Show that : $$\int_0^1\frac{(\sin ^{-1}x)^2}{x}\text{d}x=\frac{\pi ^2\ln 2}{4}-\frac78\zeta(3)$$ This integral is in "irresistible integrals" on page 122. I can't prove this one.


Using the change of variables $ x=\cos(t) $ gives $$\int_0^1\frac{(\sin ^{-1}x)^2}{x}\text{d}x= \frac{1}{4}\,\int _{0}^{\pi/2 }\!{\frac { \left( \pi -2\,t \right) ^{2}\sin\left( t \right) }{\cos \left( t \right) }}{dt}.$$

Integrating the last integral using by parts techniques with $u=(\pi-2t)^2$ yields

$$ \frac{1}{4}\,\int _{0}^{\pi/2 }\!{\frac { \left( \pi -2\,t \right) ^{2}\sin \left( t \right) }{\cos \left( t \right) }}{dt}=-\pi\int _{0}^{\pi/2 }\ln\left(\cos \left( t \right)\right) \,dt + 2\int_{0}^{\pi/2}\,t\,\ln \left( \cos \left( t \right) \right) \,{dt}$$

$$ = \frac{{\pi }^{2}\ln\left( 2 \right)}{2} + 2\int_{0}^{\pi/2}\,t\,\ln \left( \cos \left( t \right) \right) \,{dt}. $$

Now, I leave it here for you to finish the task. You may need to use some techniques from a previous problem.

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  • $\begingroup$ Thx ! I got it. $\endgroup$ – Ryan Feb 13 '13 at 3:21
  • $\begingroup$ @Ryan: You are welcome. $\endgroup$ – Mhenni Benghorbal Feb 13 '13 at 13:17

This is not an answer, but you might be interested in knowing why this can work. I thought of writing this as a comment, but it's quite big.

  • Note that $\displaystyle \ln(1+x) = x-\frac{x^2}{2} + \frac{x^3}{3} + \cdots$

  • So $\ln(2) = \displaystyle 1-\frac{1}{2} + \frac{1}{3} + \cdots$

Now I tried the standard integration by parts method. Put $x =\sin\theta$. Then the integral becomes $$I =\int_{0}^{\pi/2} \theta^{2} \cdot \cot\theta \ d\theta$$ Using the standard ILATE rule take $dv=\cot\theta \ d\theta$ so that you have $v=\ln(\sin\theta)$. Take $u=x^{2} \implies du = 2\theta \ d\theta$. So we have $$I = \underbrace{\bigl( u\cdot v\bigr)_{0}^{\pi/2}} - 2\int_{0}^{\pi/2} \theta \cdot \ln(\sin\theta)\ d\theta .$$

The underbrace term $\to 0$ so we are left with $$-2\int_{0}^{\pi/2} \theta \cdot \ln(\sin\theta) \ d\theta$$

Now \begin{align*} \int_{0}^{\pi/2} \theta \cdot \ln(\sin\theta) \ d\theta &=\int_{0}^{\pi/2} \theta \cdot \ln\bigl(1+ (\sin\theta-1)\bigr)\ d\theta \\ &= \int_{0}^{\pi/2} \theta \cdot \biggl[ (\sin\theta-1) - \frac{(\sin\theta -1)^{2}}{2}+\frac{(\sin\theta-1)^{3} }{3}- \cdots \biggr] \ d\theta \end{align*}

Now collecting the $\theta$ involving terms we see that its: \begin{align*} \int_{0}^{\pi/2} \biggl( -\theta + \frac{\theta}{2} -\frac{\theta}{3} + \cdots \biggr) \ d\theta &=-\biggl(1-\frac{1}{2}+\frac{1}{3} - \cdots\biggr)\cdot \int_{0}^{\pi/2} \theta \ d\theta \\ &= -\ln(2) \cdot \frac{\pi^{2}}{8} \end{align*}

Note that we have to multiply by $-2$ to complete the solution.

  • $\begingroup$ For getting the $\zeta(3)$ part, i think we have evaluate $\int_{0}^{\pi/2} \sin^{n}\theta d\theta$ for different values of $n$ and that leads to Wallis formula. I am not sure of it though :( $\endgroup$ – user61846 Feb 11 '13 at 16:03
  • $\begingroup$ I am not able to community wikitize this answer as I haven't registered yet, can anyone do it for me please $\endgroup$ – user61846 Feb 11 '13 at 16:05

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