This is the integral one can find in the Introduction of 'Special Integrals of Gradshteyn and Ryzhik the Proofs - Volume I' by Victor H. Moll:

$$\int_0^{+\infty} \frac{dx}{(1+x^2)^{3/2} \left[ \phi(x) + \sqrt{\phi(x)} \right]^{1/2}}, \quad \phi(x) = 1 + \frac{4x^2}{3(1+x^2)^2} \;.$$ The author doesn't know the final answer. It is claimed that it is $\pi / 2 \sqrt{6}$, though numerical integration contradicts this.

Any ideas how to solve it or where to find clues?

  • $\begingroup$ This is a weird question, but it could be interesting if we knew where does this beast come from...Otherwise it just looks like an exercise made up by someone getting bored bad. $\endgroup$ – DonAntonio Aug 6 '16 at 14:20
  • 1
    $\begingroup$ I agree with you that there is a problem somewhere since numerical integration leads to $0.66637711426883385640$ and inverse symbolic calculators do not find anything. $\endgroup$ – Claude Leibovici Aug 6 '16 at 14:45
  • $\begingroup$ The $4/3$ ratio hints that the triple angle formula might come at play. Hence i would make a hyperbolic substitution and adjust the function $\phi(x)$ accordingly. I'll try that when i get home. $\endgroup$ – nospoon Aug 6 '16 at 16:01
  • $\begingroup$ the conjectured closed form seems to be not correct :( $\endgroup$ – tired Aug 7 '16 at 1:28
  • $\begingroup$ I have not been able to find the book you refer to. Could you e-mail a copy of this page (my address is in my profile). Thanks. $\endgroup$ – Claude Leibovici Aug 7 '16 at 4:31

Just out of curiosity, I worked numerically$$I(k)=\int_0^{\infty } \frac{dx}{\left(x^2+1\right)^{3/2} \sqrt{\phi (x)+\sqrt{\phi (x)}}}\qquad, \qquad \phi(x) = 1 + \frac{kx^2}{(1+x^2)^2} $$ and tried to find $k$ such that $$I(k)=\frac{\pi }{2 \sqrt{6}}$$ The closest value I found is $k_*=0.5923509316314110643$ which inverse symbolic calculators did not find any equivalent. $$I(k_*)\approx 0.6412749150809320477747$$ $$\frac{\pi }{2 \sqrt{6}}\approx0.6412749150809320477720$$

| cite | improve this answer | |

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