rigliztetbf

2022-06-22

Integral ${\int }_{0}^{\mathrm{\infty }}\frac{1}{\sqrt[3]{x}}\left(1+\mathrm{log}\frac{1+{e}^{x-1}}{1+{e}^{x}}\right)dx$
Is it possible to evaluate this integral in a closed form?

Govorei9b

We first remark the following identity:
$\begin{array}{rl}{\int }_{0}^{\mathrm{\infty }}\frac{{x}^{s-1}}{z{e}^{x}-1}\phantom{\rule{thinmathspace}{0ex}}dx& ={\int }_{0}^{\mathrm{\infty }}\frac{{z}^{-1}{x}^{s-1}{e}^{-x}}{1-{z}^{-1}{e}^{-x}}\phantom{\rule{thinmathspace}{0ex}}dx\\ & =\sum _{n=1}^{\mathrm{\infty }}{z}^{-n}{\int }_{0}^{\mathrm{\infty }}{x}^{s-1}{e}^{-nx}\phantom{\rule{thinmathspace}{0ex}}dx\\ & =\mathrm{\Gamma }\left(s\right)\sum _{n=1}^{\mathrm{\infty }}\frac{{z}^{-n}}{{n}^{s}}=\mathrm{\Gamma }\left(s\right){\mathrm{L}\mathrm{i}}_{s}\left({z}^{-1}\right)\end{array}$
which initially holds for $|z|>1$, and then extends holomorphically for ${z}^{-1}\notin \left(1,\mathrm{\infty }\right]$ since both sides define holomorphic functions on this region. Now, by integrating by parts,
$\begin{array}{rl}{\int }_{0}^{\mathrm{\infty }}\frac{1}{\sqrt[3]{x}}\left(1+\mathrm{log}\frac{1+{e}^{x-1}}{1+{e}^{x}}\right)\phantom{\rule{thinmathspace}{0ex}}dx& =\frac{3}{2}{\int }_{0}^{\mathrm{\infty }}{x}^{2/3}\left(\frac{1}{\left(-1\right){e}^{x}-1}-\frac{1}{\left(-{e}^{-1}\right){e}^{x}-1}\right)\phantom{\rule{thinmathspace}{0ex}}dx\\ & =\frac{3}{2}\mathrm{\Gamma }\left(\frac{5}{3}\right)\left\{{\mathrm{L}\mathrm{i}}_{5/3}\left(-1\right)-{\mathrm{L}\mathrm{i}}_{5/3}\left(-e\right)\right\}\\ & =-\frac{3}{2}\mathrm{\Gamma }\left(\frac{5}{3}\right)\left\{\left(1-{2}^{-2/3}\right)\zeta \left(5/3\right)+{\mathrm{L}\mathrm{i}}_{5/3}\left(-e\right)\right\}.\end{array}$
I can hardly believe that ${\mathrm{L}\mathrm{i}}_{5/3}\left(-e\right)$ can be simplified further.

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