Evaluate triple integrals $\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sqrt[3]{\log{(xyz)}}dxdydz$ I am trying to evaluate this integral:
$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sqrt[3]{\log{(xyz)}}dxdydz$$
Honestly, I have no ideas to deal with this.
I hope I can be helped by everyone.
I just need a hint to process; thank you.

My second solution:
Using the identity ($q > 0, \alpha \in (0, 1)$)
$$q^\alpha = \frac{\alpha}{\Gamma(1 - \alpha)}\int_0^\infty \frac{1 - \mathrm{e}^{-qs}}{s^{1 + \alpha}}\,\mathrm{d} s,$$
we have
$$\sqrt[3]{\ln \frac{1}{xyz}}
 = \frac{1}{3\Gamma(2/3)}\int_0^\infty \frac{1 - (xyz)^s}{s^{4/3}}\,\mathrm{d} s.$$
Denote the integral by $I$.
We have
\begin{align*}
 I &= - \int_0^1 \int_0^1 \int_0^1 \frac{1}{3\Gamma(2/3)}\int_0^\infty \frac{1 - (xyz)^s}{s^{4/3}}\,\mathrm{d} s \,\mathrm{d}x \,\mathrm{d} y\, \mathrm{d} z \\
 &= - \frac{1}{3\Gamma(2/3)}\int_0^\infty 
 \frac{1}{s^{4/3}}\left(\int_0^1 \int_0^1 \int_0^1 (1 -x^s y^s z^s) \,\mathrm{d}x \,\mathrm{d} y\, \mathrm{d} z\right)\,\mathrm{d} s\\
 &= - \frac{1}{3\Gamma(2/3)}\int_0^\infty 
 \frac{1}{s^{4/3}}\left(1 - \frac{1}{(s + 1)^3}\right)\,\mathrm{d} s \\
 &= -\frac{1}{3\Gamma(2/3)}\int_0^\infty 
 \frac{s^2 + 3s + 3}{s^{1/3}(s + 1)^3}\,\mathrm{d} s\\
 &= -\frac{1}{3\Gamma(2/3)}\int_0^\infty 
 \frac{3u(u^6 + 3u^3 + 3)}{(u^3 + 1)^3}\,\mathrm{d} u \qquad\qquad (s = u^3)\\
 &= - \frac{1}{3\Gamma(2/3)}\cdot \frac{28}{27}\pi\sqrt 3
\end{align*}
where we have used Fubini's theorem to interchange the order of integration; and the integral $\int_0^\infty 
\frac{3u(u^6 + 3u^3 + 3)}{(u^3 + 1)^3}\,\mathrm{d} u$ is easily evaluated using partial fraction decomposition.
(Note: $\frac{1}{3\Gamma(2/3)}\cdot \frac{28}{27}\pi\sqrt 3 = \frac12 \Gamma(10/3)$.)
