How to prove $\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{dx dy dz}{1-xyz}=\zeta(3)$

Question

The triple integral $$\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{dx dy dz}{1-xyz}=\zeta(3) \dots (1)$$ is not separable in $x,y,z$ and the integral representation of reciprocal: $\frac{1}{1-xyz}=\int_{0}^{1} t^{-xyz} dt$ also doesn't help separation of the integrand. I want help to prove (1).

Answer 1

Expand out as a geometric progression. You get $$\sum_{n=0}^\infty\int_0^1\int_0^1\int_0^1x^n y^nz^n\,dx\,dy\,dz.$$ Now do the integral.

Answer 2

$$I=\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{dx dy dz}{1-xyz}= \int_{0}^{1} \int_{0}^{1} \frac{ln(1-yz)}{yz} dy dz =\int_{0}^{1} \int_{0}^{z} \frac{\ln(1-u)}{u} \frac{du dz}{z}, u=yz$$ $$ \implies I=\int_{0}^{1}\int_{0}^{z}[1+u/2+u^2/3+u^3/4+....]\frac{ du dz}{z}= \int_{0}^{1} \sum_{k=1}^{\infty}\frac{z^{k-1}}{k^2} dz= \sum_{k=1}^{\infty} \frac{1}{k^3}=\zeta(3).$$

Answer 3

\begin{align}\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{dx dy dz}{1-xyz}&=-\int_0^1\int_0^1 \frac{\ln u}{1-uz}dudz\\ &=-\frac{1}{2}\int_0^1\int_0^1 \frac{\ln (uz)}{1-uz}dudz\\ &=\frac{1}{2}\int_0^1 \frac{\ln^2 v}{1-v}dv\\ &=\frac{1}{2}\times 2\zeta(3)\\ &=\boxed{\zeta(3)} \end{align}

NB: i have used the following results: If $f$ is a continuous function on $[0;1]$ then $\displaystyle \int_0^1 \int_0^1 f(xy)dxdy=-\int_0^1 f(z)\ln zdz$

$\displaystyle \int_0^1 \frac{\ln^2 x}{1-x}dx=2\zeta(3)$