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The material presented at this link on Zeta function values at even integers proposes a method to compute these that is based on Euler's work. I would like to present a short proof for your consideration (the goal being to keep it as simple as possible) and I'd be grateful to get confirmation of its correctness/admissibility e.g. by giving a reference.

We seek to prove that $$\zeta(2n) = \sum_{m\ge1} \frac{1}{m^{2n}} = (-1)^{n+1} \frac{B_{2n} 2^{2n}}{2(2n)!} \pi^{2n}$$ where $B_{2n}$ are Bernoulli numbers. Rewrite this as $$\zeta(2n) = - \frac{1}{2} (2\pi i)^{2n} \frac{B_{2n}}{(2n)!}.$$ Now using the method presented here we introduce $$ f(z) = \frac{1}{z^{2n}} \pi \cot(\pi z)$$ and compute the integral of $f(z)$ along a circle of radius $R$ in the complex plane, where $R$ goes to infinity. We certainly have $|\pi \cot(\pi z)|<2\pi$ for $z$ on the circle and $R$ large enough. The term $1/z^{2n}$ is $\theta(1/R^{2n})$ so that the integral along the circle is $\theta(1/R^{2n-1})$ and vanishes in the limit. Using the Cauchy Residue theorem we thus obtain $$ 2 \zeta(2n) + \operatorname{Res}_{z=0} f(z) = \lim_{R\to\infty} \frac{1}{2\pi i} \int_{|z|=R} f(z) dz = 0.$$ The conclusion is that $$ \zeta(2n) = - \frac{1}{2} \operatorname{Res}_{z=0} f(z)$$ For the exponential generating function of the Bernoulli numbers we have $$\sum_{m=0}^\infty B_m \frac{t^m}{m!} = \frac{t}{e^t-1}$$ so that $$ \sum_{m=0}^\infty B_{2m} \frac{t^{2m}}{(2m)!} = \frac{1}{2} \left( \frac{t}{e^t-1} - \frac{t}{e^{-t}-1}\right) = \frac{1}{2} t \frac{e^{t/2}+e^{-t/2}}{e^{t/2}-e^{-t/2}}.$$ Setting $t=2\pi i z$, we obtain $$ \sum_{m=0}^\infty B_{2m} (2\pi i)^{2m} \frac{z^{2m}}{(2m)!} = \pi i z \frac{e^{\pi i z}+e^{-\pi i z}}{e^{\pi i z}-e^{-\pi i z}} = \pi z \cot(\pi z).$$ Putting it all together, we find $$\zeta(2n) = - \frac{1}{2} [z^{-1}] f(z) = - \frac{1}{2} [z^{2n-1}] \pi \cot(\pi z) = - \frac{1}{2} [z^{2n}] \pi z \cot(\pi z) = - \frac{1}{2} (2\pi i)^{2n} \frac{B_{2n}}{(2n)!},$$ where $[z^q]$ is the coefficient extraction operator for power series. This concludes the proof.

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It looks okay, although I don't believe that $\cot (\pi z)$ has the bound you mentioned. You probably need to work harder to show that the integral goes to 0 as $R \to \infty$, by splitting up the circle into two pieces: one that $\cot$ has a nice bound, and one that is very short. – user27126 Dec 1 '12 at 3:07
Thanks for commenting. At this link I do show that $\cot(z)$ is bounded by a constant for $R$ large enough, by considering the upper and lower half plane separately. Of course as always in these kinds of summations and also in integral transform inversions the contours do not pass through the singularities. In the present case we could choose $R$ so that it passes halfway between consecutive poles on the real axis. – Marko Riedel Dec 1 '12 at 3:23
The bound I refer to applies to $|\cot(z)|$. – Marko Riedel Dec 1 '12 at 3:58
Sorry I don't quite see it. No matter what $R$ you choose there would be a small region, where $t$ (as in your link) is very close to 0, making your bound blow up. It's true though that such $t$ does not take up much space. – user27126 Dec 1 '12 at 7:17
As an observation, if I were to post this now, I would use a rectangular rather than a circular contour. – Marko Riedel Sep 19 '13 at 22:00

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