Evaluate the integral $\int_0^{\infty}\frac{x^4e^x}{(e^x-1)^2} \, dx$ How do I integrate the following:
$$\int_0^{\infty}\frac{x^4e^x}{(e^x-1)^2} \, dx$$
I have tried everything from integrating by parts to simply expanding the denominator, but it always makes it much more complex. No idea how to solve this.
 A: If you are familiar with the Riemann Zeta Function, and also note that the integral can be written as:
$$\displaystyle\int_0^{\infty} \dfrac{x^4e^{-x}}{(1-e^{-x})^2} \: dx = \displaystyle\sum_{k=1}^{\infty} k\displaystyle\int_0^{\infty}x^4 e^{-kx} \:dx = \displaystyle\sum_{k=1}^{\infty}k \cdot \dfrac{24}{k^5} = 24 \zeta(4) = \dfrac4{15} \pi^4$$
A: This is just a sequitur to Yagna Patel's answer. To prove $\zeta(4)=\frac{\pi^4}{90}$ is equivalent to proving:
$$ \sum_{n\geq 0}\frac{1}{(2n+1)^4}=\frac{\pi^4}{96},\tag{1} $$
since $\zeta(4)=\sum_{n\geq 0}\frac{1}{(2n+1)^4}+\sum_{n\geq 1}\frac{1}{(2n)^4}$ gives $\zeta(4)=\frac{16}{15}\sum_{n\geq 0}\frac{1}{(2n+1)^4}$.
If we recall that the Fourier (cosine) series of $f(x)=\frac{\pi}{2}-|x|$ over $(-\pi,\pi)$ is given by:
$$ f(x) = \frac{4}{\pi}\sum_{n\geq 0}\frac{\cos((2n+1)x)}{(2n+1)^2}\tag{2} $$
Parseval's identity then gives:
$$ \int_{-\pi}^{\pi}f(x)^2\,dx = \frac{16}{\pi}\sum_{n\geq 0}\frac{1}{(2n+1)^4}\tag{3}$$
hence:
$$ \sum_{n\geq 0}\frac{1}{(2n+1)^4}=\frac{\pi}{8}\int_{0}^{\pi}f(x)^2\,dx = \frac{\pi}{8}\cdot\frac{\pi^3}{12}=\frac{\pi^4}{96}\tag{4}$$
as wanted.
A: As I have once heard, there is more than one way to skin an integral.  And so herein, we present an approach that relies on the Polylogarithm Functions.  We will engage in a series of four integration by parts that will culminate in the answer $24\text{Li}_4(1)=24\zeta(4)$.  
To that end, we begin with the integral of interest $I$, which is given by
$$I=\int_0^{\infty}\frac{x^4e^x}{(e^x-1)^2}\,dx$$
Enforcing the substitution $x\to -\log (1-x)$ yields 
$$I=\int_0^1\frac{\log^4 (1-x)}{x^2}\,dx \tag 1$$
We now begin a series of four straightforward integration by parts applications.  

First, letting $u=\log^4(1-x)$ and $v=-1/x$ in $(1)$ yields
$$I=-4\int_0^1\frac{\log^3 (1-x)}{x}\,dx \tag 2$$
where we tacitly applied $\lim_{x\to 1}\left(\log^4(1-x)(1-1/x)\right)=0$ to arrive at $(2)$.

Second, letting $u=-4\log^3(1-x)$ and $v=\log x$ in $(2)$ yields
$$I=-12\int_0^1 \log^2(1-x)\frac{\log x}{1-x}\,dx \tag 3$$

Third, letting $u=-12\log^2(1-x)$ and $v=\text{Li}_2(1-x)$ in $(3)$ yields
$$I=-24\int_0^1 \log(1-x)\frac{\text{Li}_2(1-x)}{1-x}\,dx \tag 4$$

Fourth, letting $u=-24\log(1-x)$ and $v=-\text{Li}_3(1-x)$ in $(4)$ yields
$$\begin{align}
I&=-24\int_0^1 \frac{\text{Li}_3(1-x)}{1-x}\,dx\\\\ \tag 5
&=24\text{Li}_4(1)\\\\
&=24\zeta(4)\\\\
&=24\frac{\pi^4}{90}
\end{align}$$

Thus, we have 
$$\bbox[5px,border:2px solid #C0A000]{I=\frac{4\pi^4}{15}}$$
as expected and as easy as one, two, three, four!
A: I am going to evaluate the integral with $x^n$ in general.
$$
\begin{aligned}
\int_{0}^{\infty} \frac{x^{n} e^{x}}{\left(e^{x}-1\right)^{2}} d x =&-\int_{0}^{\infty} x^{n} d\left(\frac{1}{e^{x}-1}\right) \\
=&-\left[\frac{x^{n}}{e^{x}-1}\right]_{0}^{\infty}+n \int_{0}^{\infty} \frac{x^{n-1}}{e^{x}-1} d x \\
=& n \int_{0}^{\infty} \frac{x^{n-1} e^{-x}}{1-e^{-x}} d x \\
=& n \sum_{k=0}^{\infty} \int_{0}^{\infty} x^{n-1} e^{-(k+1) x} d x \\
=& n \sum_{k=0}^{\infty} \int_{0}^{\infty} x^{n-1} d\left(\frac{e^{-(k+1) x}}{-(k+1)}\right) \\
=& n \sum_{k=0}^{\infty}\left[-\frac{x^{n} e^{-(k+1) x}}{k+1}\right]_{0}^{\infty}+\frac{n(n-1)}{k+1} \sum_{k=0}^{\infty} \int_{0}^{\infty} x^{n-2} e^{-(k+1) x} d x
\end{aligned}
$$
Keeping on the integration by parts, we get
$$
\begin{aligned}
\int_{0}^{\infty} \frac{x^{n} e^{x}}{\left(e^{x}-1\right)^{2}} d x &=n ! \sum_{k=0}^{\infty} \frac{1}{(k+1)^{n}} =n ! \zeta(n)
\end{aligned}
$$
Back to our integral, putting $n=4$ yields
$$\int_{0}^{\infty} \frac{x^{4} e^{x}}{\left(e^{x}-1\right)^{2}} d x =4 ! \sum_{k=0}^{\infty} \frac{1}{(k+1)^{4}} =4 ! \zeta(4)=\frac{4 \pi^{4}}{15} $$
