.
Note. Two other derivations of this identity, using Fourier series, were given in Book I.
Solution
Let \[ f(z)=\frac{\pi\cot(\pi z)}{(u+z)^2},\qquad u\notin\mathbb Z, \] and let \(C_N\) be \(|z|=R_N=N+\tfrac12\), with \(N\ge |u|\).
At each integer \(n\) with \(|n|\le N\), \(\pi\cot(\pi z)\) has residue \(1\), so \[ \operatorname{Res}_{z=n}f(z)=\frac1{(u+n)^2}. \] At \(z=-u\) there is a double pole. If \(g(z)=\pi\cot(\pi z)\), then \[ \operatorname{Res}_{z=-u}\frac{g(z)}{(z+u)^2}=g'(-u) =-\pi^2\csc^2(\pi u). \] Therefore the residue theorem gives \[ \int_{C_N}f(z)\,dz =2\pi i\left( \sum_{n=-N}^N\frac1{(u+n)^2} -\frac{\pi^2}{\sin^2(\pi u)} \right). \tag{1} \]
It remains to show the contour integral tends to \(0\). On the circles \(|z|=N+\tfrac12\), the distance to every integer is bounded below by a positive absolute constant in the horizontal direction, and the standard formula \[ \cot(x+iy)=\frac{\sin 2x-i\sinh 2y}{\cosh 2y-\cos 2x} \] shows \(|\cot(\pi z)|\) is uniformly bounded on all these circles. Also \(|z+u|\ge R_N-|u|\). Hence \[ \left|\int_{C_N}f(z)\,dz\right| \le 2\pi R_N\,\frac{C}{(R_N-|u|)^2}\longrightarrow0. \] Letting \(N\to\infty\) in (1) yields \[ \sum_{n=-\infty}^{\infty}\frac1{(u+n)^2} =\frac{\pi^2}{\sin^2(\pi u)}. \]