Evaluate triple integrals $\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sqrt[3]{\log{(xyz)}}dxdydz$

Question

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.

Answer 1

Let $f(u) := \sqrt[3]{\ln \frac{1}{u}}$. Let $$I := \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sqrt[3]{\log{(xyz)}}dxdydz.$$

We use the trick in Evaluate the triple integral $\int _0^1\int _0^1\int _0^1\frac{\ln ^n\left(xyz\right)}{1-xyz}dxdydz$.

We have \begin{align*} I &= -\int_0^1\int_0^1 \int_0^1 f(xyz)\,\mathrm{d} x\,\mathrm{d}y\,\mathrm{d} z\\ & = -\int_0^1 \int_0^1 \left(\frac{1}{yz}\int_0^{yz} f(t)\,\mathrm{d} t\right)\,\mathrm{d} y\,\mathrm{d} z \qquad\qquad\qquad (xyz = t)\\ &= - \int_0^1 \frac{1}{z} \left[\int_0^1 \left((\ln y)' \int_0^{yz} f(t)\,\mathrm{d} t \right)\,\mathrm{d} y\right]\,\mathrm{d} z\\ & = - \int_0^1 \frac{1}{z} \left[\left(\ln y \int_0^{yz} f(t)\,\mathrm{d} t \right)\Bigg\vert_0^1 - \int_0^1 f(yz) z \ln y \,\mathrm{d} y \right]\,\mathrm{d} z \qquad (\mathrm{IBP})\\ &= \int_0^1 \left(\int_0^1 f(yz) \ln y \,\mathrm{d} y\right) \,\mathrm{d} z \\ &= \int_0^1 \ln y\left(\int_0^1 f(yz) \,\mathrm{d} z\right)\,\mathrm{d} y \\ &= \int_0^1 \ln y \left(\frac{1}{y}\int_0^y f(s)\,\mathrm{d}s \right)\,\mathrm{d} y \qquad \qquad \qquad (yz = s)\\ &= \int_0^1 \left((2^{-1}\ln^2 y)' \int_0^y f(s)\,\mathrm{d} s\right)\,\mathrm{d} y\\ &= \left(2^{-1}\ln^2 y \int_0^y f(s)\,\mathrm{d} s\right)\Bigg\vert_0^1 - \int_0^1 2^{-1}\ln^2 y\, \, f(y)\,\mathrm{d} y \qquad\qquad (\mathrm{IBP})\\ &= -\int_0^1 \frac{1}{2}\ln^2 y\,\, f(y)\,\mathrm{d}y \\ &= -\int_0^1 \frac{1}{2}\ln^2 y \,\, \sqrt[3]{\ln \frac{1}{y}} \, \mathrm{d} y\\ &= -\int_0^\infty \frac12 v^{7/3}\mathrm{e}^{-v}\,\mathrm{d} v \qquad\qquad (y = \mathrm{e}^{-v})\\ &= - \frac12\, \Gamma(10/3). \end{align*}

Answer 2

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: $\frac{1}{3\Gamma(2/3)}\cdot \frac{28}{27}\pi\sqrt 3 = \frac12 \Gamma(10/3)$.)

Answer 3

We can at least compute two integrals $$I_1=\int_{0}^{1}\sqrt[3]{\log{(xyz)}}\,dx=\sqrt[3]{\log (y z)}+\frac{\sqrt[3]{-1}\, \Gamma \left(\frac{1}{3},-\log (yz)\right)}{3 y z}$$ $$I_2=\int_{0}^{1}I_1 \,dy=\frac{\sqrt[3]{\log (z)} \left(4 E_{\frac{2}{3}}(-\log (z))+3 \log (z) E_{\frac{2}{3}}(-\log (z))+12 z\right)}{9 z}$$ For the third integral, we can expand the integrand as a series around $z=1$ $$I_2=\frac{1+i \sqrt{3}}{2}\,\Gamma \left(\frac{1}{3}\right)\sum_{n=0}^\infty a_n (z-1)^n$$ where the $a_n$ form the sequence $$\left\{\frac{4}{9},-\frac{1}{9},-\frac{1}{18},\frac{1}{6},-\frac{1}{4},\frac{19}{60}, -\frac{67}{180},\frac{529}{1260},-\frac{1163}{2520},\frac{3769}{7560},-\frac{4021} {7560},\frac{46751}{83160},-\frac{49061}{83160},\cdots\right\}$$ giving $$I_3=\frac{1+i \sqrt{3}}{2} \Gamma \left(\frac{1}{3}\right)\sum_{n=0}^\infty \frac{(-1)^n }{n+1}a_n$$ The problem is that many terms would be required to arrive close to the solution.