TITLE: How to evaluate $\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^5\left(1+x\right)}{1+x}\:dx$
QUESTION [8 upvotes]: Before you think I haven't tried anything, please read.
I've been trying to evaluate $$\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^5\left(1+x\right)}{1+x}\:dx$$
But I can't find a way to simplify it. Integration by parts is not valid since we face convergence issues. Subbing the $1-x$ term is also not quite helpful here. I also tried to use the sub $\frac{1}{1+x}$ but this is not useful either. Using algebraic identities isn't that useful either, since we run into similar difficulty integrals. Since these attempts do not lead to anything, how can I approach it?

REPLY [3 votes]: Bonus:
$$\int _0^1\frac{\ln ^4\left(1-x\right)\ln ^3\left(1+x\right)}{1+x}\:dx$$
Let's make use of the algebraic identity
$a^4b^3=\frac{1}{70}\left(a+b\right)^7-\frac{1}{70}\left(a-b\right)^7-\frac{1}{5}a^6b-\frac{3}{5}a^2b^5-\frac{1}{35}b^7$.

$$=\int _0^1\frac{1}{70}\frac{\ln ^7\left(1-x^2\right)}{1+x}-\frac{1}{70}\frac{\ln ^7\left(\frac{1-x}{1+x}\right)}{1+x}-\frac{1}{5}\frac{\ln ^6\left(1-x\right)\ln \left(1+x\right)}{1+x}$$
$$-\frac{3}{5}\frac{\ln ^2\left(1-x\right)\ln ^5\left(1+x\right)}{1+x}-\frac{1}{35}\frac{\ln ^7\left(1+x\right)}{1+x}\:dx$$

Here we can use the same strategy found in the book (Almost) Impossible integrals, Sums, and Series page $80$.
$$\frac{1}{70}\int _0^1\frac{\ln ^7\left(1-x^2\right)}{1+x}\:dx=\frac{1}{70}\underbrace{\int _0^1\frac{\ln ^7\left(1-x^2\right)}{1-x^2}\left(1-x\right)\:dx}_{t=x^2}$$
$$=\frac{1}{140}\int _0^1\frac{\ln ^7\left(1-t\right)}{1-t}\frac{\left(1-\sqrt{t}\right)}{\sqrt{t}}\:dt=-\frac{1}{2240}\int _0^1\:\frac{\ln ^8\left(1-t\right)}{t^{\frac{3}{2}}}\:dt$$
$$=-\frac{1}{2240}\lim_{\alpha\rightarrow -1/2\\\beta\rightarrow 0}\frac{\partial ^8}{\partial \beta ^8}\operatorname{B}\left(\alpha ,\beta \right)$$
$$=-\frac{7017}{16}\zeta \left(8\right)+1296\ln \left(2\right)\zeta \left(7\right)-\frac{711}{2}\ln ^2\left(2\right)\zeta \left(6\right)-72\zeta \left(2\right)\zeta ^2\left(3\right)+432\zeta \left(3\right)\zeta \left(5\right)$$
$$+\frac{96}{5}\ln ^5\left(2\right)\zeta \left(3\right)-96\ln ^3\left(2\right)\zeta \left(2\right)\zeta \left(3\right)+288\ln ^3\left(2\right)\zeta \left(5\right)+144\ln ^2\left(2\right)\zeta ^2\left(3\right)$$
$$-324\ln \left(2\right)\zeta \left(3\right)\zeta \left(4\right)-432\ln \left(2\right)\zeta \left(2\right)\zeta \left(5\right)+\frac{8}{35}\ln ^8\left(2\right)-\frac{16}{5}\ln ^6\left(2\right)\zeta \left(2\right)-54\ln ^4\left(2\right)\zeta \left(4\right)$$

$$-\frac{1}{70}\underbrace{\int _0^1\frac{\ln ^7\left(\frac{1-x}{1+x}\right)}{1+x}\:dx}_{t=\frac{1-x}{1+x}}=-\frac{1}{70}\int _0^1\frac{\ln ^7\left(t\right)}{1+t}\:dt=72\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k^8}$$
$$=\frac{1143}{16}\zeta \left(8\right)$$

