TITLE: How do I start this integral problem: $\int_0^1 \frac{\ln^3 u}{1-u} du = -\frac{\pi^4}{15} $?
QUESTION [6 upvotes]: How do I prove this?
$$\int_0^1 \frac{\ln^3 u}{1-u} du = -\frac{\pi^4}{15} $$ 
I'm guessing using Riemann zeta function? But then how do I start?

REPLY [13 votes]: Using $x=-\ln u$ gives
\begin{align}
\int_0^1\frac{(\ln u)^r}{1-u}\,du
&=(-1)^r\int_0^\infty\frac{x^r e^{-x}}{1-e^{-x}}\,dx\\
&=(-1)^r\sum_{n=1}^\infty\int_0^\infty x^r e^{-nx}\,dx\\
&=(-1)^r\sum_{n=1}^\infty\frac 1{n^{r+1}}\int_0^\infty x^r e^{-x}\,dx\\
&=(-1)^r\zeta(r+1)\Gamma(r+1).
\end{align}
When $r$ is a nonnegative integer, $\Gamma(r+1)=r!$. So when $r=3$ this
is $-6\zeta(4)=-\pi^4/15$.