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?

I will write your integral as
$$\int_0^{\infty} dx \: x^2 \log{(1-e^{-a x})} = \frac{1}{3} \int_0^{\infty} d(x^3) \log{(1-e^{-a x})}  $$
Now integrate by parts:
$$ =  \frac{1}{3} [x^3 \log{(1-e^{-a x})}]_0^{\infty} - \frac{1}{3} \int_0^{\infty} dx\: x^3 \frac{d}{dx} \log{(1-e^{-a x})}  $$
The term in the brackets goes to zero at $\infty$ and $0$ (why?), so we get
$$ = -\frac{a}{3} \int_0^{\infty} dx\: x^3 e^{-a x} (1-e^{-a x})^{-1} $$
For some values of $a$ (which ones?), we may Taylor expand the term in parentheses inside the integral and get
$$ = -\frac{a}{3} \int_0^{\infty} dx\: x^3 e^{-a x} \sum_{k=0}^{\infty} e^{-k a x} $$
Because integral and sum are absolutely convergent (application of Fubini's theorem), we may interchange order of sum and integral and get:
$$ -\frac{a}{3} \sum_{k=0}^{\infty} \int_0^{\infty} dx\: x^3 e^{-(k+1) a x} $$
These integrals are well-known:
$$ \int_0^{\infty} dx\: x^3 e^{-(k+1) a x} = \frac{3!}{a^4 (k+1)^4}  $$
so we now have
$$\int_0^{\infty} dx \: x^2 \log{(1-e^{-a x})} = -\frac{2}{a^3} \sum_{k=0}^{\infty} \frac{1}{(k+1)^4} = -\frac{\pi^4}{45 a^3} $$
