Question

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?

Answer 1

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

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