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?
 A: 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
$$\int_0^{\infty} dx \: x^2 \log{(1-e^{-a x})} = -\frac{2}{a^3} \sum_{k=0}^{\infty} \frac{1}{(k+1)^4} = -\frac{\pi^4}{45 a^3} $$
A: Using the power series for $\log(1-x)$: $$\int_{0}^{\infty}n^2\log(1-e^{-\alpha n})\,\mathrm{d}n=-\int_{0}^{\infty}n^2\sum_{k=1}^{\infty}\frac{e^{-k\alpha n}}{k}\,\mathrm{d}n\\=-\int_{0}^{\infty}\sum_{k=1}^{\infty}\frac{\partial^2}{\partial\alpha^2}\left(\frac{e^{-kn\alpha}}{k^3}\right)\,\mathrm{d}n\\=-\frac{\partial^2}{\partial\alpha^2}\sum_{k=1}^{\infty}\int_{0}^{\infty}\frac{e^{-kn\alpha}}{k^3}\,\mathrm{d}n\\=-\frac{\partial^2}{\partial\alpha^2}\sum_{k=1}^{\infty}\left(-\frac{e^{-kn\alpha}}{\alpha k^4}\right)_0^{\infty}\\=-\frac{\partial^2}{\partial\alpha^2}\sum_{k=1}^{\infty}\frac{1}{\alpha k^4}\\=-\sum_{k=1}^{\infty}\frac{\partial^2}{\partial\alpha^2}\left(\frac{\alpha^{-1}}{ k^4}\right)\\=-2\sum_{k=1}^{\infty}\frac{\alpha^{-3}}{k^4}=-\frac{2\zeta(4)}{\alpha^3}\\=-\frac{\pi^4}{45\alpha^3}.$$
A: Integration by parts yields
$$
\begin{align}
\int_0^\infty n^2 e^{-n}\,\mathrm{d}n
&=\left.-n^2e^{-n}\right]_0^\infty+\int_0^\infty2n\,e^{-n}\,\mathrm{d}n\\
&=0+\left.-2n\,e^{-n}\vphantom{n^2}\right]_0^\infty+\int_0^\infty 2\,e^{-n}\,\mathrm{d}n\\
&=2\tag{1}
\end{align}
$$
Apply $(1)$ and $\log(1-x)=-\sum\limits_{k=1}^\infty\dfrac{x^k}{k}$ to
$$
\begin{align}
\int_0^\infty n^2\log(1-e^{-an})\,\mathrm{d}n
&=-\int_0^\infty n^2\sum_{k=1}^\infty\frac1ke^{-kan}\,\mathrm{d}n\\
&=-\sum_{k=1}^\infty\frac1k\frac2{(ka)^3}\\
&=-\frac2{a^3}\zeta(4)\\
&=-\frac{\pi^4}{45a^3}\tag{2}
\end{align}
$$
Using $\zeta(4)$ from this answer.
A: Partial integration:
$$\int_0^{\infty}n^2\cdot \ln(1-e^{-an})dn=\underbrace{\left[\frac{x^3}{3}\ln(1-e^{-an})\right]_0^{\infty}}_{=\;0}-\frac{1}{3}\int_0^{\infty}n^3\left(\frac{a}{e^{an}-1}\right)dn$$
Let $s=an:$
$$\int_0^{\infty}n^2\cdot \ln(1-e^{-an})dn=-\frac{1}{3a^3}\int_0^{\infty}\frac{s^3}{e^{s}-1}ds$$
Using the well-known relation
$$\Gamma(z)\zeta (z)=\int_0^{\infty}\frac{s^{z-1}}{e^s-1}ds\qquad(\Re (z)>1)$$
We get,
$$\int_0^{\infty}n^2\cdot \ln(1-e^{-an})dn=-\frac{1}{3a^3}\Gamma(4)\zeta(4)=-\frac{\pi^4}{45a^3}$$
A: This is a generalization. By expanding the logarithm in its Taylor series (the same method in the other answers) you'll get that
$$ \int_0^\infty x^{n-1}\log(1-e^{-\alpha x})dx=-\frac{1}{\alpha^n}\zeta(n+1)\Gamma(n)$$
Your case is $n=3$
