Is there a closed form for $\int_0^1\frac{\ln(x) \sin^{-1}(x)}{x\sqrt{1-x^2}}dx\ ?$

Question

Is there a closed form this integral:

$$I=\int_0^1\frac{\ln(x) \sin^{-1}(x)}{x\sqrt{1-x^2}}dx\ ?$$

I came across this integral while I was working on $\int_0^1\frac{\operatorname{Li}_2(x)}{\sqrt{1-x^2}}dx$.

Here is what I did:

First I tried subbing $x=\sin(\theta)$ which yields

$$I=\int_0^{\pi/2}\theta\csc(\theta)\ln(\sin(\theta))d\theta$$ and I do not think using Fourier series for $\ln(\sin(\theta)$ is useful here.

Second I tried the identity

$$\frac{\sin^{-1}(z)}{\sqrt{1-z^2}}=\sum_{k=1}^\infty \frac{(2z)^{2k-1}}{k{2k\choose k}},\quad |z|<1$$

From which, it follows

$$I=\sum_{k=1}^\infty \frac{(2)^{2k-1}}{k{2k\choose k}}\int_0^1 x^{2k-2}\ln(x)dx=-\sum_{k=1}^\infty \frac{(2)^{2k-1}}{k(2k-1)^2{2k\choose k}}$$

and I do not know if I made it more complicated or easier here. any idea?

Answer 1

Let $x = \frac{2 u}{1+u^2}$ to find \begin{align} -I &= \int \limits_0^1 \frac{-\ln(x) \arcsin(x)}{x \sqrt{1-x^2}} \, \mathrm{d} x = 2 \int \limits_0^1 \ln \left(\frac{1+u^2}{2u}\right) \frac{\arctan(u)}{u} \, \mathrm{d} u \\ &= 2 \left[- \ln(2) \int \limits_0^1 \frac{\arctan(u)}{u} \, \mathrm{d} u + \int \limits_0^1 \frac{- \ln(u) \arctan(u)}{u} \, \mathrm{d} u + \int \limits_0^1 \frac{\ln(1+u^2) \arctan(u)}{u} \, \mathrm{d} u \right] \\ &\equiv 2 \left[- \ln(2) \mathrm{G} + \sum \limits_{k=0}^\infty \frac{(-1)^k}{(2k+1)^3} + J\right] , \end{align} where $\mathrm{G}$ is Catalan's constant. The series is well-known and equals $\frac{\pi^3}{32}$. The remaining integral can be found by writing \begin{align} J &= \int \limits_0^1 \frac{\ln(1+u^2) \arctan(u)}{u} \, \mathrm{d} u = \operatorname{Im} \int \limits_0^1 \frac{\ln^2(1 + \mathrm{i} u)}{u} \, \mathrm{d} u \\ &= \operatorname{Im} \left[\ln(-\mathrm{i})\ln^2(1+\mathrm{i}) + 2 \mathrm{i} \int \limits_0^1 \frac{-\ln(1 - (1 + \mathrm{i}u))}{1+\mathrm{i}u} \ln(1+\mathrm{i}u) \, \mathrm{d} u \right] \\ &= \operatorname{Im} \left[\ln(-\mathrm{i})\ln^2(1+\mathrm{i}) + 2 \operatorname{Li}_2(1+\mathrm{i}) \ln(1+\mathrm{i}) - 2 \mathrm{i} \int \limits_0^1 \frac{\operatorname{Li}_2(1+\mathrm{i}u)}{1+\mathrm{i} u} \, \mathrm{d} u\right] \\ &= \operatorname{Im} \left[\ln(-\mathrm{i})\ln^2(1+\mathrm{i}) + 2 \operatorname{Li}_2(1+\mathrm{i}) \ln(1+\mathrm{i}) - 2 \operatorname{Li}_3(1+\mathrm{i}) + 2 \operatorname{Li}_3(1)\right] \\ &= \frac{\pi^3}{16} + \frac{\pi}{8} \ln^2(2) + \ln(2) \mathrm{G} - 2 \operatorname{Im} \operatorname{Li}_3(1+\mathrm{i}) \, , \end{align} where we have used $ \operatorname{Li}_2(1+\mathrm{i}) = \frac{\pi^2}{16} + \mathrm{i} \left(\mathrm{G} + \frac{\pi}{4} \ln(2)\right)$.

Combining these results we obtain $$ -I = \frac{3 \pi^3}{16} + \frac{\pi}{4} \ln^2(2) - 4 \operatorname{Im} \operatorname{Li}_3(1+\mathrm{i}) \, . $$