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For integration of $ \int_0^1 (\log(1-u)^5)( \log(u)^5)/(u-1) du ,$ I have tried integral-by-parts, change of variable $ \log u =x $, and also expand it in series. But I didn't get an answer. Could any one help me out?

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    $\begingroup$ I suppose that you want $[\log(1-u)]^5$, not $\log(1-u)^5=5\log(1-u)$, etc.... $\endgroup$ – Kelenner Nov 27 '20 at 9:41
  • $\begingroup$ @Kelenner. I took is like $[\log(1-u)]^5$ too ! $\endgroup$ – Claude Leibovici Nov 27 '20 at 11:01
  • $\begingroup$ @YU MU What happened you lost interest in your problem! $\endgroup$ – Z Ahmed Nov 28 '20 at 4:39
  • $\begingroup$ @Kelenner Yes, it is like $ [log(1-u)]^5$. $\endgroup$ – YU MU Nov 28 '20 at 6:41
  • $\begingroup$ @ZAhmed It is the first time I use this. I didn't quite familiar to see the updates. Sorry about this, I will pay more attention to this next time $\endgroup$ – YU MU Nov 28 '20 at 6:43
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Use $\int_{0}^{a} f(x) dx=\int_{0}^{a} f(a-x) dx.$ $$I=\int_{0}^{1} \frac{\ln(1-x)^5 \ln x^5}{x-1}dx=-25\int_{0}^{1} \frac{\ln(1-x) \ln x}{x}dx.~~~~(1)$$ Let us use polylog functions defined as $$\text{Li}_s(x)=\sum_{k=1}^{\infty} \frac{x^k}{k^s}$$ See https://en.wikipedia.org/wiki/Polylogarithm $$\int \frac{\ln(1-x)}{x} dx= -\text{Li}_2(x), \int \frac{\text{Li}_2(x)}{x} dx= \text{Li}_3(x), \lim_{|x|\to 0} \text{Li}_s(x)=x$$ Let us do integration by parts of (1) taking $\ln x$ as first function, then $$I=-25\left(\left . -\ln x \text{Li}_2(x)\right|_{0}^{1}+\int_{0}^{1} \frac{1}{x} \text{Li}_2(x) dx\right)=-25~\text{Li}_3(x)|_{0}^{1}=-25 \sum_{k=1}^{\infty} \frac{1}{k^3}=-25 \zeta(3)$$

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  • $\begingroup$ This is because you are integrating a different function :$ \int_{0}^{1} \frac{\log^5(1-x)\log^5 x}{x-1} dx$. Mind that the power 5 is on $\log$. but your question is different, it puts the power 5 on $(1-x)$ and $x$.. $\endgroup$ – Z Ahmed Nov 27 '20 at 10:49
  • $\begingroup$ Sorry for that ! Being almost blind, I have too often this kind of mistake. I delete my answer and comment. Cheers :-( $\endgroup$ – Claude Leibovici Nov 27 '20 at 10:58
  • $\begingroup$ Very detailed answer, thanks @ZAhmed $\endgroup$ – YU MU Nov 28 '20 at 7:02
  • $\begingroup$ @ZAhmed I am wondering if we can do the integral using $\Li_s(x)$ in this more complicated question: math.stackexchange.com/questions/3925889/… $\endgroup$ – YU MU Nov 28 '20 at 7:05

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