Prove $\int_{0}^{1} \frac{\ln\left ( 1+\sqrt{1-x^2} \right ) \ln(1+x)}{ x} \text{d}x =-\frac{5}{16}\zeta(3)+\frac{\pi^2}{8}\ln(2)$

Question

I recently got an integral accidently($\zeta(s)$ is Riemann $\zeta$ function), $$ \int_{0}^{1} \frac{\ln\left ( 1+\sqrt{1-x^2} \right ) \ln(1+x)}{ x} \text{d}x =-\frac{5}{16}\zeta(3)+\frac{\pi^2}{8}\ln(2), $$ which seems slightly harmonious. So I put it here to ask for an elegant proof for the integral. Besides, let $\operatorname{Li}_n(z)$ be polylogarithms, do similar expressions hold true for $$ \int_{0}^{1} \frac{\ln\left ( 1+\sqrt{1-x^2} \right ) \operatorname{Li}_n(x)}{ x} \text{d}x $$ and $$ \int_{0}^{1} \frac{\ln\left ( 1+\sqrt{1-x^2} \right ) \operatorname{Li}_n(-x)}{ x} \text{d}x? $$ Thanks for helping and enlightening me.

Answer 1

Substitute $x=\frac{1-t^2}{1+t^2}$ and then utilize $2\ln a \ln b= \ln^2 a+\ln^2b -\ln^2\frac ab$ to break up the integrand \begin{align} I=&\int_{0}^{1} \frac{\ln( 1+\sqrt{1-x^2} ) \ln(1+x)}{ x} {d}x\\ = &\int_0^1 \frac{4t \ln\frac2{1+t^2}\bigg(\ln2+\ln\frac{(1+t)^2}{2(1+t^2)}\bigg)}{1-t^4}dt\\ =&\int_0^1 \frac{2 t \left( 2\ln2 \ln\frac2{1+t^2}+ \ln^2\frac2{1+t^2}+\ln^2\frac{(1+t)^2}{2(1+t^2)}-\ln^2\frac{(1+t)^2}{4}\right)}{1-t^4}dt\\ \end{align} where the four resulting integrals are evaluated below \begin{align} &\int_0^1 \frac{4 t \ln2 \ln\frac2{1+t^2}}{1-t^4}{dt} \overset{t^2\to t} = 2 \ln2\int_0^1 \frac{\ln\frac2{1+t}}{1-t^2}{dt} = \frac{\pi^2}{12}\ln2\\ \\ &\int_0^1 \frac{2 t \ln^2\frac2{1+t^2}}{1-t^4}{dt} \overset{t^2\to \frac{1-t}{1+t}} = \int_0^1 \frac{\ln^2(1+t)}{2t}{dt} = \frac{1}{8}\zeta(3)\\ \\ &\int_0^1 \frac{2 t \ln^2\frac{(1+t)^2}{2(1+t^2)}}{1-t^4}{dt} \overset{t\to \frac{1-\sqrt t}{1+\sqrt t}} = \int_0^1 \frac{\ln^2(1+t)}{4t}- \frac{\ln^2(1+t)}{2(1+t)}\ {dt}\\ &\hspace{40mm}= \frac{1}{16}\zeta(3)-\frac1{6}\ln^32\\ \\ &\int_0^1 \frac{2 t \ln^2\frac{(1+t)^2}{4}}{1-t^4}dt =\int_0^1 \frac{4t\ln^2\frac{1+t}2}{1+t^2}+ \frac{4t\ln^2\frac{1+t}2}{1-t^2}\ dt\\ &\hspace{35mm}= \left( \frac{7}{6}\ln^32 -\frac{\pi^2}{24}\ln2\right)+\left(\frac12\zeta(3)-\frac{4}{3}\ln^32\right) \end{align} As a result $$I=-\frac{5}{16}\zeta(3)+\frac{\pi^2}{8}\ln2$$