<s>[INST] Prove $\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$ Proof of the integral
$$\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$$
I try to substitute $u = \ln x$. Then $x = e^u,\>\mathrm{d}x = e^u\mathrm{d}u$ and the limits $(0,\infty)\to (-\infty,\infty)$.
The integral becomes $$\int_{-\infty}^\infty \frac{ue^u}{e^{3u}-1}\mathrm{d}u.$$
 [/INST] I thought it might be instructive to present an approach that relies on straightforward contour integration.  To that end, we now proceed.
Let $I$ be the integral given by
$$I=\int_0^\infty \frac{\log(x)}{x^3-1}\,dx$$
Now, moving to the complex plane, we analyze the contour integral $J$ given by
$$J=\oint_C \frac{\log^2(z)}{z^3-1}\,dz$$
where $C$ is the classical keyhole contour with a semi-circular deformation at $z=1+i0^-$.  That is, $J$ is given by
$$\begin{align}
J&=\int_0^\infty \frac{\log^2(x)}{x^3-1}\,dx-\text{PV}\int_0^\infty\frac{(\log(x)+i2\pi)^2}{x^3-1}\,dx+i\frac{4\pi^3}3\\\\
&=-i4\pi I+4\pi^2\text{PV}\int_0^\infty \frac1{x^3-1}\,dx+i\frac{4\pi^3}3\\\\
&=-i4\pi I-\frac{4\pi^3}{3\sqrt3}+i\frac{4\pi^3}3\tag1
\end{align}$$
Rearranging $(1)$ reveals that the integral of interest can by written in terms of $J$ as
$$I=i\frac1{4\pi}J+i \frac{\pi^2}{3\sqrt 3}+\frac{\pi^2}{3}\tag2$$
Now, applying the residue theorem we see that $\frac{iJ}{4\pi}$ is given by
$$\begin{align}
\frac{iJ}{4\pi}&=-\frac12 \left(\text{Res}\left(\frac{\log^2(z)}{z^3-1}, z=e^{i2\pi/3}\right)+\text{Res}\left(\frac{\log^2(z)}{z^3-1}, z=e^{i4\pi/3}\right)\right)\\\\
&=-\frac{4\pi^2}{27}\left(\frac54+i\frac{3\sqrt{3}}{4}\right)\tag3
\end{align}$$
Using $(3)$ in $(2)$ yields the coveted result
$$I=\frac{4\pi^2}{27}$$
as expected!
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