4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

FWIW, Maple evaluates it using a hypergeometric function:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.