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?
The starting point is the integral $$ \Gamma(s) = \int_{0}^{\infty}e^{x}x^{s1}dx $$ for the gamma function. Make the change of variable $x = nu$ with $n$ an arbitrary positive integer. Then $$ \Gamma(s)n^{s} = \int_{0}^{\infty}e^{nu}u^{s1}du $$ and summing over $n$ from $n = 1$ yields $$ \Gamma(s)\zeta(s) = \int_0^{\infty}\frac{1}{e^u  1}u^{s1}du. $$ 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 $$ \zeta(2) = \int_{0}^{\infty}\frac{ue^u}{e^{2u}  e^u}du = \int_{0}^{\infty}\frac{u(e^u  1) + u}{e^{2u}  e^u}du = \int_{0}^{\infty}\left(ue^{u} + \frac{u}{e^{2u}  e^u}\right)du = 1 + \int_{0}^{\infty}\frac{u}{e^{2u}  e^u}du. $$ 

