TITLE: Evaluate $\int_{0}^{1} \frac{\left[\rm{Li}_2\left(\frac{1}{2} \right)-\rm{Li}_2\left(\frac{1 + x}{2}\right)\right]\ln( 1 - x)}{1 + x}\,dx$
QUESTION [7 upvotes]: $\def\Li{{\rm{Li}}}$How to evaluate the following integral$${\large\int_0^1} {\frac{{\left[ {\Li_2\left( {\frac{1}{2}} \right) - \Li_2\left( \frac{1 + x}{2} \right)} \right]\ln \left( {1 - x} \right)}}{{1 + x}}}\, dx$$

REPLY [6 votes]: $\def\Li{{\rm{Li}}}$As suggested by M.N.C.E, let $t=\frac{1+x}{2}$, then we have
\begin{align}
&{\large\int_0^1} {\frac{{\left[\Li_2\left(\frac{1}{2}\right) - \Li_2\left(\frac{1+x}{2}\right)\right]\ln \left( {1 - x} \right)}}{{1 + x}}}\, dx={\large\int_\frac{1}{2}^1} {\frac{{\left[\Li_2\left(\frac{1}{2}\right) - \Li_2\left(t\right)\right]\ln \left( {2 - 2t} \right)}}{{t}}}\, dt\\[10pt]
I&=\Li_2\left(\frac{1}{2}\right)\int_\frac{1}{2}^1\frac{\ln(2-2t)}{t}\,dt-\int_\frac{1}{2}^1\frac{\Li_2(t)\ln(2-2t)}{t}\,dt\\[10pt]
I&=I_1+I_2-I_3-I_4
\end{align}
where
\begin{align}
I_1&=\Li_2\left(\frac{1}{2}\right)\ln2\int_\frac{1}{2}^1\frac{dt}{t}\\
&=\left.\Li_2\left(\frac{1}{2}\right)\ln2\ln\left( t\right)\right|_\frac{1}{2}^1\\
&=\Li_2\left(\frac{1}{2}\right)\ln^22\\
&=\frac{\pi^2}{12}\ln^22-\frac{\ln^42}{2}
\end{align}

\begin{align}
I_2&=\Li_2\left(\frac{1}{2}\right)\int_\frac{1}{2}^1\frac{\ln(1-t)}{t}\,dt\\
&=-\left.\Li_2\left(\frac{1}{2}\right)\Li_2\left( t\right)\right|_\frac{1}{2}^1\\
&=\Li_2^2\left(\frac{1}{2}\right)-\Li_2\left(\frac{1}{2}\right)\Li_2\left(1\right)\\
&=\frac{\ln^42}{4}-\frac{\pi^4}{144}
\end{align}

\begin{align}
I_3&=\ln2\int_\frac{1}{2}^1\frac{\Li_2(t)}{t}\,dt\\
&=\ln2\,\,\Li_3(t)\bigg|_\frac{1}{2}^1\\
&=\left[\Li_3(1)-\Li_3\left(\frac{1}{2}\right)\right]\ln2\\
&=\frac{\zeta(3)}{8}\ln2+\frac{\pi^2}{12}\ln^22-\frac{1}{6}\ln^42
\end{align}

\begin{align}
I_4&=\int_\frac{1}{2}^1\frac{\Li_2(t)\ln(1-t)}{t}\,dt\\
&=\left[\Li_2(t)\int\frac{\ln(1-t)}{t}\,dt\right]_\frac{1}{2}^1-\int_\frac{1}{2}^1\left[\Li_2'(t)\int\frac{\ln(1-t)}{t}\,dt\right]\,dt\\
&=-\Li_2^2(t)\bigg|_\frac{1}{2}^1-\int_\frac{1}{2}^1\frac{\Li_2(t)\ln(1-t)}{t}\,dt\\
&=-\frac{\Li_2^2(t)}{2}\bigg|_\frac{1}{2}^1\\
&=\frac{\ln^42}{8}-\frac{\pi^4}{96}-\frac{\pi^2}{24}\ln^22
\end{align}

Therefore
\begin{align}
{\large\int_0^1} {\frac{{\left[\Li_2\left(\frac{1}{2}\right) - \Li_2\left(\frac{1+x}{2}\right)\right]\ln \left( {1 - x} \right)}}{{1 + x}}}\, dx&=\frac{\pi^4}{288}-\frac{5}{24}\ln^42+\frac{\pi^2}{24}\ln^22-\frac{\zeta(3)}{8}\ln2
\end{align}
and the result numerically agrees.
$$I\approx0.3835634533628654520513648150132866552816792868280275$$