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How can we evaluate the following integrals: $$\int_0^1\frac{\ln(1-x)}{x}\text{Li}_3\left(\frac{1 + x}{2} \right)\,dx\\ .\\ \int_0^1\frac{\ln^2(1-x)}{x}\text{Li}_2\left(\frac{1 + x}{2} \right)\,dx$$

Infiniticism
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xuce1234
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    This kind of OP will get no answer on Math SE. Ask one integral problem only & share your attempt here – Venus Oct 12 '14 at 13:38
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    @Venus: Nice sentiment, but wishful thinking. Someone will put up an attempt if there's a plausible one to make. That said, the OP may want to rephrase his question in terms that are not a command that is clearly copied from an exercise. – Ron Gordon Oct 12 '14 at 14:17
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    @Cleo You should re-post your answer here. I'll vote to reopen the OP. BTW, nice to see you on I&S. Finally, I can contact you. To be honest, I'm your fan & I admire every single post of your answers ♥(ˆ⌣ˆԅ) – Anastasiya-Romanova 秀 Oct 12 '14 at 18:56
  • @Cleo Now you can. Please post your answer. I'll upvote it (ʃƪ ˘ ³˘) – Anastasiya-Romanova 秀 Oct 12 '14 at 19:10
  • @RonGordon and which textbook would that be? :-) – Jan M. Jun 19 '23 at 07:47

1 Answers1

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$$\begin{align*}{\large\int}_0^1\frac{\ln(1-x)\,\operatorname{Li}_3\left(\frac{1+x}2\right)}xdx&=\frac{29\,\zeta(5)}{16}-\frac{19\pi^2}{96}\zeta(3)+\frac{5\,\zeta(3)}{16}\ln^22+\frac{\ln^52}{40}\\&-\frac{5\pi^2}{72}\ln^32+\frac{11\pi^4}{1440}\ln2-3\operatorname{Li}_5\left(\tfrac12\right).\\ \\ {\large\int}_0^1\frac{\ln^2(1-x)\,\operatorname{Li}_2\left(\frac{1+x}2\right)}xdx&=\frac{81\,\zeta(5)}{32}+\frac{5\pi^2}{16}\zeta(3)-\frac{\zeta(3)}8\ln^22+\frac1{15}\ln^52\\&-\frac{\pi^2}{18}\ln^32-\frac{\pi^4}{15}\ln2+2\operatorname{Li}_5\left(\tfrac12\right)+2\operatorname{Li}_4\left(\tfrac12\right)\ln2. \end{align*}$$

Cleo
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