Problem: Prove that $$\int_{0}^{\infty}\sqrt[3]{(e^x-1)^2}\cdot \frac{\mathrm dx}{1-2\cosh x}=-\frac{4}{3}\pi\sin\left(\frac{2}{9}\pi\right)\,.$$
What method can be use to tackle this prove this problem?
Attempt.
By letting $z=e^x$, I get $$I:=\int_{0}^{\infty}\,\sqrt[3]{(e^x-1)^2}\, \frac{\mathrm dx}{1-2\cosh x}=\int_{1}^{\infty}\,\frac{(z-1)^{\frac{2}{3}}}{1-z-\frac{1}{z}}\,\frac{\text{d}z}{z}.$$ Therefore, $$I=-\int_{1}^{\infty}\,\frac{(z-1)^{\frac23}}{z^2-z+1}\,\text{d}z\,.$$ This doesn't seem to be going anywhere.
The substitution $t=e^x-1$ transforms the integral into:
\begin{align*} I &= -\int_0^\infty \frac{t^{2\over 3}}{1+t+t^2}dt \end{align*}
Now, consider the principal branch of $f(z) = \dfrac{(-z)^{\frac{2}{3}}}{1+z+z^2}$ and integrate it around a key-hole contour (see the picture below) which consists of a small circle about the origin of radius $\epsilon$ say, extending to a line segment parallel and close to the positive real axis but not touching it, to an almost full circle of radius $R$, returning to a line segment parallel, close, and below the positive real axis in the negative sense, returning to the small circle in the middle. As $R\to \infty$ and $\epsilon\to 0^+$, only the integrals above and below the positive real axis survive.
The residues at the poles $z=e^{2i\pi/3}, \; e^{-2i\pi/3}$ are computed as follows: \begin{align*} \mathop{\text{Res}}\limits_{z=e^{2i\pi/3}} \; f(z) &= \lim_{z\to e^{2i\pi/3}}(z-e^{2i\pi/3}) \frac{(-z)^{\frac{2}{3}}}{1+z+z^2} = \frac{e^{\frac{2}{3}\log (-\exp(\frac{2i\pi}{3}))}}{2i\sin\left(\frac{2\pi}{3}\right)} = \frac{e^{-\frac{2i\pi}{9}}}{i\sqrt{3}} \\ \mathop{\text{Res}}\limits_{z=e^{-2i\pi/3}} \; f(z) &= \lim_{z\to e^{-2i\pi/3}}(z-e^{-2i\pi/3}) \frac{(-z)^{\frac{2}{3}}}{1+z+z^2} = -\frac{e^{\frac{2}{3}\log (-\exp(-\frac{2i\pi}{3}))}}{2i\sin\left(\frac{2\pi}{3}\right)} = -\frac{e^{\frac{2i\pi}{9}}}{i\sqrt{3}} \end{align*}
Therefore, using the Residue Theorem,
\begin{align*} \int_0^\infty \frac{e^{\frac{2}{3}\ln(x)-\frac{2i\pi}{3}}}{1+x+x^2}dx - \int_0^\infty \frac{e^{\frac{2}{3}\ln(x)+\frac{2i\pi}{3}}}{1+x+x^2}dx &= 2\pi i \left(\mathop{\text{Res}}\limits_{z=e^{2i\pi/3}} \; f(z) + \mathop{\text{Res}}\limits_{z=e^{-2i\pi/3}} \; f(z) \right) \\ \implies -2i\sin\left(\frac{2\pi}{3}\right)\int_0^\infty \frac{x^{2\over 3}}{1+x+x^2}dx &= -2\pi i \left(\frac{2}{\sqrt{3}}\sin\left(\frac{2\pi}{9}\right) \right) \\ \int_0^\infty \frac{x^{2\over 3}}{1+x+x^2}dx &= \frac{4\pi}{3} \sin\left(\frac{2\pi}{9}\right) \end{align*}
Therefore, $I=-\dfrac{4\pi}{3} \sin\left(\dfrac{2\pi}{9}\right)$, as desired.