The following exponential generating function of the Bernoulli Numbers is the following:
$$\frac{t}{e^t-1}=\sum_{n=0}^{\infty} B_n \frac{t^n}{n!}$$ The following result, which is a well known special case of the Bose Integral: $$\zeta(2)=\int_0^\infty \frac{x^{2-1}}{e^x-1}dx$$ Hence $$\frac{\pi^2}{6}=\int_0^\infty \frac{x}{e^x-1}dx$$ But, using the exponential generating function of the Bernoulli Numbers, one obtains: $$\int_0^\infty \frac{x}{e^x-1}dx=\int_0^\infty\sum_{n=0}^{\infty} B_n \frac{x^n}{n!}dx$$ Now, we can interchange the sum and the integral (As it is satisfying Tolleni & Fubini's Theorems), and finally obtain: $$\sum_{n=0}^{\infty} B_n \frac{x^{n+1}}{(n+1)!}$$, where x is taken from 0 to infinity. It it easy to prove that the overall sum is equal to 0-Thus the question: Where did I go wrong?