Integral $\int_0^\infty\frac{1}{\sqrt[3]{x}}\left(1+\log\frac{1+e^{x-1}}{1+e^x}\right)dx$ Is it possible to evaluate this integral in a closed form?
$$\int_0^\infty\frac{1}{\sqrt[3]{x}}\left(1+\log\frac{1+e^{x-1}}{1+e^x}\right)dx$$
 A: We first remark the following identity:
\begin{align*}
\int_{0}^{\infty} \frac{x^{s-1}}{ze^{x} - 1} \, dx
&= \int_{0}^{\infty} \frac{z^{-1}x^{s-1}e^{-x}}{1 - z^{-1}e^{-x}} \, dx \\
&= \sum_{n=1}^{\infty} z^{-n} \int_{0}^{\infty} x^{s-1} e^{-nx} \, dx \\
&= \Gamma(s) \sum_{n=1}^{\infty} \frac{z^{-n}}{n^{s}}
 = \Gamma(s)\mathrm{Li}_{s}(z^{-1})
\end{align*}
which initially holds for $|z| > 1$, and then extends holomorphically for $z^{-1} \notin (1, \infty]$ since both sides define holomorphic functions on this region. Now, by integrating by parts,
\begin{align*}
\int_{0}^{\infty} \frac{1}{\sqrt[3]{x}} \left(1+\log\frac{1+e^{x-1}}{1+e^{x}}\right) \, dx
&= \frac{3}{2} \int_{0}^{\infty} x^{2/3} \left( \frac{1}{(-1)e^{x} - 1} - \frac{1}{(-e^{-1})e^{x} - 1} \right) \, dx \\
&= \frac{3}{2} \Gamma\left(\frac{5}{3} \right) \left\{ \mathrm{Li}_{5/3}(-1) - \mathrm{Li}_{5/3}(-e) \right\} \\
&= -\frac{3}{2} \Gamma\left(\frac{5}{3} \right) \left\{ (1 - 2^{-2/3})\zeta(5/3) + \mathrm{Li}_{5/3}(-e) \right\}.
\end{align*}
I can hardly believe that $\mathrm{Li}_{5/3}(-e)$ can be simplified further.
