# Integration of product of logarithm

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?

• I suppose that you want $[\log(1-u)]^5$, not $\log(1-u)^5=5\log(1-u)$, etc.... – Kelenner Nov 27 '20 at 9:41
• @Kelenner. I took is like $[\log(1-u)]^5$ too ! – Claude Leibovici Nov 27 '20 at 11:01
• @YU MU What happened you lost interest in your problem! – Z Ahmed Nov 28 '20 at 4:39
• @Kelenner Yes, it is like $[log(1-u)]^5$. – YU MU Nov 28 '20 at 6:41
• @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 – YU MU Nov 28 '20 at 6:43

## 1 Answer

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

• 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 $x$.. – Z Ahmed Nov 27 '20 at 10:49
• Sorry for that ! Being almost blind, I have too often this kind of mistake. I delete my answer and comment. Cheers :-( – Claude Leibovici Nov 27 '20 at 10:58
• Very detailed answer, thanks @ZAhmed – YU MU Nov 28 '20 at 7:02
• @ZAhmed I am wondering if we can do the integral using $\Li_s(x)$ in this more complicated question: math.stackexchange.com/questions/3925889/… – YU MU Nov 28 '20 at 7:05