We want to prove that: $$\frac{I}{J}=\frac{\int_0^\pi x^3\ln(\sin x)dx} {\int_0^\pi x^2\ln\left(\sqrt 2\sin x\right)dx}=\frac{3\pi}2$$
First, let's take the $I$ integral and perform the $x\to \pi-x$ substitution:
$$I=\int_0^\pi x^3\ln(\sin x)dx\overset{x\to\pi-x}=\int_0^\pi (\pi^3-3\pi^2x+3\pi x^2-x^3)\ln(\sin x)dx$$
In the $J$ integral we had an additional $\ln \sqrt 2$ for the $x^2$ term, thus we can also add it here:
$$\Rightarrow I=\int_0^\pi (\pi^3-3\pi^2x+3\pi x^2-x^3)\ln(\sin x)dx+ 3\pi(\underbrace{\ln \sqrt 2-\ln \sqrt 2}_{=0})\int_0^\pi x^2 dx$$
$$=\pi^3 \underbrace{\int_0^\pi \ln(\sin x)dx}_{=\mathcal L}-3\pi^2 \underbrace{\int_0^\pi x\ln(\sin x)dx}_{=\mathcal K}+3\pi J-I-{\pi^4}\ln \sqrt 2$$
$$\Rightarrow 2I=\left(\pi^3-\frac{3\pi^3}{2}\right)\int_0^\pi \ln(\sin x)dx+3\pi J-{\pi^4}\ln \sqrt 2$$
$$\require{cancel} 2I=\cancel{\frac{\pi^3}{2}\cdot 2\pi \ln \sqrt 2}+3\pi J-\cancel{\pi^4 \ln \sqrt 2}\Rightarrow I=\frac{3\pi}2J$$
Note that above we used:
$$\mathcal K=\int_0^\pi x\ln(\sin x)dx\overset{x\to \pi-x}=\int_0^\pi (\pi-x)\ln(\sin x)dx$$
$$\Rightarrow \mathcal K=\frac{\pi}{2}\underbrace{\int_0^\pi \ln(\sin x)dx}_{=\mathcal L}=\frac{\pi}{2}\mathcal L$$
$$\mathcal L=\int_0^\pi \ln(\sin x)dx=\int_0^\frac{\pi}{2} \ln(\sin x)dx+\int_0^\frac{\pi}{2} \ln(\cos x)dx$$
$$=\int_0^\frac{\pi}{2} \ln\left(\sin x\cos x\right)=\int_0^\frac{\pi}{2} \ln(\sin 2x)dx-\int_0^\frac{\pi}{2} \ln 2dx$$
$$=\frac12 \underbrace{\int_0^\pi \ln(\sin x) dx}_{=\mathcal L}-\ln\sqrt 2 \int_0^{\pi} dx\Rightarrow \mathcal L=-2\pi \ln\sqrt 2$$
Best Answer
According to Table of Integrals, Series, and Products Seventh Edition, I.S. Gradshteyn and I.M. Ryzhik, 3.326.1,
$\int_0^{\infty} e^{-x^u}dx = \dfrac1{u}\Gamma(\dfrac1{u}) = \Gamma(1+\dfrac1{u}) $ when $\Re(u) > 0$.
Letting $x = cy$, this becomes $c\int_0^{\infty} e^{-c^uy^u}dy = \Gamma(1+\dfrac1{u}) $.
Letting $c^u=a$, $a^{1/u}\int_0^{\infty} e^{-ay^u}dy = \Gamma(1+\dfrac1{u}) $, so $\int_0^{\infty} e^{-ay^u}dy = \dfrac1{a^{1/u}}\Gamma(1+\dfrac1{u}) $.
Putting $a = 1+i$, as Claude Leibovici suggested, and not worrying about this being legal,
$\begin{array}\\ \dfrac1{(1+i)^{1/u}}\Gamma(1+\dfrac1{u}) &=\int_0^{\infty} e^{-(1+i)y^u}dy\\ &=\int_0^{\infty} e^{-y^u}e^{-iy^u}dy\\ &=\int_0^{\infty} e^{-y^u}(\cos(-y^u)+i\sin(-y^u))dy\\ &=\int_0^{\infty} e^{-y^u}\cos(-y^u)dy+i\int_0^{\infty} e^{-y^u}\sin(-y^u)dy\\ \end{array} $
Since $1+i =\sqrt{2}e^{\pi i/4} $,
$\begin{array}\\ (1+i)^{-1/u} &=\sqrt{2^{-1/u}}e^{-\pi i/(4u)}\\ &=2^{-1/(2u)}(\cos(-\pi /(4u))-i\sin(-\pi /(4u)))\\ &=2^{-1/(2u)}(\cos(\pi /(4u))+i\sin(\pi /(4u)))\\ \end{array} $
$\int_0^{\infty} e^{-y^u}\sin(-y^u)dy =2^{-1/(2u)}\Gamma(1+\dfrac1{u})\sin(\pi /(4u)) $
and
$\int_0^{\infty} e^{-y^u}\cos(-y^u)dy =2^{-1/(2u)}\Gamma(1+\dfrac1{u})\cos(\pi /(4u)) $.
Putting $u=3$, $\int_0^{\infty} e^{-y^3}\sin(-y^3)dy =2^{-1/6}\Gamma(\frac43)\sin(\pi /12) $.
Wolfy says $\int_0^∞ e^{-y^3} \sin(-y^3) dy = -\frac{(\sqrt{3} - 1) Γ(4/3)}{2^{5/3}} ≈-0.205905 $ and $\sin(\pi/12) =\dfrac{\sqrt{3} - 1}{2 \sqrt{2}} $ so the results agree.