The integral $\int\limits_0^\infty\frac{x^4e^x}{(e^x-1)^2} \mathrm{d}x$ How to calculate the following integral $$\int_0^\infty\frac{x^4e^x}{(e^x-1)^2}\mathrm{d}x$$
I would like to solve this integral by means of two different ways: for example, integration by parts and using Residue Theorem.  
 A: Using
$$ \frac1{(1-x)^2}=\sum_{n=0}^\infty(n+1)x^n $$
\begin{eqnarray}
\int_0^\infty\frac{x^4e^x}{(e^x-1)^2}\mathrm{d}x&=&\int_0^\infty\frac{x^4e^{-x}}{(1-e^{-x})^2}\mathrm{d}x\\
&=&\int_0^\infty x^4e^{-x}\sum_{n=0}^\infty(n+1)e^{-nx}\mathrm{d}x\\
&=&\sum_{n=0}^\infty\int_0^\infty(n+1)x^4e^{-(n+1)x}\mathrm{d}x\\
&=&\sum_{n=0}^\infty \frac{24}{(n+1)^4}\\
&=&24\zeta(4)\\
&=&\frac{4\pi^4}{15}.
\end{eqnarray}
A: $$I=\int_0^\infty\frac{x^4e^x}{(e^x-1)^2}\ dx\overset{e^{-x}=y}{=}\int_0^1\frac{\ln^4x}{(1-x)^2}\ dx=\sum_{n=1}^\infty n\int_0^1x^{n-1}\ln^4x\ dx=\sum_{n=1}^\infty\frac{4!}{n^4}=4!\zeta(4)$$
A: I will generalise this integral to solve
$$I_n=\int_0^\infty\frac{x^ne^x}{(e^x-1)^2}\mathrm{d}x$$
Applying integration by parts and L'Hôpital's rule gives
$$\begin{align}
I_n
&=\left[-\frac{x^n}{e^x-1}\right]_0^\infty+n\int_0^\infty\frac{x^{n-1}}{e^x-1}\mathrm{d}x\\
&=n\int_0^\infty\frac{x^{n-1}}{e^x-1}\mathrm{d}x\\
&=n\,\zeta(n)\,\Gamma(n)\\
&=\zeta(n)\,\Gamma(n+1)\\
\end{align}$$
where $\Re{(n)}\gt1$. The last equality follows from the famous Bose integral (See proofs at 1, 2 and 3). The answer required is then given by
$$I_4=\zeta(4)\,\Gamma(5)=\frac{\pi^4}{90}\cdot4!=\frac{4\pi^4}{15}$$