$$-\frac{1}{5}\underbrace{\int _0^1\frac{\ln ^6\left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx}_{t=1-x}$$
$$=-\frac{1}{10}\int _0^1\frac{\ln ^6\left(t\right)\ln \left(1-\frac{t}{2}\right)}{1-\frac{t}{2}}\:dt-\frac{1}{10}\ln \left(2\right)\int _0^1\frac{\ln ^6\left(t\right)}{1-\frac{t}{2}}\:dt$$
$$=144\sum _{k=1}^{\infty }\frac{H_k}{k^7\:2^k}-144\sum _{k=1}^{\infty }\frac{1}{k^8\:2^k}-144\ln \left(2\right)\sum _{k=1}^{\infty }\frac{1}{k^7\:2^k}$$
$$=144\sum _{k=1}^{\infty }\frac{H_k}{k^7\:2^k}-144\operatorname{Li}_8\left(\frac{1}{2}\right)-144\ln \left(2\right)\operatorname{Li}_7\left(\frac{1}{2}\right)$$

Thanks to Ali Shather we may use the value of this integral which he evaluated nicely in this same thread.
$$-\frac{3}{5}\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^5\left(1+x\right)}{1+x}\:dx$$
$$=180\zeta \left(8\right)-144\zeta \left(3\right)\zeta \left(5\right)-576\ln \left(2\right)\zeta \left(7\right)+144\ln \left(2\right)\zeta \left(2\right)\zeta \left(5\right)+144\ln \left(2\right)\zeta \left(3\right)\zeta \left(4\right)$$
$$+198\ln ^2\left(2\right)\zeta \left(6\right)-36\ln ^2\left(2\right)\zeta ^2\left(3\right)-120\ln ^3\left(2\right)\zeta \left(5\right)+24\ln ^3\left(2\right)\zeta \left(2\right)\zeta \left(3\right)+\frac{51}{2}\ln ^4\left(2\right)\zeta \left(4\right)$$
$$-\frac{36}{5}\ln ^5\left(2\right)\zeta \left(3\right)-144\sum _{k=1}^{\infty }\frac{H_k}{k^7\:2^k}+144\operatorname{Li}_8\left(\frac{1}{2}\right)+\frac{6}{5}\ln ^6\left(2\right)\zeta \left(2\right)$$
$$+144\ln \left(2\right)\operatorname{Li}_7\left(\frac{1}{2}\right)-\frac{1}{10}\ln ^8\left(2\right)$$

$$-\frac{1}{35}\underbrace{\int _0^1\frac{\ln ^7\left(1+x\right)}{1+x}\:dx}_{t=\ln\left(1+x\right)}=-\frac{1}{280}\ln ^8\left(2\right)$$

Collecting the results we find
$$\int _0^1\frac{\ln ^4\left(1-x\right)\ln ^3\left(1+x\right)}{1+x}\:dx=-\frac{1497}{8}\zeta \left(8\right)+288\zeta \left(3\right)\zeta \left(5\right)+720\ln \left(2\right)\zeta \left(7\right)$$
$$-288\ln \left(2\right)\zeta \left(2\right)\zeta \left(5\right)-180\ln \left(2\right)\zeta \left(3\right)\zeta \left(4\right)-72\zeta \left(2\right)\zeta ^2\left(3\right)-\frac{315}{2}\ln ^2\left(2\right)\zeta \left(6\right)$$
$$+108\ln ^2\left(2\right)\zeta ^2\left(3\right)+168\ln ^3\left(2\right)\zeta \left(5\right)-72\ln ^3\left(2\right)\zeta \left(2\right)\zeta \left(3\right)-\frac{57}{2}\ln ^4\left(2\right)\zeta \left(4\right)$$
$$+12\ln ^5\left(2\right)\zeta \left(3\right)-2\ln ^6\left(2\right)\zeta \left(2\right)+\frac{1}{8}\ln ^8\left(2\right)$$
Notice how the harmonic series canceled out giving us an actual closed form.