taoigas

2022-01-22

A closed form for ${\int }_{0}^{1}\frac{\mathrm{ln}\left(-\mathrm{ln}x\right)l{i}^{2}x}{x}dx$

stamptsk

Expert

rakije2v

Expert

Change the integration variable $z=\mathrm{ln}x$. The integral becomes
$I={\int }_{\left\{-\mathrm{\infty }\right\}}^{0}\mathrm{ln}\left(-z\right)E{i}^{2}zdz,$
where $Eiz$ is the exponential integral:
$Eiz=-{\int }_{\left\{-z\right\}}^{\mathrm{\infty }}\frac{{e}^{-t}}{t}dt$
In this form the integral can be evaluated by Mathematica:

$I=-\frac{{\pi }^{2}}{6}-{\mathrm{ln}}^{2}2-2\left(1+\gamma \right)\mathrm{ln}2,$

where $\gamma$ is the Euler-Mascheroni Constant.

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