Show \[ \sum_{n \geq 1} \frac{1}{n^{2}}=\frac{\pi^{2}}{6} \] by integrating \(\pi \cot(\pi z)z^{-2}\).
Solution
1. Let \(f(z) = \pi \cot(\pi z) z^{-2}\).
definition.
2. \(f\) has a pole of order \(3\) at \(z = 0\) and simple poles at each nonzero integer \(z = n\).
\(\cot(\pi z)\) has simple poles at the integers with residue \(1/\pi\), and \(z^{-2}\) adds a pole of order \(2\) at \(0\).
3. \(\operatorname{Res}(f, n) = 1/n^2\) for \(n \neq 0\).
\(\pi \cot(\pi z)\) has residue \(1\) at \(z = n\), so \(f\) has residue \(1 \cdot n^{-2}\).
4. \(\operatorname{Res}(f, 0) = -\pi^2/3\).
\(\pi \cot(\pi z) = \frac{1}{z} - \frac{\pi^2 z}{3} - \frac{\pi^4 z^3}{45} - \cdots\), so \(f(z) = \frac{1}{z^3} - \frac{\pi^2}{3z} - \frac{\pi^4 z}{45} - \cdots\), and the coefficient of \(z^{-1}\) is \(-\pi^2/3\).
5. Let \(\gamma_N\) be the square with vertices \((\pm (N + \tfrac12), \pm (N + \tfrac12))\).
choose a contour enclosing the poles \(-N, \ldots, N\).
6. \(\int_{\gamma_N} f(z)\,dz = 2\pi i \left( \sum_{n=-N, n \neq 0}^{N} \frac{1}{n^2} - \frac{\pi^2}{3} \right) = 2\pi i \left( 2 \sum_{n=1}^{N} \frac{1}{n^2} - \frac{\pi^2}{3} \right)\).
residue theorem, using 3 and 4.
7. \(\int_{\gamma_N} f(z)\,dz \to 0\) as \(N \to \infty\).
\(|\cot(\pi z)|\) is bounded on \(\gamma_N\) (the contour avoids the poles), \(|z^{-2}| \le C/N^2\), and the perimeter is \(O(N)\), so the integral is \(O(1/N)\).
8. Hence \(0 = 2\pi i \left( 2 \sum_{n=1}^{\infty} \frac{1}{n^2} - \frac{\pi^2}{3} \right)\).
6 and 7, taking the limit.
9. Therefore \(\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}\).
8, solving for the sum.
10. Q.E.D.
9.