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?



Integral expression for zeta(2)
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


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


