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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.


marked as duplicate by Carl Mummert, David H, Frank W., Community Jan 16 at 4:37

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  • $\begingroup$ Have you tried letting $x=\sin\theta$ or some trigonometric substitution? $\endgroup$ – Frank W. Jan 15 at 15:02
  • $\begingroup$ @FrankW. yes,but failed $\endgroup$ – Renascence_5. Jan 15 at 15:04
  • $\begingroup$ @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. $\endgroup$ – TheSimpliFire Jan 15 at 15:08
  • $\begingroup$ 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). $\endgroup$ – GEdgar Jan 15 at 15:51
  • $\begingroup$ @TheSimpliFire $-d \left(\frac{1}{x}-1\right)$ integration by parts and use that substitution $\endgroup$ – Renascence_5. Jan 15 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.


A start

$$\frac{\arcsin(x)^2}{x^2}=\sum_{n\geq1}\frac{2^{2n-1}}{n^2{2n\choose n}}x^{2n-2}$$ 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.

That's all I've got for now.


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