By computing \[ {1\over 2\pi i}\oint {\cot(\pi z)\over z^2}\dz ,\] say using a large rectangle, show that \[ \zeta(2) = \sum_{k\geq 0} {1\over k^2} = {\pi^2\over 6} .\]
Solution
1. Let \(f(z) = \frac{\cot(\pi z)}{z^2}\).
definition.
2. \(f\) has a pole of order \(3\) at \(z = 0\) and simple poles at each nonzero integer \(z = k\).
\(\cot(\pi z)\) has simple poles at the integers with residue \(1/\pi\), and \(z^{-2}\) adds a pole of order \(2\) at \(0\).
3. \(\operatorname{Res}(f, k) = \frac{1}{\pi k^2}\) for \(k \neq 0\).
\(\cot(\pi z)\) has residue \(1/\pi\) at \(z = k\), so \(f\) has residue \(\frac{1}{\pi} \cdot k^{-2}\).
4. \(\operatorname{Res}(f, 0) = -\frac{\pi}{3}\).
\(\cot(\pi z) = \frac{1}{\pi z} - \frac{\pi z}{3} - \frac{\pi^3 z^3}{45} - \cdots\), so \(f(z) = \frac{1}{\pi z^3} - \frac{\pi}{3z} - \frac{\pi^3 z}{45} - \cdots\), and the coefficient of \(z^{-1}\) is \(-\pi/3\).
5. Let \(\gamma_N\) be the rectangle with vertices \((\pm (N + \tfrac12), \pm (N + \tfrac12))\).
choose a contour enclosing the poles \(-N, \ldots, N\).
6. \(\frac{1}{2\pi i}\oint_{\gamma_N} f(z)\,dz = \sum_{k=-N, k \neq 0}^{N} \frac{1}{\pi k^2} - \frac{\pi}{3} = \frac{2}{\pi}\sum_{k=1}^{N} \frac{1}{k^2} - \frac{\pi}{3}\).
residue theorem, using 3 and 4.
7. \(\oint_{\gamma_N} f(z)\,dz \to 0\) 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 \(0 = \frac{2}{\pi}\sum_{k=1}^{\infty} \frac{1}{k^2} - \frac{\pi}{3}\).
6 and 7, taking the limit.
9. Therefore \(\zeta(2) = \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}\).
8, solving for the sum.
10. Q.E.D.
9.