Evaluate $\int_0^{\infty} \frac{x^2}{e^x-1} dx$ The problem is to show that the improper integral $I = \int_0^{\infty} \frac{x^2}{e^x-1} dx$ converges to $2\sum_1^{\infty} \frac1{n^3}$.
Previously, I computed the following integral:
$$f(x) = -\int_0^x\frac{\log(1-t)}{t}dt = \sum_1^{\infty} \frac{x^n}{n^2}$$
which holds for $-1 \le x \le 1$.  From this, I deduce the following:
$$g(x) = \int_0^x \frac{f(t)}{t}dt = \sum_1^{\infty} \frac{x^n}{n^3}$$
which also holds for $-1 \le x \le 1$.  I am attempting to show that $I = g(1)$.
The first thing I do with $I$ is make the substitution $t = 1 - e^{-x}$, which yields:
$$I = \int_0^1 \frac{\log^2(1-t)}{t} dt$$
Then I attempt to integrate by parts, differentiating $\log(1-t)$ and integrating $\frac{\log(1-t)}{t}$, which yields:
$$I = -\log(1-t)f(t)|_0^1-\int_0^1\frac{f(t)}{1-t}dt $$
That doesn't work.  The first term doesn't converge.  The upper bound evaluates to $-\log(0)\frac{pi^2}6$ and the second term is not g(1).  The denominator needs to be $t$, not $t-1$.  So I am not sure what to do next.
 A: So it's well known that $\sum_{n=0}^\infty x^n = \frac{1}{1-x}$ provided $|x|\leq 1$ but a perhaps less often used variant is that if $|x|>1$ then  $$\frac{1}{x-1}= \frac{1}{x}\frac{1}{1-\frac{1}{x}} = \frac{1}{x}\sum_{n=0}^\infty x^{-n} = \sum_{n=1}^\infty x^{-n}$$ This means that, for our integral, since $x\in (0,\infty)$ that means $e^x \in (1,\infty)$. So we can rewrite our integral as: $$ \int_0^\infty \frac{x^2}{e^x-1}dx = \int_0^\infty\left(\sum_{n=1}^{\infty}x^2e^{-nx}\right)dx$$ and since our sum converges within the boundary of the integral we can interchange the sum and the integral. So we consider: $$\int x^2e^{-nx}dx  = \frac{-e^{-nx}(n^2x^2+2nx+2)}{n^3}$$ which is obtained by using integration by parts twice. Then plugging in boundaries gives our desired result:
$$ \int_0^\infty \frac{x^2}{e^x-1}dx = 2\sum_{n=1}^\infty \frac{1}{n^3}$$
A: Let's proceed along a similar line as in the OP.  We enforce the substitution $x\to -\log(x)$ so that 
$$\int_0^\infty \frac{x^2}{e^x-1}\,dx=\int_0^1\frac{\log^2(x)}{1-x}\,dx \tag 1$$
Integrating by parts the right-hand side of $(1)$ with $u=\log^2(x)$ and $v=\log(1-x)$ yields
$$\begin{align}
\int_0^1\frac{\log^2(x)}{1-x}\,dx&=2\int_0^1\frac{\log(x)\log(1-x)}{x}\,dx\\\\
&=2\int_0^1\left(\int_1^x \frac{1}{t}\,dt\right)\frac{\log(1-x)}{x}\,dx\\\\
&=2\int_0^1\frac{1}{t}\left(-\int_0^t\frac{\log(1-x)}{x}\,dx\right)\,dt\\\\
&=2\int_0^1\frac{\text{Li}_2(t)}{t}\,dt\\\\
&=2\text{Li}_3(1)\\\\
&=2\zeta(3)
\end{align}$$
where have made use of the relationships between $(i)$ the dilogarithm function $\text{Li}_2(x)$ and the trilogarithm function $\text{Li}_3(x)$ and $(ii)$ the polylogarithm function $\text{Li}_s(1)=\zeta(s)$ (for $\text{Re}(s)>1$).
A: $$-\int_0^\infty x^2 * (1-e^{x}) dx = -\int_0^\infty x^2 \sum_{n=1}^\infty\begin{equation*}
\binom{-1}{n}
\end{equation*}(-e^{x})^n dx = \int_0^\infty x^2\sum_{n=1}^\infty  e^{nx} dx= \sum_{n=1}^\infty \int_0^\infty x^2 * e^{nx}dx $$
put nx=-y
dx=-1/n dy 
$$\sum_{n=1}^\infty 1/n^3 \int_0^\infty e^{-y}*y^2 dy= \sum_{n=1}^\infty 1/n^3 \Gamma(3)=2\sum_{n=1}^\infty 1/n^3   $$ 
