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Question: Evaluate$$I=\int\limits_0^{1/2}dx\,\frac {\log x}{\sqrt{1-x^2}}$$

I've given it the best I could. The first thing I did was use differentiation under the integral sign, but I wasn't sure where to continue after that.

The next thing I did was try substituting $x\mapsto 2x$ to get the limits in terms of zero and one and transform $I$ into the beta function (no luck because the denominator is $1-4x^2$).

Then I tried integration by parts. It seemed promising at first because I got$$I=-\frac {\pi}6-\int\limits_0^{1/2}dx\,\frac {\arcsin x}{x}$$But I'm not sure how to evaluate the second integral. I even tried using a Laplace Transform, but I'm again not sure how to evaluate the integral$$\mathcal{L}(f(t))=\int\limits_0^{\infty}dt\, e^{-st}\arcsin t$$

Can anybody give me a few hints as to where to begin and what to do? I'm out of ideas.

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  • $\begingroup$ What convinces you that there exists a closed-form solution? $\endgroup$ – Mark Viola May 15 '18 at 0:45
  • $\begingroup$ I am not sure if a closed form exist if the interval of integration is $[0,1/2]$, but I know there is a closed form for the interval $[0,1]$, and you can use differentiation under the integral sign. Would you be interested in that intergral? $\endgroup$ – Zachary May 15 '18 at 1:01
  • $\begingroup$ I was trying to show that the integral$$\int\limits_0^1dx\,\frac {\log^2x}{\sqrt{4-x^2}}=\frac {7\pi^3}{216}$$And arrived at$$\int\limits_0^{1/2}dx\,\frac {\log^2x}{\sqrt{1-x^2}}+\log 4\int\limits_0^{1/2}dx\,\frac {\log x}{\sqrt{1-x^2}}=\frac {7\pi^3}{216}-\frac {\pi}6\log^22$$ $\endgroup$ – Crescendo May 15 '18 at 1:16
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Hint. By the change of variable, $$ \theta=\arcsin x, \qquad d\theta=\frac {dx}{\sqrt{1-x^2}}, $$one gets $$ I=\int_0^{1/2}dx\,\frac {\log x}{\sqrt{1-x^2}}=\int_{0}^{\large\frac \pi6}\ln(\sin \theta)\,d\theta $$ then one may use the standard identity, $$\log(\sin \theta)=-\ln 2-\sum_{k\geq 1}\frac{\cos(2k\theta)}{k} \qquad \left(0<\theta<\pi \right), $$ to obtain, by a valid termwise integration, $$ I=-\frac{\pi\ln 2}6-\frac12\sum_{k\ge 1}\frac{\sin(\pi k/3)}{k^2}=-\frac{\pi\ln 2}6+\frac12\Im\:\text{Li}_2\left(\frac12-i\frac{\sqrt{3}}2\right), $$ that is $$\bbox[15px,border:1px solid #ff6600]{ \int_0^{1/2}dx\,\frac {\log x}{\sqrt{1-x^2}}=-\frac{\pi\ln 2}6+\frac12\Im\:\text{Li}_2\left(\frac12-i\frac{\sqrt{3}}2\right)} $$ where $\text{Li}_2(\cdot)$ is the dilogarithm function.

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  • $\begingroup$ Is there a similar expansion for $\log^2\sin x$? I might be able to directly use that in evaluating the next integral$$I=\int\limits_0^{1/2}dx\,\frac {\log^2x}{\sqrt{1-x^2}}$$ $\endgroup$ – Crescendo May 16 '18 at 0:00
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FWIW, Maple evaluates it using a hypergeometric function:

$$ -\frac{\pi}6-\frac{\pi}{6}\,\ln \left( 2 \right) +{\frac { {\mbox{$_3$F$_2$}(3/2,3/2,3/2;\,5/2,5/2;\,1/4)}}{72}}$$

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