How to evaluate $\int_0^\infty\frac{\frac{\pi^2}{6}-\operatorname{Li}_2\left(e^{-x}\right)-\operatorname{Li}_2\left(e^{-\frac{1}{x}}\right)}{x}dx$ I need to evaluate the following integral with a high precision:
$$
I=\int_{0}^{\infty}\left[%
{\pi^{2} \over 6} - {\rm Li}_2\left({\rm e}^{-x}\right)
-{\rm Li}_2\left({\rm e}^{-1/x}\right)\right]\,{{\rm d}x \over x},
$$
where ${\rm Li}_{2}$ denotes the
dilogarithm
$\displaystyle{%
\left(~\mbox{note that}\ {\rm Li}_{2}\left(1\right) = {\pi^{2} \over 6}~\right)}$.
Unfortunately, a numerical integration in my CAS is only able to produce $3$ stable digits $I \approx 3.77\ldots$ that I do not even sure to be provably correct. 
So, if only I were so lucky that a closed form existed for this integral, then, hopefully, it could be used to easily evaluate $I$ with a much higher precision. Could you suggest how to find a closed form ( if one exists )?
 A: $\newcommand{\+}{^{\dagger}}
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 \newcommand{\ic}{{\rm i}}
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 \newcommand{\isdiv}{\,\left.\right\vert\,}
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$\ds{\bbox[10px,#ffd]{I \equiv \int_{0}^{\infty}{\pi^{2}/6 -
{\rm Li}_{2}\pars{\expo{-x}}
-{\rm Li}_{2}\pars{\expo{-1/x}} \over x}\,\dd x}:\
{\large ?}}$

\begin{align}
\color{#c00000}{I} & =
-\int_{0}^{\infty}\ln\pars{x}\left[%
-{\rm Li}_{2}'\pars{\expo{-x}}\pars{-\expo{-x}} -\right.
\\ & \phantom{\color{#c00000}{I}  =
-\int_{0}^{\infty}\ln\left(x\right)\left[\right.}
\left.{\rm Li}_{2}'\pars{\expo{-1/x}}\pars{\expo{-1/x} \over x^{2}}\right]\!\dd x
\\[5mm] & =
-\int_{0}^{\infty}\ln\pars{x}{\rm Li}_{2}'\pars{\expo{-x}}\expo{-x}\,\dd x
\\[2mm] &\ +\int_{\infty}^{0}\ln\pars{1/x}{\rm Li}_{2}'\pars{\expo{-x}}x^{2}\expo{-x}
\,\pars{-\,{\dd x \over x^{2}}}
\\[5mm]&=-2\int_{0}^{\infty}\ln\pars{x}{\rm Li}_{2}'\pars{\expo{-x}}\expo{-x}\,\dd x
\\[5mm] & =
-2\int_{0}^{\infty}\ln\pars{x}\,
{{\rm Li}_{1}\pars{\expo{-x}} \over \expo{-x}}\expo{-x}\,\dd x
\\[5mm] & =
\color{#c00000}{2\int_{0}^{\infty}\ln\pars{x}
\ln\pars{1 - \expo{-x}}\,\dd x}
\end{align}
where we used $\ds{{\rm Li}_{s + 1}'\pars{z} = {{\rm Li}_{s}\pars{z} \over z}}$
and $\ds{{\rm Li}_{1}\pars{z} = -\log\pars{1 - z}}$. See
[this web page](http://en.wikipedia.org/wiki/Polylogarithm).
\begin{align}
\color{#00f}{\large I}&=2\int_{0}^{\infty}\ln\pars{x}
\ln\pars{1 - \expo{-x}}\,\dd x =
2\sum_{\ell = 1}^{\infty}
\pars{-\,{1 \over \ell}}\lim_{\mu \to 0}\partiald{}{\mu}\int_{0}^{\infty}x^{\mu}\expo{-\ell x}\,\dd x
\\[3mm]&=2\sum_{\ell = 1}^{\infty}
\pars{-\,{1 \over \ell}}\lim_{\mu \to 0}\partiald{}{\mu}
\bracks{\Gamma\pars{\mu + 1} \over \ell^{\mu + 1}}
\\[3mm]&=2\sum_{\ell = 1}^{\infty}
\pars{-\,{1 \over \ell}}\lim_{\mu \to 0}
\bracks{{\Gamma\pars{\mu + 1}\Psi\pars{\mu + 1} \over \ell^{\mu + 1}}
-{\Gamma\pars{\mu + 1}\ln\pars{\ell} \over \ell^{\mu + 1}}}
\\[3mm]&=2\sum_{\ell = 1}^{\infty}\pars{-\,{1 \over \ell}}\bracks{%
{-\gamma \over \ell} - {\ln\pars{\ell} \over \ell}}
=2\gamma\sum_{\ell = 1}^{\infty}{1 \over \ell^{2}}
+2\sum_{\ell = 1}^{\infty}{\ln\pars{\ell} \over \ell^{2}}
\\[3mm]&=\color{#00f}{\large 2\bracks{\gamma\zeta\pars{2} - \zeta\,'\pars{2}}}
\approx {\tt 3.7741}
\end{align}
A: $$
I=\int^{\infty}_0 x^{-1}\sum^{\infty}_{k=1}\frac{1-e^{-kx}-e^{-kx^{-1}}}{k^2}dx\\
=\sum^{\infty}_{k=1}k^{-2}\int^{\infty}_0 \frac{1-e^{-kx}-e^{-kx^{-1}}}{x}dx\\
=2\sum^{\infty}_{k=1}k^{-2}(\gamma+\log k)\\
=2\gamma\zeta(2)-2\zeta'(2)\\
=2\zeta(2)(12\log A-\log2\pi).
$$
Here $A$ is the Glaisher-Kinkelin constant.
Edit: 
$$\int^{\infty}_0 (1-e^{-kx}-e^{-kx^{-1}})\frac{dx}{x}\\
=\int^{1}_0 \frac{1-e^{-kx}}{x}dx-\int^{1}_0 \frac{e^{-kx^{-1}}}{x}dx+\int^{\infty}_1 \frac{1-e^{-kx^{-1}}}{x}dx-\int^{\infty}_1 \frac{e^{-kx}}{x}dx\\
=\int^{1}_0 \frac{1-e^{-kx}}{x}dx-\int^{\infty}_1 \frac{e^{-ky}}{y}dy+\int^{1}_0 \frac{1-e^{-ky}}{y}dy-\int^{\infty}_1 \frac{e^{-kx}}{x}dx\\
=2\int^{1}_0 \frac{1-e^{-kx}}{x}dx-2\int^{\infty}_1 \frac{e^{-kx}}{x}dx\\
=2\int^{k}_0 \frac{1-e^{-x}}{x}dx-2\int^{\infty}_k \frac{e^{-x}}{x}dx\\
=2(\operatorname{Ein}(k)-E_1(k))=2(\gamma+\log k).
$$
Here $E_1$ and $\operatorname{Ein}$ are exponential integrals, see §6.2 of DLMF.
