Calculating $\int_0^{\infty} \frac{\log^2(1 - e^{-x})\:x^5}{e^x - 1} \: dx $ I am having trouble calculating the following improper integral:
$$\displaystyle \int\limits_0^{\infty} \frac{\log^2(1 - e^{-x})x^5}{e^x - 1} \, dx $$
Can someone give me a way that I can calculate this?
 A: Hint. Here is an approach using the Euler beta function. 
By the change of variable, $u=e^{-x}$, you get
$$
\begin{align}
\int_0^{\infty} \frac{\log^2(1 - e^{-x})x^5}{e^x - 1} \, dx&=-\int_0^1\frac{\log^2(1-u)\log^5u}{1-u}\:du\\\\
&=-\left.\partial_b^2\partial_a^5\left(\int_{0}^{1} {u}^{a-1}(1-u)^{b-1}du\right)\right|_{a=1,b=0}\\\\
&=-\left.\partial_b^2\partial_a^5\left(\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\right)\right|_{a=1,b=0}\\\\
&=\frac{4}{3} \pi ^4 \zeta(3)+20 \pi ^2 \zeta(5)-360 \zeta (7).
\end{align}
$$
A: Given
$$I = \displaystyle \int\limits_0^{\infty} \frac{\log^2(1 - e^{-x})x^5}{e^x - 1} \, dx $$
then let $t = e^{-x}$ to obtain
\begin{align}
I &= - \int_{1}^{0} ( - \ln t)^{5} \, \ln^{2}(1-t) \, \frac{dt}{1-t} \\
&= - \int_{0}^{1} \frac{\ln^{5}t \, \ln^{2}(1-t)}{1-t} \, dt \\
&= \int_{0}^{1} \ln^{5}t \, \left( \frac{1}{3} \, \frac{d}{dt} \ln^{3}(1-t)\right) \, dt \\
&= \left[ \frac{1}{3} \ln^{5}t \, \ln^{3}(1-t) \right]_{0}^{1} - \frac{5}{3} \, \int_{0}^{1} \frac{\ln^{3}(1-t) \, \ln^{4}t}{t} \, dt \\
&= - \frac{5}{3} \, \int_{0}^{1} \frac{\ln^{3}(1-t) \, \ln^{4}t}{t} \, dt
\end{align}
From here consider the Beta function in the form
\begin{align}
B(x,y) = \int_{0}^{1} t^{x-1} \, (1-t)^{y-1} \, dt
\end{align}
for which
\begin{align}
I &= - \frac{5}{3} \, \partial_{x}^{4} \partial_{y}^{3} \, \left. B(x,y) \right|_{x=0, y=1} \\
&= 5! \, (\zeta(4) \, \zeta(3) + \zeta(2) \, \zeta(5) - 3 \, \zeta(7))
\end{align}
