= Solution
Use the positive orientation shown in the source: the upper bank runs outwards and the lower bank returns towards one. On those banks the <logarithm branch> has values $\ell-i\pi$ and $\ell+i\pi$, respectively, with $\ell=\log(x-1)$. Their squared difference is $-4i\pi\ell$. The <residue theorem> therefore gives
$$
2\pi i=\lim\oint_\Gamma\frac{L(z)^2}{z^3}\,dz=-4\pi i\int_1^\infty\frac{\log(x-1)}{x^3}\,dx,
$$
and hence
$$
\boxed{\int_1^\infty\frac{\log(x-1)}{x^3}\,dx=-\frac12.}
$$
The large-circle contribution is $O(R^{-2}\log^2R)$ and the small circle about one contributes $O(\varepsilon|\log\varepsilon|^2)$, both tending to zero. These estimates justify replacing the closed <contour integral> by its two cut-bank <integrals>.
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