i am trying to compute the integral given below using Wolfram alpha $$\int_0^{\sqrt2}\sin^{-1}\left(\frac{\sqrt{2-x^2}}{2}\right)dx$$ However it can be solved by numerical method but i don't know why Wolfram alpha doesn't compute it. could somebody explain or help me solve this integral? thank u very much

  • $\begingroup$ Because Wolfram alpha has been made up by people and those people don't know how to compute it? $\endgroup$ – Han de Bruijn Sep 3 '16 at 8:25
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    $\begingroup$ @HandeBruijn Wolfram Mathematica computes the integral just fine. $\endgroup$ – Math1000 Sep 3 '16 at 8:26
  • $\begingroup$ it can be solved by using numerical method. $\endgroup$ – Bhaskara-III Sep 3 '16 at 8:26
  • $\begingroup$ Mathematica gives a nice closed form $2\pi^{3/2} / \Gamma(\frac{1}{4})^2$, though I have no idea why (without using elliptic functions). $\endgroup$ – Sangchul Lee Sep 3 '16 at 8:35
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    $\begingroup$ @SangchulLee no elliptic integrals are necessary it is all quite straightforward please see my answer $\endgroup$ – tired Sep 3 '16 at 9:32

The most direct way of calculating this integral starts as already explained by @Claude Leibovici: We integrate by parts and observe that the boundary term vanishs, so we are down to

$$ I=\int_0^{\sqrt{2}}dx\frac{x^2}{2\sqrt{1-(x^4/4)}} $$

now setting $r=x^4/4$ we obtain

$$ I=\frac{1}{2\sqrt{2}}\int_0^{1}dr\frac{1}{r^{1/4}(1-r)^{1/2}} $$

this integrals equals an representation of the Eulerian Beta function which, may in turn be expressed in terms of Gamma functions

$$ I=\frac{1}{2\sqrt{2}}\frac{\Gamma(1/2)\Gamma(3/4)}{\Gamma(5/4)} $$

using Gamma duplication as well as $\Gamma(1/2)=\sqrt{\pi}$ this boils down to

$$ I=\frac{1}{2\sqrt{2}}\frac{4 \sqrt{2}\pi^{3/2}}{\Gamma(1/4)^2}=\frac{2\pi^{3/2}}{\Gamma(1/4)^2} $$

as expected

  • $\begingroup$ thank you for the help but please tell me how to find out the value of $\Gamma(\frac 14)$? $\endgroup$ – Bhaskara-III Sep 3 '16 at 11:26
  • $\begingroup$ en.wikipedia.org/wiki/Gauss%27s_constant you might enjoy this wikipedia entry $\endgroup$ – tired Sep 3 '16 at 12:00
  • $\begingroup$ please tell me how you reduced $\Gamma(\frac34)$ to $\Gamma(\frac14)$ from second to last step? $\endgroup$ – Bhaskara-III Sep 3 '16 at 12:42
  • $\begingroup$ you can relate both values by the gamma duplication theorem $\endgroup$ – tired Sep 3 '16 at 12:46
  • $\begingroup$ please provide a helpful link to go through the gamma duplication theorem $\endgroup$ – Bhaskara-III Sep 3 '16 at 13:24

I do not know why Wolfram alpha does not compute this integral.

Let use consider the antiderivative $$I=\int\sin^{-1}\left(\frac{\sqrt{2-x^2}}{2}\right)dx$$ Integrating by parts lead to $$I=x \sin ^{-1}\left(\frac{\sqrt{2-x^2}}{2}\right)+\int\frac{x^2}{\sqrt{4-x^4}}dx$$ The second term involves elliptic integrals $$\int\frac{x^2}{\sqrt{4-x^4}}dx=\sqrt{2} \left(E\left(\left.\sin ^{-1}\left(\frac{x}{\sqrt{2}}\right)\right|-1\right)-F\left(\left.\sin ^{-1}\left(\frac{x}{\sqrt{2}}\right)\right|-1\right)\right)$$ Concerning the definite integral, the first term cancels because of the given bounds and $$J=\int_0^t\frac{x^2}{\sqrt{4-x^4}}dx=\sqrt{2} \left(E\left(\left.\sin ^{-1}\left(\frac{t}{\sqrt{2}}\right)\right|-1\right)-F\left(\left.\sin ^{-1}\left(\frac{t}{\sqrt{2}}\right)\right|-1\right)\right)$$ which tends to $$\sqrt{2} \left(E(-1)-K(-1)\right)$$ when $t \to \sqrt{2}$. In fact this can simplify to $$\frac{2 \pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}\approx 0.847213$$

  • $\begingroup$ Alpha actually does compute it given proper input. +1 for writing down solution. $\endgroup$ – Ennar Sep 3 '16 at 8:52
  • $\begingroup$ No offense, but don't you think elliptic integrals are a bit too much here? given the limits of integration you can straightforwardly transform, after the integration by parts, it into an euler beta like expression by using $r=x^4/4$ $\endgroup$ – tired Sep 3 '16 at 9:00
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    $\begingroup$ @tired. You are right and may I confess that I upvoted your good answer ? Cheers. $\endgroup$ – Claude Leibovici Sep 3 '16 at 9:29

WolframAlpha isn't really that good at reading $\TeX$; you need to simplify the input a bit.

  • $\begingroup$ Actually, your input is not what is intended. It should be this. $\endgroup$ – Ennar Sep 3 '16 at 8:36
  • $\begingroup$ @Ennar You mean arcsin instead of sin^(-1)? WA handles that. Actually, note that in your link, WA actually converts back to $\sin^{-1}$. Also, you still get $\frac{2\pi^{3/2}}{\Gamma(1/4)^2}$ $\endgroup$ – Arthur Sep 3 '16 at 8:43
  • $\begingroup$ No, arcsin/sin^{-1} is irrelevant. What you link gives is: "Using closest Wolfram|Alpha interpretation: -1 sqrt 2-x^2 /2" i.e. it is parsed completely wrong way by alpha. $\endgroup$ – Ennar Sep 3 '16 at 8:50
  • $\begingroup$ @Ennar When I click my link, I get this result, which to me looks like it's parsed correctly. True, it first gets some strange complex value, then corrects it, but that's the same for yours. $\endgroup$ – Arthur Sep 3 '16 at 8:54
  • $\begingroup$ I'm getting this, and have no idea what is going on... $\endgroup$ – Ennar Sep 3 '16 at 9:01

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