How to prove $\int_0^1x\ln^2(1+x)\ln(\frac{x^2}{1+x})\frac{dx}{1+x^2}$

Question

How to prove $$\int_0^1x\ln^2(1+x)\ln\left(\frac{x^2}{1+x}\right)\frac{dx}{1+x^2}=-\frac{7}{32}\cdot\zeta{(3)}\ln2+\frac{3\pi^2}{128}\cdot\ln^22-\frac{1}{64}\cdot\ln^42-\frac{13\pi^4}{46080}$$ The substitution $$x=\frac{1-y}{1+y}$$ leads to calculate the integrals that are unknown: $$\int_0^1y\ln(1-y)\ln^2(1+y)\frac{dy}{1+y^2}, \int_0^1\frac{y\ln^3(1+y)}{1+y^2}dy$$ For the moment,I do not see how to calculate this integral.

Answer 1

Note that: $$\int _0^1\frac{x\ln ^3\left(1+x\right)}{1+x^2}\:dx=\frac{1}{2}\ln ^4\left(2\right)-\frac{3}{2}\int _0^1\frac{\ln ^2\left(1+x\right)\ln \left(1+x^2\right)}{1+x}\:dx$$

Now I'll evaluate that integral in large steps. $$\int _0^1\frac{\ln ^2\left(1+x\right)\ln \left(1+x^2\right)}{1+x}\:dx$$ $$=\int _{\frac{1}{2}}^1\frac{\ln ^2\left(x\right)\ln \left(1-2x+2x^2\right)}{x}\:dx-2\int _{\frac{1}{2}}^1\frac{\ln ^3\left(x\right)}{x}\:dx$$ $$=2\operatorname{\mathfrak{R}} \left\{\int _{\frac{1}{2}}^1\frac{\ln ^2\left(x\right)\ln \left(1-\left(1+i\right)x\right)}{x}\:dx\right\}+\frac{1}{2}\ln ^4\left(2\right)$$ $$=-2\operatorname{\mathfrak{R}} \left\{-2\operatorname{Li}_4\left(\frac{1+i}{2}\right)+2\operatorname{Li}_4\left(1+i\right)-2\ln \left(2\right)\operatorname{Li}_3\left(\frac{1+i}{2}\right)-\ln ^2\left(2\right)\operatorname{Li}_2\left(\frac{1+i}{2}\right)\right\}$$ $$+\frac{1}{2}\ln ^4\left(2\right)$$ The real part of these polylogarithm are well-known, using their closed-form yields: $$\int _0^1\frac{\ln ^2\left(1+x\right)\ln \left(1+x^2\right)}{1+x}\:dx$$ $$=-\frac{627}{256}\zeta \left(4\right)+\frac{5}{2}\operatorname{Li}_4\left(\frac{1}{2}\right)+\frac{35}{16}\ln \left(2\right)\zeta \left(3\right)-\frac{21}{32}\ln ^2\left(2\right)\zeta \left(2\right)+\frac{41}{96}\ln ^4\left(2\right)$$ Replacing this into the first line yields the closed-form for that other integral. $$\int _0^1\frac{x\ln ^3\left(1+x\right)}{1+x^2}\:dx$$ $$=\frac{1881}{512}\zeta \left(4\right)-\frac{15}{4}\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{105}{32}\ln \left(2\right)\zeta \left(3\right)+\frac{63}{64}\ln ^2\left(2\right)\zeta \left(2\right)-\frac{9}{64}\ln ^4\left(2\right)$$

---

Now for your next integral: $$\int _0^1\frac{x\ln \left(1-x\right)\ln ^2\left(1+x\right)}{1+x^2}\:dx$$ $$=\frac{1}{6}\int _0^1\frac{x\ln ^3\left(1-x^2\right)}{1+x^2}\:dx+\frac{1}{6}\int _0^1\frac{x\ln ^3\left(\frac{1-x}{1+x}\right)}{1+x^2}\:dx-\frac{1}{3}\int _0^1\frac{x\ln ^3\left(1-x\right)}{1+x^2}\:dx$$ $$=\frac{1}{12}\int _0^1\frac{\ln ^3\left(1-x\right)}{1+x}dx+\frac{1}{6}\int _0^1\frac{\ln ^3\left(x\right)}{1+x}\:dx-\frac{1}{6}\int _0^1\frac{x\ln ^3\left(x\right)}{1+x^2}\:dx$$ $$-\frac{1}{3}\operatorname{\mathfrak{R}} \left\{\frac{1-i}{2}\int _0^1\frac{\ln ^3\left(x\right)}{1-\frac{1-i}{2}x}\:dx\right\}$$ $$=-\frac{1}{2}\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{105}{128}\zeta \left(4\right)+2\operatorname{\mathfrak{R}} \left\{\operatorname{Li}_4\left(\frac{1-i}{2}\right)\right\}$$ Thus: $$\int _0^1\frac{x\ln \left(1-x\right)\ln ^2\left(1+x\right)}{1+x^2}\:dx=-\frac{77}{512}\zeta \left(4\right)+\frac{1}{8}\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{5}{64}\ln ^2\left(2\right)\zeta \left(2\right)+\frac{1}{48}\ln ^4\left(2\right)$$