Integration by parts of $\int_0^\infty \! n^2 \ln(1-e^{-an}) \, \mathrm{d} n$ Im trying to do the following integration by hand, 
$\int_0^\infty \! n^2 \ln(1-e^{-an}) \, \mathrm{d} n$
I have tried to use integration by parts and substitution, but each time it gets to complicated and messy (polylog terms etc...), is there any straight forward way to solve this?

Using the power series for $\log(1-x)$: $$\int_{0}^{\infty}n^2\log(1-e^{-\alpha n})\,\mathrm{d}n=-\int_{0}^{\infty}n^2\sum_{k=1}^{\infty}\frac{e^{-k\alpha n}}{k}\,\mathrm{d}n\\=-\int_{0}^{\infty}\sum_{k=1}^{\infty}\frac{\partial^2}{\partial\alpha^2}\left(\frac{e^{-kn\alpha}}{k^3}\right)\,\mathrm{d}n\\=-\frac{\partial^2}{\partial\alpha^2}\sum_{k=1}^{\infty}\int_{0}^{\infty}\frac{e^{-kn\alpha}}{k^3}\,\mathrm{d}n\\=-\frac{\partial^2}{\partial\alpha^2}\sum_{k=1}^{\infty}\left(-\frac{e^{-kn\alpha}}{\alpha k^4}\right)_0^{\infty}\\=-\frac{\partial^2}{\partial\alpha^2}\sum_{k=1}^{\infty}\frac{1}{\alpha k^4}\\=-\sum_{k=1}^{\infty}\frac{\partial^2}{\partial\alpha^2}\left(\frac{\alpha^{-1}}{ k^4}\right)\\=-2\sum_{k=1}^{\infty}\frac{\alpha^{-3}}{k^4}=-\frac{2\zeta(4)}{\alpha^3}\\=-\frac{\pi^4}{45\alpha^3}.$$
