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

$\def\Li{{\rm{Li}}}$How to evaluate the following integral$${\large\int_0^1} {\frac{{\left[ {\Li_2\left( {\frac{1}{2}} \right) - \Li_2\left( \frac{1 + x}{2} \right)} \right]\ln \left( {1 - x} \right)}}{{1 + x}}}\, dx$$

Answer 1

$\def\Li{{\rm{Li}}}$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}

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\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}

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\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}

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\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}

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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. $$I\approx0.3835634533628654520513648150132866552816792868280275$$