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?

  • 1
    $\begingroup$ 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

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 $$I=-25\left(\left . -\ln x \text{Li}_2(x)\right|_{0}^{1}+\int_{0}^{1} \frac{1}{x} \text{Li}_2(x) dx\right)=-25~\text{Li}_3(x)|_{0}^{1}=-25 \sum_{k=1}^{\infty} \frac{1}{k^3}=-25 \zeta(3)$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.