Evaluating the definite integral $\int_0^\infty \frac{\mathrm{e}^x}{\left(\mathrm{e}^x-1\right)^2}\,x^n \,\mathrm{d}x$ I am having difficulty evaluating definite integrals of the form $\int_0^\infty \frac{\mathrm{e}^x}{\left(\mathrm{e}^x-1\right)^2}\,x^n \,\mathrm{d}x$. I would appreciate any guidance that could be offered. I am aware that these evaluate to constant powers of $\pi$, but find the integration challenging. Thank you.
 A: $$ \int_0^\infty \frac{e^x}{(e^x-1)^2} x^n dx =  \int_0^\infty \frac{e^{-x}}{(1-e^{-x} )^2} x^n dx .$$ 
For $|z|<1 $ we have $\displaystyle \frac{1}{1-z} =\sum_{k=0}^{\infty} z^k $ so by differentiating we get $\displaystyle  \frac{1}{(1-z)^2} = \sum_{k=1}^{\infty} k z^{k-1}. $ Thus for $x\in (0,\infty),$ $$  \frac{e^{-x}}{(1-e^{-x} )^2} = \sum_{k=1}^{\infty} k e^{-kx}.$$ Hence, (after applying the monotone convergence theorem) the desired integral is equal to $$ \sum_{k=1}^{\infty} k \int^{\infty}_0 x^n e^{-kx} dx .$$
If we let $u=kx $ we find that $$  \int^{\infty}_0 x^n e^{-kx} dx = \int^{\infty}_0 \left( \frac{u}{k} \right)^n e^{-u} \cdot \frac{1}{k} du = \frac{1}{k^{n+1}} \Gamma(n+1) = \frac{n!}{k^{n+1}}.$$
Thus, $$\int^{\infty}_0 \frac{e^x}{(e^x-1)^2} x^n dx = n! \zeta(n) .$$
Your suspicion that the value of the integral is a constant times a power of $\pi$ is correct for even integers, as $$(2n)! \cdot \zeta(2n) = (2n)! \cdot (-1)^{n+1} \frac{B_{2n} (2\pi)^{2n} }{2 (2n)!} = (-1)^{n+1} \frac{B_{2n} (2\pi)^{2n} }{2} $$
but for odd n no simpler form for $\zeta(n)$ is known (and it probably isn't simple powers of $\pi.$)
