Show that \[ \sum_{k\in \ZZ} {1\over k^2 + a^2} = {\pi \coth(\pi a) \over a} \qquad\text{for } a>0 .\]
Solution
Let \(f(z) = \frac{\pi \cot(\pi z)}{z^2 + a^2}\), and for \(N>a\) let \(\gamma_N\) be the positively oriented square with vertices \((\pm (N + \tfrac12), \pm (N + \tfrac12))\). It encloses the poles \(-N, \ldots, N\) and \(\pm ia\).
1. \(f\) has simple poles at each integer \(z = k\) and at \(z = \pm i a\).
\(\cot(\pi z)\) has simple poles at the integers, and \(z^2 + a^2 = (z - ia)(z + ia)\) gives simple poles at \(\pm ia\).
2. \(\operatorname{Res}(f, k) = \frac{1}{k^2 + a^2}\) for each integer \(k\).
\(\pi \cot(\pi z)\) has residue \(1\) at \(z = k\), so \(f\) has residue \(\frac{1}{k^2 + a^2}\).
3. \(\operatorname{Res}(f, ia) = \frac{\pi \cot(\pi i a)}{2ia} = -\frac{\pi \coth(\pi a)}{2a}\).
\(\cot(i\pi a) = -i\coth(\pi a)\), so \(\pi \cot(\pi i a) = -i\pi \coth(\pi a)\), and dividing by \(2ia\) gives \(-\frac{\pi \coth(\pi a)}{2a}\).
4. \(\operatorname{Res}(f, -ia) = \frac{\pi \cot(-\pi i a)}{-2ia} = -\frac{\pi \coth(\pi a)}{2a}\).
\(\cot(-\pi i a) = i\coth(\pi a)\), so \(\pi \cot(-\pi i a) = i\pi \coth(\pi a)\), and dividing by \(-2ia\) gives \(-\frac{\pi \coth(\pi a)}{2a}\).
5. \(\oint_{\gamma_N} f(z)\,dz = 2\pi i \left( \sum_{k=-N}^{N} \frac{1}{k^2 + a^2} - \frac{\pi \coth(\pi a)}{a} \right)\).
This is the residue theorem with the residues of steps 2 to 4.
6. \(\oint_{\gamma_N} f(z)\,dz \to 0\) as \(N \to \infty\).
\(|\cot(\pi z)|\) is bounded on \(\gamma_N\) independently of \(N\), and \(|z^2 + a^2| \ge (N+\tfrac12)^2-a^2\) there, so \(|f(z)| \le C/N^2\) on \(\gamma_N\). The perimeter is \(O(N)\), so the integral is \(O(1/N)\).
7. Q.E.D.
Letting \(N\to\infty\) in step 5 and using step 6 gives \(\sum_{k \in \ZZ} \frac{1}{k^2 + a^2} = \frac{\pi \coth(\pi a)}{a}\).