Integration of $\int_0^1 \int_0^1 \int_0^1 \frac{\ln(1-xyz)}{xyz} dx dy dz = - \frac{\pi^4}{90}$ Using wolfram alpha,
$$-\int_0^1 \int_0^1 \int_0^1 \frac{\ln(1-xyz)}{xyz} dx dy dz = \frac{\pi^4}{90}$$
How do you solve this integral, and why is it equal to $\zeta(4)$? I can see that
$$-\frac{\ln(1-z)}{z} = 1 + \frac{z}{2} + \frac{z^2}{3} + \ldots$$
 A: $$I=\int_0^1 \int_0^1 \int_0^1 \frac{\ln(1-xyz)}{xyz} dx dy dz $$
$$I=-\int_{0}^{1} \int_0^1 \int_0^1 \int_0^1 \frac{1}{1-xyzt} dx dy dz dt$$
$$I=-\int_{0}^{1} \int_0^1 \int_0^1 \int_0^1 \sum_{n=0}^{\infty} x^ny^nz^nt^n dx dy dz dt$$
$$\implies I=-\sum_{n=0}^{\infty}\left(\int_{0}^{1} u^n du \right)^4=-\sum_{n=0}^{\infty}\frac{1}{(n+1)^4}=-\zeta(4)=-\frac{\pi^4}{90}.$$
A: If $f$ is a continuous function on $[0,1]$ then $\displaystyle \int_0^1\int_0^1 f(xy)dxdy=-\int_0^1 f(u)\ln u du$
(proof: integration by parts)
Application:
\begin{align}I&=\int_0^1 \int_0^1 \int_0^1 \frac{\ln(1-xyz)}{xyz} dx dy dz\\
&\overset{f(x,y)=\frac{\ln(1-xyz)}{xyz}}=-\int_0^1 \left(\int_0^1 \frac{\ln u\ln(1-uz)}{uz}du\right)dz\\
&=-\frac{1}{2}\int_0^1 \int_0^1\frac{\ln(uz)\ln(1-uz)}{uz}dudz\\
&\overset{f(u,z)=\frac{\ln(uz)\ln(1-uz)}{uz}}=\frac{1}{2}\int_0^1\frac{\ln^2 v\ln(1-v)}{v}dv\\
&\overset{\text{IBP}}=\frac{1}{6}\Big[\ln^3 v\ln(1-v)\Big]_0^1+\frac{1}{6}\int_0^1 \frac{\ln^3 v}{1-v}dv\\
&=\frac{1}{6}\int_0^1 \frac{\ln^3 v}{1-v}dv\\
&=\frac{1}{6}\times -6\zeta(4)\\
&=-\zeta(4)\\
&=\boxed{-\frac{\pi^4}{90}}
\end{align}
NB:
I assume that: $\displaystyle \int_0^1 \dfrac{\ln^3 v}{1-v}dv=-6\zeta(4)$ and $\displaystyle \zeta(4)=\dfrac{\pi^4}{90}$
