I want to evaluate $$\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^3\left(1+x\right)}{1+x}\:dx$$ Im not sure if this has a closed form, integration by parts is out of the question since there would be divergence issues.
I also cant turn any of the terms in the numerator into taylor series since they have powers, expanding the denominator is also useless, could you at least give me a hint on how to tackle it, I've really got no clue.
On
With subbing $1+x\to x$ we have
$$\int_0^1\frac{\ln^2(1-x)\ln^3(1+x)}{1+x}dx=\int_1^2\frac{\ln^2(2-x)\ln^3x}{x}dx$$
$$=\ln^22\underbrace{\int_1^2\frac{\ln^3x}{x}dx}_{I_1}+2\ln2\underbrace{\int_1^2\frac{\ln(1-\frac x2)\ln^3x}{x}dx}_{I_2}+\underbrace{\int_1^2\frac{\ln^2(1-\frac x2)\ln^3x}{x}dx}_{I_3}$$
$$I_1=\frac14\ln^42$$
$$I_2=-\sum_{n=1}^\infty\frac1n\int_1^2 2^{-n}x^{n-1}\ln^3xdx$$
$$=-\sum_{n=1}^\infty\frac1n\left(\frac{\ln^32}{n}-\frac{3\ln^22}{n^2}+\frac{6\ln2}{n^3}-\frac{6}{n^4}+\frac{6}{n^42^n}\right)$$
$$=-\ln^32\zeta(2)+3\ln^22\zeta(3)-6\ln2\zeta(4)+6\zeta(5)-6\text{Li}_5\left(\frac12\right)$$
$$I_3=2\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac1{n^2}\right)\int_1^2 2^{-n}x^{n-1}\ln^3xdx$$
$$=2\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac1{n^2}\right)\left(\frac{\ln^32}{n}-\frac{3\ln^22}{n^2}+\frac{6\ln2}{n^3}-\frac{6}{n^4}+\frac{6}{n^42^n}\right)$$
$$=2\ln^32\sum_{n=1}^\infty\frac{H_n}{n^2}-6\ln^22\sum_{n=1}^\infty\frac{H_n}{n^3}+12\ln2\sum_{n=1}^\infty\frac{H_n}{n^4}-12\sum_{n=1}^\infty\frac{H_n}{n^5}+12\sum_{n=1}^\infty\frac{H_n}{n^52^n}$$
$$-2\ln^32\zeta(3)+6\ln^22\zeta(4)-12\ln2\zeta(5)+12\zeta(6)-12\text{Li}_6\left(\frac12\right)$$
$$=2\ln^32\zeta(3)-\frac32\ln^22\zeta(4)+24\ln2\zeta(5)-12\ln2\zeta(2)\zeta(3)$$ $$-9\zeta(6)+6\zeta^2(3)-12\text{Li}_6\left(\frac12\right)+12\sum_{n=1}^\infty\frac{H_n}{n^52^n}$$
Combine all integrals we get
$$\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^3\left(1+x\right)}{1+x}dx=12\sum _{n=1}^{\infty }\frac{H_n}{n^52^n}-12\text{Li}_6\left(\frac{1}{2}\right)-12\ln 2\text{Li}_5\left(\frac{1}{2}\right)-9\zeta(6)$$
$$+6\zeta ^2(3)+36\ln2\zeta(5)-12\ln2\zeta(2)\zeta(3)-\frac{27}{2}\ln ^22\zeta(4)+8\ln ^32\zeta(3)-2\ln ^42\zeta(2)+\frac{1}{4}\ln ^62.$$
Note that in $I_2$ we use $\ln(1-z)=-\sum_{n=1}^\infty\frac{z^n}{n}$ and in $I_3$ we use $\ln^2(1-z)=2\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac{1}{n^2}\right)z^n.$
If you use the algebraic identity $a^2b^3=\frac{1}{20}\left(a+b\right)^5-\frac{1}{20}\left(a-b\right)^5-\frac{1}{2}a^4b-\frac{1}{10}b^5$ the integral turns into $$\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^3\left(1+x\right)}{1+x}\:dx$$ $$=\frac{1}{20}\int _0^1\frac{\ln ^5\left(1-x^2\right)}{1+x}\:dx-\frac{1}{20}\int _0^1\frac{\ln ^5\left(\frac{1-x}{1+x}\right)}{1+x}\:dx-\frac{1}{2}\int _0^1\frac{\ln ^4\left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx$$ $$-\frac{1}{10}\int _0^1\frac{\ln ^5\left(1+x\right)}{1+x}\:dx$$
