LLCoolDave wrote:[math]\int_{0}^{\infty} \frac{x^3}{e^x-1} dx = \frac{\pi^4}{15}[/math]

On second thought, probably the most natural way to do this is contour integration, since this identity seems closely related to a connection between the

Bernoulli numbers, whose generating function is [imath]\frac{x}{e^x - 1}[/imath], and the values of the

zeta function at the even integers.

Edit: Define

[math]F(s) = \int_{0}^{\infty} \frac{te^{-st}}{1 - e^{-t}} \, dt[/math]

i.e. the Laplace transform of [imath]\frac{t}{1 - e^{-t}}[/imath]. Since the Laplace transform of [imath]te^{-kt}[/imath] is equal to [imath]\frac{1}{(s+k)^2}[/imath] it follows that, after expanding the denominator of the integrand as a geometric series, the above is equal to

[math]F(s) = \sum_{k=0}^{\infty} \frac{1}{(s+k)^2}.[/math]

Setting [imath]s = 1[/imath] we obtain

[math]\int_{0}^{\infty} \frac{t}{e^t - 1} \, dt = \zeta(2) = \frac{\pi^2}{6}.[/math]

Taking two more derivatives we obtain

[math]F^{(2)}(s) = \sum_{k=0}^{\infty} \frac{6}{(s+k)^4}[/math]

and setting [imath]s = 1[/imath] we obtain

[math]\int_{0}^{\infty} \frac{t^3}{e^t - 1} \, dt = 6 \zeta(4) = \frac{\pi^4}{15}.[/math]

The generalization is clear. In fact it may even be the case that

[math]\int_{0}^{\infty} \frac{t^a}{e^t - 1} \, dt = \Gamma(a+1) \zeta(a+1).[/math]