MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

By computing the sum of all Bernoulli numbers via Borel summation (I learned this technique from Varadarajan's excellent book Euler through time. A new look at old themes, 2006) I found that $$\sum B_n = \int_0^\infty \frac{t}{e^{2t}-e^t} dt$$ and discovered numerically that this expression equals $\zeta(2)-1$. The web is not very good for finding out where this can be found in print. Where should I look, and how can equations such as $$\zeta(2) = 1 + \int_0^\infty \frac{t}{e^{2t}-e^t}\ dt$$ be proved?

share|cite|improve this question
up vote 13 down vote accepted

The starting point is the integral

$$ \Gamma(s) = \int_{0}^{\infty}e^{-x}x^{s-1}dx $$

for the gamma function. Make the change of variable $x = nu$ with $n$ an arbitrary positive integer. Then

$$ \Gamma(s)n^{-s} = \int_{0}^{\infty}e^{-nu}u^{s-1}du $$

and summing over $n$ from $n = 1$ yields

$$ \Gamma(s)\zeta(s) = \int_0^{\infty}\frac{1}{e^u - 1}u^{s-1}du. $$

This formula was the starting point of one of Riemann's two proofs of the functional equation. I am not certain who discovered it first, but it may have been Abel.

Substituting $s = 2$ gives

$$ \zeta(2) = \int_{0}^{\infty}\frac{u}{e^u - 1}du $$

and so

$$ \zeta(2) = \int_{0}^{\infty}\frac{ue^u}{e^{2u} - e^u}du = \int_{0}^{\infty}\frac{u(e^u - 1) + u}{e^{2u} - e^u}du = \int_{0}^{\infty}\left(ue^{-u} + \frac{u}{e^{2u} - e^u}\right)du = 1 + \int_{0}^{\infty}\frac{u}{e^{2u} - e^u}du. $$

share|cite|improve this answer
See also my answer to Why does the Gamma-function complete the Riemann Zeta-function?… – Harald Hanche-Olsen Feb 6 '10 at 17:44
I did have the general formula involving zeta(n), but stupidly put the n! on the wrong side of the equation and concluded that the formula for Gamma(s) zeta(s) would not help me -(. I am not aware of any work on zeta(s) done by Abel - did you mean Euler? He had a formula equivalent to that for Gamma(n) zeta(n) for n=2, 3, 4, 5. \@engelbrekt: If you care about being referred to by your real name, please send me an email : hb3 at – Franz Lemmermeyer Feb 6 '10 at 18:32
I now have a reference (from page 9 of H. M. Edwards' book Riemann's Zeta Function): N. H. Abel, Solution de quelques problemes a l'aide d'integrales définies . Mag. Naturvidenskaberne 2 (1823). Abel does the case when $s = 2m$ is an even positive integer. According to W. Narkiewicz on page 133 of The Development of Prime Number Theory, the early history of $\zeta(s)$ is covered by G. Schuppener in Geschichte der Zeta-Funktion von Oresme bis Poisson. Deutsche Hochschulschriften 533. Hänsel-Hohenhausen, Egersbach. – engelbrekt Feb 6 '10 at 19:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.