$$\frac{1}{20}\int _0^1\frac{\ln ^5\left(1-x^2\right)}{1+x}\:dx$$ $$=\frac{1}{20}\underbrace{\int _0^1\frac{\ln ^5\left(1-x^2\right)}{1-x^2}\left(1-x\right)\:dx}_{t=x^2}=\frac{1}{40}\underbrace{\int _0^1\frac{\ln ^5\left(1-t\right)}{1-t}\frac{\left(1-\sqrt{t}\right)}{\sqrt{t}}\:dt}_{\text{IBP}}$$ $$=-\frac{1}{480}\int _0^1\frac{\ln ^6\left(1-t\right)}{t^{\frac{3}{2}}}\:dt=-\frac{1}{480}\lim_{\alpha\rightarrow -1/2 \\\beta\rightarrow 1}\frac{\partial^6}{\partial \beta^6}\text{B}\left(\alpha ,\beta \right)$$ $$=-\frac{237}{16}\zeta \left(6\right)+6\zeta ^2\left(3\right)-12\ln \left(2\right)\zeta \left(2\right)\zeta \left(3\right)-2\ln ^4\left(2\right)\zeta \left(2\right)+36\ln \left(2\right)\zeta \left(5\right)+8\ln ^3\left(2\right)\zeta \left(3\right)$$ $$-\frac{27}{2}\ln ^2\left(2\right)\zeta \left(4\right)+\frac{4}{15}\ln ^6\left(2\right)$$
$$-\frac{1}{20}\underbrace{\int _0^1\frac{\ln ^5\left(\frac{1-x}{1+x}\right)}{1+x}\:dx}_{t=\frac{1-x}{1+x}}=-\frac{1}{20}\int _0^1\frac{\ln ^5\left(t\right)}{1+t}\:dt=6\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k^6}$$ $$=\frac{93}{16}\zeta \left(6\right)$$
$$-\frac{1}{2}\int _0^1\frac{\ln ^4\left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx$$
$$-\frac{1}{2}\underbrace{\int _0^1\frac{\ln ^4\left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx}_{t=1-x}=-\frac{1}{2}\int _0^1\frac{\ln ^4\left(t\right)\ln \left(2-t\right)}{2-t}\:dt$$ $$=-\frac{1}{4}\int _0^1\frac{\ln ^4\left(t\right)\ln \left(1-\frac{t}{2}\right)}{1-\frac{t}{2}}\:dt-\frac{1}{4}\ln \left(2\right)\int _0^1\frac{\ln ^4\left(t\right)}{1-\frac{t}{2}}\:dt$$ $$=12\sum _{k=1}^{\infty }\frac{H_k}{k^5\:2^k}-12\sum _{k=1}^{\infty }\frac{1}{k^6\:2^k}-12\ln \left(2\right)\sum _{k=1}^{\infty }\frac{1}{k^5\:2^k}$$ $$=12\sum _{k=1}^{\infty }\frac{H_k}{k^5\:2^k}-12\text{Li}_6\left(\frac{1}{2}\right)-12\ln \left(2\right)\text{Li}_5\left(\frac{1}{2}\right)$$ Where I used the generating function $\displaystyle \sum _{k=1}^{\infty }H_k\:x^k=-\frac{\ln \left(1-x\right)}{1-x}$ on the left integral.
$$-\frac{1}{10}\int _0^1\frac{\ln ^5\left(1+x\right)}{1+x}\:dx$$ $$-\frac{1}{10}\underbrace{\int _0^1\frac{\ln ^5\left(1+x\right)}{1+x}\:dx}_{t=\ln\left(1+x\right)}=-\frac{1}{60}\ln ^6\left(2\right)$$
Collecting the results we get $$\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^3\left(1+x\right)}{1+x}\:dx=-9\zeta \left(6\right)+6\zeta ^2\left(3\right)-12\ln \left(2\right)\zeta \left(2\right)\zeta \left(3\right)+12\sum _{k=1}^{\infty }\frac{H_k}{k^5\:2^k}$$ $$-12\text{Li}_6\left(\frac{1}{2}\right)-12\ln \left(2\right)\text{Li}_5\left(\frac{1}{2}\right)-2\ln ^4\left(2\right)\zeta \left(2\right)+36\ln \left(2\right)\zeta \left(5\right)$$ $$+8\ln ^3\left(2\right)\zeta \left(3\right)-\frac{27}{2}\ln ^2\left(2\right)\zeta \left(4\right)+\frac{1}{4}\ln ^6\left(2\right)$$ I couldnt find a closed form represented by known functions for that sum so it appears the integral doesnt have a closed form.