Proof that $\int_0^\infty\frac{\ln x}{x^3 - 1} \, dx = \frac{4 \pi^2}{27}$ I realise this question was asked here, but I'm not able to work with any of the answers. The hint given by my professor is

Integrate around the boundary of an indented sector of aperture $\frac{2 \pi}{3}$

but when I try that I can't figure out how to deal with the (divergent) integral along the radial line at angle $2 \pi / 3$. My issue with the accepted answer is that it uses the residue theorem where it doesn't apply, at least as we've learned it, since
$$z \mapsto \frac{\log^2z}{z^3 - 1}$$
has non-isolated singularities on the closed region bounded by the proposed contour (due to branch cuts), and I am not sure how to relate the integral along the real axis to one over a contour modified to avoid the branch cut.
For a fixed $\varepsilon > 0$, and for any $\delta \in (0, 1 - \varepsilon)$, we could let $\log_{-\delta / 2}$ be the branch of the logarithmic function with a cut along the ray $\operatorname{arg}z = -\delta / 2$ and define a contour which goes along the positive real axis from $\varepsilon$ to $1 - \delta$, a semicircle in the upper half plane of radius $\delta$ around $1$, the positive real axis from $1 + \delta$ to $2$, an arc of radius $2$ around $0$ with central angle $2 \pi - \delta$, the ray $\operatorname{arg}z = 2 \pi - \delta$ from $r = 2$ to $r = \varepsilon$, and finally an arc of radius $\varepsilon$ around $0$ back to $\varepsilon$. But then, for example, I don't know how to calculate the limit of integral along the arc of radius $\varepsilon$
$$\lim_{\delta \to 0}\int_0^{2 \pi - \delta}\frac{\log_{-\delta / 2}^2(\varepsilon e^{i \theta})}{\varepsilon^3 e^{3 i \theta} - 1} \varepsilon i e^{i \theta} \, d\theta.$$
If I instead try to first use the substitution $x = e^u$ on the real integral and then compute a contour integral, I still get a divergent integral that I don't know how to handle, this time along the top of an indented rectangle.

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\cal C}$ is a key-hole contour which "takes care" of the $\ds{\ln}$-branch cut with $\ds{0 < \arg\pars{z} < 2\pi}$.
Integrand poles are $\ds{\expo{2\pi\ic/3}}$ and
$\ds{\expo{4\pi\ic/3}}$.

Residue Theorem yields:
\begin{align}
\oint_{\cal C}{\ln^{2}\pars{z} \over z^{3} - 1}\,\dd z& =
2\pi\ic\bracks{%
\pars{{2\pi \over 3}\,\ic}^{2}\,{\expo{2\pi\ic/3} \over 3} +
\pars{{4\pi \over 3}\,\ic}^{2}\,{\expo{4\pi\ic/3}  \over 3}}
\\[5mm] & = -\,{4 \over 27}\pars{3\root{3} - 5\ic}\pi^{3}
\end{align}

Note that the singularity at $\ds{z = 1}$ lies on the
$\ds{\ln}$-branch cut. Hereafter we properly deals with this fact ( see below ).
Namely,
\begin{align}
&\bbox[5px,#ffd]{\oint_{\cal C}{\ln^{2}\pars{z} \over z^{3} - 1}\,\dd z}
\\[5mm] = &\
\int_{0}^{\infty}{\ln^{2}\pars{x} \over
\pars{x - 1 + \ic 0^{+}}\pars{x^{2} + x + 1}}\,\dd x
\\[2mm] + &\
\int_{\infty}^{0}{\bracks{\ln\pars{x} + 2\pi\ic}^{2} \over
\pars{x - 1 - \ic 0^{+}}\pars{x^{2} + x + 1}}\,\dd x
\\[5mm] = &\
\mrm{P.V.}\int_{0}^{\infty}{\ln^{2}\pars{x} \over
\pars{x - 1}\pars{x^{2} + x + 1}}\,\dd x
\\[2mm] - &\
\mrm{P.V.}\int_{0}^{\infty}{\ln^{2}\pars{x} + 4\pi\ic\ln\pars{x} - 4\pi^{2} \over
\pars{x - 1}\pars{x^{2} + x + 1}}\,\dd x
+ {4\pi^{3} \over 3}\,\ic
\\[5mm] = &\
-4\pi\ic\int_{0}^{\infty}{\ln\pars{x} \over x^{3} - 1}\,\dd x +
4\pi^{2}\
\underbrace{\mrm{P.V.}\int_{0}^{\infty}{\dd x \over x^{3} - 1}
\,\dd x}_{\ds{-\,{\root{3} \over 9}\,\pi}}\  +\
{4\pi^{3} \over 3}\,\ic
\end{align}

\begin{align}
&\int_{0}^{\infty}{\ln\pars{x} \over x^{3} - 1}\,\dd x
\\[2mm] = &\
{-\,\pars{4/27}\pars{3\root{3} - 5\ic}\pi^{3} -
4\pi^{2}\pars{-\root{3}\pi/9} - 4\pi^{3}\ic/3 \over -4\pi\ic}
\\[2mm] & =
\bbx{4\pi^{2} \over 27} \approx 1.4622 \\ &
\end{align}
