# Finding a closed form for $\int_{0}^{1}\frac{\ln\left ( 1-x^{2} \right )\arcsin ^{2}x}{x^{2}}\mathrm{d}x\approx -0.939332$ [duplicate]

I'm attempting to find a closed form for

$$\int_{0}^{1}\frac{\ln\left ( 1-x^{2} \right )\arcsin ^{2}x}{x^{2}}\mathrm{d}x\approx -0.939332$$

I tried to use $$\displaystyle \arcsin^{2}x=\frac{1}{2}\sum_{k=1}^{\infty }\frac{\left ( 2x \right )^{2k}}{k^{2}\dbinom{2k}{k}}$$ but it didn't work and became even more complicated. Any help would be appreciated.

• Have you tried letting $x=\sin\theta$ or some trigonometric substitution? – RecordTime Jan 15 '19 at 15:02
• @FrankW. yes,but failed – Renascence_5. Jan 15 '19 at 15:04
• @Renascence_5. Can we see your attempt using that substitution? Your work may be useful to us. By the way, [W|A] can't find one. – TheSimpliFire Jan 15 '19 at 15:08
• Maple also fails. So it seems likely that the indefinite integral is not elementary. There may remain a chance for the definite integral (but contour integration has been discouraged by the OP). – GEdgar Jan 15 '19 at 15:51
• @TheSimpliFire $-d \left(\frac{1}{x}-1\right)$ integration by parts and use that substitution – Renascence_5. Jan 15 '19 at 16:01

Using series, I get $$-2\sum _{k=0}^{\infty }{\frac {{4}^{k} \left( \Psi \left( k+3/2 \right) +\gamma \right) \left( k! \right) ^{2}}{ \left( 1+2\,k \right) \left( 2\,k+2 \right) !}}$$ But I don't expect there to be a closed form.
$$\frac{\arcsin(x)^2}{x^2}=\sum_{n\geq1}\frac{2^{2n-1}}{n^2{2n\choose n}}x^{2n-2}$$ Thus $$I=\int_0^1\frac{\log(1-x^2)\arcsin(x)^2}{x^2}\mathrm dx=\sum_{n\geq1}\frac{2^{2n-1}}{n^2{2n\choose n}}\int_0^1x^{2n-2}\log(1-x^2)\mathrm dx$$ Then we look at $$J(n)=\int_0^1x^{2n-2}\log(1-x^2)\mathrm dx$$ Wolfram Alpha produces $$J(n)=\frac1{1-2n}H_{n-1/2}$$ Where $$H_n$$ is the $$n$$-th harmonic number. This can also be written as $$J(n)=\frac1{1-2n}\left[\psi(n+1/2)+\gamma\right]$$ Where $$\psi(s)$$ is the dilogarithm and $$\gamma$$ is the Euler-Mascheroni constant.