This question already has an answer here:

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?


marked as duplicate by David H, Joel Reyes Noche, Joey Zou, Henrik, Daniel W. Farlow Aug 10 '16 at 22:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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. $$I(k_*)\approx 0.6412749150809320477747$$ $$\frac{\pi }{2 \sqrt{6}}\approx0.6412749150809320477720$$


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