How do I start this integral problem: $\int_0^1 \frac{\ln^3 u}{1-u} du = -\frac{\pi^4}{15} $? 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? 
 A: $\displaystyle \int_0^1 \frac{\ln^3 u}{1-u} \, du = \int_0^1 \ln^3{u} \sum_{k \ge 0} u^k \,{du}= \sum_{k \ge 0} \int_0^1 u^k \ln^3{u} \, {du}$
Let $\displaystyle f(k) = \int_0^1 u^k \, {du} = \frac{1}{(1+k)}$. Differentiating w.r.t. $k$
We have  $\displaystyle f^{(3)}(k) = \int_0^1 u^k \ln^3(u)\,{du} = \frac{-6}{(1+k)^4} $.
Hence $\displaystyle I = -6 \sum_{k\ge 0} \frac{1}{(1+k)^4} = - 6\zeta(4) = -\frac{\pi^4}{15}. $
A: $$
\begin{align*}
\int_0^1 \frac{\ln^3(x)}{1-x}dx &= \int_0^1\ln^3(x)\sum_{n=0}^\infty x^n \; dx\\ 
&= \sum_{n=0}^\infty \int_0^1  x^n \ln^3(x) \; dx \\
&= -6\sum_{n=0}^\infty \frac{1}{(n+1)^4} \\
&= -6 \zeta(4) \\
&= -6 \times \frac{\pi^4}{90} \\ 
&= -\frac{\pi^4}{15}
\end{align*}
$$
A: 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 $-6\zeta(4)=-\pi^4/15$.
