Editing Residue theorem (section)

===Evaluating zeta functions===
The fact that {{math|''π'' cot(''πz'')}} has simple poles with residue 1 at each integer can be used to compute the sum
<math display="block"> \sum_{n=-\infty}^\infty f(n).</math>

Consider, for example, {{math|1=''f''(''z'') = ''z''<sup>−2</sup>}}. Let {{math|Γ<sub>''N''</sub>}} be the rectangle that is the boundary of {{math|[−''N'' − {{sfrac|1|2}}, ''N'' + {{sfrac|1|2}}]<sup>2</sup>}} with positive orientation, with an integer {{mvar|N}}. By the residue formula,

<math display="block">\frac{1}{2 \pi i} \int_{\Gamma_N} f(z) \pi \cot(\pi z) \, dz = \operatorname{Res}\limits_{z = 0} + \sum_{n = -N \atop n\ne 0}^N n^{-2}.</math>

The left-hand side goes to zero as {{math|''N'' → ∞}} since <math>|\cot(\pi z)|</math> is uniformly bounded on the contour, thanks to using <math>x = \pm \left(\frac 12 + N\right)</math> on the left and right side of the contour, and so the integrand has order <math>O(N^{-2})</math> over the entire contour. On the other hand,<ref>{{harvnb|Whittaker|Watson|1920|loc=§7.2|page=125}}. Note that the Bernoulli number <math>B_{2n}</math> is denoted by <math>B_{n}</math> in Whittaker & Watson's book.</ref>

<math display="block">\frac{z}{2} \cot\left(\frac{z}{2}\right) = 1 - B_2 \frac{z^2}{2!} + \cdots </math> where the [[Bernoulli number]] <math>B_2 = \frac{1}{6}.</math>

(In fact, {{math|1={{sfrac|''z''|2}} cot({{sfrac|''z''|2}}) = {{sfrac|''iz''|1 − ''e''<sup>−''iz''</sup>}} − {{sfrac|''iz''|2}}}}.) Thus, the residue {{math|Res{{sub|1=''z''=0}}}} is {{math|−{{sfrac|''π''<sup>2</sup>|3}}}}. We conclude:

<math display="block">\sum_{n = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}</math>
which is a proof of the [[Basel problem]].

The same argument works for all <math>f(x) = x^{-2n}</math> where <math>n</math> is a positive integer, [[Particular values of the Riemann zeta function|giving us]]<math display="block"> \zeta(2n) = \frac{(-1)^{n+1}B_{2n}(2\pi)^{2n}}{2(2n)!}.</math>The trick does not work when <math>f(x) = x^{-2n-1}</math>, since in this case, the residue at zero vanishes, and we obtain the useless identity <math>0 + \zeta(2n+1) - \zeta(2n+1) = 0</math>.