Integration of product of logarithm

Question

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?

Answer 1

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