TITLE: Calculating $ \int _{0} ^{\infty} \frac{x^{3}}{e^{x}-1}\;dx$
QUESTION [13 upvotes]: how to calculate $$\int_0^\infty \frac{x^{3}}{e^{x}-1} \; dx$$

Be $q:= e^{z}-1 , p:= z^{3}$ , then $e^{z} = 1 $ if $z= 2\pi n i $, so the residue at 0 is : $$\frac{p(z_{0})}{q'(z_{0})} = 2\pi i n ^{3}$$
problem is that this is not symmetric, so how does one find the definite integral?

REPLY [5 votes]: I had trouble seeing this until I found the following approach--late and for what it is worth. I like that it shows the summation explicitly (and reminds us of the expressions for Gamma and Zeta).
$\Gamma(x) = \int_{0}^{\infty}t^{x-1}e^{-t}dt$ for x > 0. Make the variable substitution t = ru:
$\Gamma(x) = \int_{0}^{\infty}(ru)^{x-1}e^{-ru}r\ du  =r^x \int_{0}^{\infty}u^{x-1}e^{-ru}du $
So that 
$\frac{1}{r^x} = \frac{1}{\Gamma(x)}\int_{0}^{\infty}u^{x-1}e^{-ru}du$.
$\zeta(x) = \sum_{1}^{\infty}\frac{1}{r^x}= \frac{1}{\Gamma(x)}\sum_{1}^{\infty} \int_{0}^{\infty}u^{x-1}e^{ru}du$ = $\frac{1}{\Gamma(x)}\int_{0}^{\infty}u^{x-1}\sum_{1}^{\infty}e^{-ru}du$.
Finally, 
$\zeta(x) = \frac{1}{\Gamma(x)}\int_{0}^{\infty}u^{x-1}\frac{ e^{-u}}{1-e^{-u}}du$ and so $\Gamma(x)\zeta(x) = \int_{0}^{\infty}\frac{u^{x-1}}{e^u-1}du   $