The problem is to evaluate in closed-form:


An estimate for the integral is $1.76663875028545$.

I have tried all the usual techniques, including by parts and various $u$-subs, but they all seem to make the integral more contrived. The Weierstrass substitution $x=\tan{\left(\theta/2\right)}$ 'simplifies' the integrand to one with only polynomials, with a nice factorization as well:


However, I do not know how to continue from here, primarily because the exponent $\frac{2}{3}$ in the denominator still remains.

In addition, both Maple and WolframAlpha could not find a closed-form answer (though that does not mean that one does not exist).

Any ideas? Thanks!

  • $\begingroup$ Have you tried using $f(a+b-x)=f(x)$? $\endgroup$ – Teh Rod Aug 2 '17 at 15:13
  • $\begingroup$ @TehRod : Do you mean that we can make the upper bound $\pi/4$ instead of $\pi/2$? How else can one apply this? $\endgroup$ – Ant Aug 2 '17 at 15:18
  • $\begingroup$ I don't know, that's just the first thing that came to mind when I saw this. Especially since there does not appear to be a closed form for the antiderivative $\endgroup$ – Teh Rod Aug 2 '17 at 15:19
  • $\begingroup$ Better numerical estimation: $1.766638750285906$ $\endgroup$ – Turing Aug 2 '17 at 15:36

First, enforcing the substitution $\theta=\arctan(x)$ reveals

$$\int_0^{\pi/2}\frac{1}{\left(\cos^3(\theta)+\sin^3(\theta)\right)^{2/3}}\,d\theta=\int_0^\infty \frac{1}{\left(1+x^3\right)^{2/3}}\,dx$$

Next, we let $y=x^3$ to obtain

$$\begin{align} \int_0^\infty \frac{1}{\left(1+x^3\right)^{2/3}}\,dx&=\frac13\int_0^\infty \frac{y^{-2/3}}{(1+y)^{2/3}}\,dy\\\\ &=\frac13 B\left(1/3,1/3\right)\tag 1\\\\ &=\frac13 \frac{\Gamma^2(1/3)}{\Gamma(2/3)}\tag 2 \end{align}$$

where $B(x,y)$ is the Beta function and $\Gamma(x)$ is the Gamma function. In going from $(1)$ to $(2)$ we used the relationship $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$.

  • $\begingroup$ I was going to post an answer, but yours is bloody awesome! +1 $\endgroup$ – Turing Aug 2 '17 at 15:38
  • $\begingroup$ @HenryTuring Wow! Much appreciated. $\endgroup$ – Mark Viola Aug 2 '17 at 15:38
  • $\begingroup$ (+1) Very nice! I knew about the Beta function beforehand but didn't think to apply it here. Thanks! $\endgroup$ – Ant Aug 2 '17 at 15:40
  • 2
    $\begingroup$ @JohnChessant, you should accept the answer I guess. $\endgroup$ – Zaid Alyafeai Aug 2 '17 at 20:29

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.