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? $$\int_0^\infty\frac{1}{\sqrt[3]{x}}\left(1+\log\frac{1+e^{x-1}}{1+e^x}\right)dx$$

• Is there any reason for you to expect that this has a closed form? – Sangchul Lee Oct 29 '13 at 23:10
• @sos440 One should always believe there is one :) – Vladimir Reshetnikov Oct 29 '13 at 23:19

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,
I can hardly believe that $\mathrm{Li}_{5/3}(-e)$ can be simplified further.