TITLE: Integral $\int_0^{1/2}\arcsin x\cdot\ln^2x\,dx$
QUESTION [9 upvotes]: I'm interested in this integral
$$\int_0^{1/2}\arcsin x\cdot\ln^2x\,dx$$
My idea was to first evaluate
$$\int_0^{1/2}\arcsin x\cdot x^a\,dx=\frac{2^{-a}\,\pi-6B_{1/4}\left(\frac{a}{2}+1,\frac{1}{2}\right)}{12\,(a+1)}$$
in terms of the incomplete Beta function, and then find the second derivative at $a=0$, but it ended up with ugly derivatives of hypergeometric functions w.r.t. their parameters for which I did not know how to find a closed form expression. Could you suggest a different way to evaluate this integral?

REPLY [8 votes]: There is a closed-form anti-derivative corresponding to this integral:

$$\int\arcsin x\cdot\ln^2x\,dx=
\sqrt{1-x^2}\cdot\left(\ln^2x-4\ln x+6\right)\\+x\cdot\arcsin x\cdot\left(\ln^2x-2\ln x+2\right)-\ln^2\alpha+\left(\ln4-4\right)\cdot\ln\alpha-\operatorname{Li}_2\left(-\alpha^{-2}\right),$$

where
$$\alpha=\frac{1+\sqrt{1-x^2}}x,$$
that can be proved by differentiation. It enables us to evaluate a definite integral over any interval.