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Is it possible to evaluate this integral in a closed form? $$\int_0^\infty\frac{1}{\sqrt[3]{x}}\left(1+\log\frac{1+e^{x-1}}{1+e^x}\right)dx$$

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

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