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?

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    $\begingroup$ Have you tried parts, you can get quite far: for example $\int\arcsin x \ln x dx $ can be evaluated. $\endgroup$ Dec 23 '15 at 23:24
  • $\begingroup$ Rene's idea works, but you'll need dilogarithms. $\endgroup$
    – Lucian
    Dec 24 '15 at 1:30

One may find the following closed form.

Proposition.$$ \begin{align} \int_0^{1/2}\!\!\arcsin x\cdot\ln^2x\,dx=& \small{\frac{\pi^2}{12}+\frac{\pi}6+3 \sqrt{3}-6+\left(\frac{\pi }{6}+2 \sqrt{3}+4 \right)\ln 2+\left(\frac{\pi }{12}+\frac{\sqrt{3}}{2}-1\right)\ln^2 2}\\&\small{-4 \ln\left(2+\sqrt{3}\right)+\frac12 \text{Li}_2\!\left(\frac{2-\sqrt{3}}4\right)-\frac12 \text{Li}_2\!\left(\frac{2+\sqrt{3}}4\right)}, \end{align} $$

where $\text{Li}_2(\cdot)$ stands for the dilogarithm function.

Hint. We may start with an integration by parts and by making two successive changes of variable.

$$ \begin{align} &\int_0^{1/2}\arcsin x\cdot\ln^2x\,dx \\&=\left[\arcsin x\cdot\left(2 x-2 x \ln x+x \ln^2 x\right)\right]_0^{1/2}-\int_0^{1/2}\frac{2x}{\sqrt{1-x^2}}\cdot\left(\frac12\: \ln^2 x-\ln x+1\right)dx \\&=\frac{\pi}6 \left(\frac{\ln^22}2+\ln 2+1\right) -\int_0^{1/4}\frac1{\sqrt{1-u}}\cdot\left(\frac18\: \ln^2 u-\frac12\:\ln u+1\right)du \quad (u=x^2) \\&=\frac{\pi}6 \left(\frac{\ln^22}2+\ln 2+1\right) -\int_{\sqrt{3}/2}^1\left(\frac14\: \ln^2 (1-v^2)-\ln (1-v^2)+2\right)dv \quad (v=\sqrt{1-u}). \end{align} $$ Then we are led to consider $$ \begin{align} \int_{\sqrt{3}/2}^1&\left(\frac14\: \ln^2 (1-v^2)-\ln (1-v^2)+2\right)dv \\&=\frac14\int_{\sqrt{3}/2}^1\ln^2 (1-v^2)\:dv+\int_{\sqrt{3}/2}^1\left(-\ln (1-v^2)+2\right)dv. \end{align} $$ The latter integral is easily evaluated being equal to $$ \begin{align} \int_{\sqrt{3}/2}^1\left(-\ln (1-v)-\ln(1+v)+2\right)dv =4-2\sqrt{3}-(2+\sqrt{3})\ln2+2\ln(2+\sqrt{3}). \end{align} $$ On the other hand, by using an integration by parts, we have $$ \int_{\sqrt{3}/2}^1\ln^2 (1-v^2)\:dv=\left[v\:\ln^2 (1-v^2)\right]_{\sqrt{3}/2}^1 +4\int_{\sqrt{3}/2}^1\frac{v^2}{1-v^2}\:\left(\ln (1-v)+\ln (1+v)\right)dv$$ Then we evaluate the latter integral by a partial fraction decomposition and by performing two changes of variable allowing us to use the standard identity $$ \text{Li}_2(z)=-\int_0^z\frac{\log (1-u)}u\:du. $$ Bringing all the steps together gives the announced result.

By using a similar path one may find a closed form of all integrals$$\int_0^a\!\!\arcsin x\cdot\ln^2x\,dx,\quad 0\leq a\leq1,$$ I omit the expression which is a little bit long.

We have

Example $1$.$$ \begin{align} \int_0^1\!\!\arcsin x\cdot\ln^2x\,dx=& \frac{\pi^2}{12}+\pi-\ln^2 2+4\ln 2-6. \end{align} $$

By using some special values of the dilogarithm function $\text{Li}_2(\cdot)$, one obtains the following integral given in terms of elementary constants.

Example $2$.$$ \begin{align} &\int_0^{2\sqrt{\sqrt{5}-2}}\!\!\arcsin x\cdot\ln^2x\,dx \\&=-18+6 \sqrt{5}+\frac{\pi ^2}{15}+4 \sqrt{\sqrt{5}-2} \arctan\left(2 \sqrt{\sqrt{5}+2}\right) \\&+2 \ln 2-\frac{\ln^2 2}{2}-4 \sqrt{\sqrt{5}-2} \arctan\left(2 \sqrt{\sqrt{5}+2}\right) \ln 2 \\&+2\sqrt{\sqrt{5}-2} \arctan\left(2 \sqrt{\sqrt{5}+2}\right) \ln^2 2 -\sqrt{5} \ln\left(144-64 \sqrt{5}\right) \\&+\frac{\sqrt{5}-3}4 \ln^2\left(3-\sqrt{5}\right)-\ln^2\left(\frac{1}{2}+\frac{\sqrt{5}}{2}\right) \\&+\frac{1}{4} \sqrt{5} \ln\left(14-6 \sqrt{5}\right) \ln\left(\sqrt{5}-1\right) -\frac{1}{4} \ln^2\left(\sqrt{5}-1\right)+\frac{1}{4} \sqrt{5} \ln^2\left(\sqrt{5}-1\right) \\&+2 \sqrt{\sqrt{5}-2} \arctan\left(2 \sqrt{\sqrt{5}+2}\right) \ln\left(\sqrt{5}+2\right) \\&-2\sqrt{\sqrt{5}-2} \arctan\left(2 \sqrt{\sqrt{5}+2}\right)\cdot\ln 2\cdot\ln\left(\sqrt{5}+2\right) \\&+\frac{1}{2} \sqrt{\sqrt{5}-2} \arctan\left(2 \sqrt{\sqrt{5}+2}\right) \ln^2\left(\sqrt{5}+2\right) \\&-\frac12 \ln\left(3-\sqrt{5}\right)\ln\left(4 \sqrt{5}-4\right)+2 \ln\left(208 \sqrt{5}-464\right). \end{align} $$

  • $\begingroup$ Very nice job! @Olivier Oloa $\endgroup$
    – user266519
    Dec 24 '15 at 2:08

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.

  • $\begingroup$ I think it would be interesting to see the steps of the given the reduction. Bravo! (+1) $\endgroup$ Dec 26 '15 at 11:41

Alternately, letting $~a=2+\sqrt3,~$ we have

$$\begin{align} I~&=~\dfrac\pi6~(\pi+1)~+ \\\\ &+~\bigg(\dfrac\pi{12}+\dfrac a2-4\bigg)\ln^22~+~\bigg[\dfrac\pi6+(a-1)^2\bigg]\ln2~+ \\\\ &+~\dfrac{\ln^2a}2-4\ln a-\dfrac3a-\text{Li}_2\bigg(\dfrac a4\bigg). \end{align}$$

This simpler expression is meant as a supplement to Olivier Oloa's wonderful answer. The apparent discrepancies are explained by the properties of the polylogarithm.


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