Evaluate $$\int_{0}^{2\pi}\int_{0}^{\pi} {\cos\phi \sin\phi \over \sqrt{R^2+r^2-2Rr(\cos\phi \cos\theta+\sin\phi \sin\theta \cos\psi )}} d\phi\ d\psi$$ where $R,r,\theta$ are all constants.

Sorry for all those distracting constants. (This integral came up from physics calculation.) My first idea was substitution $$\cos\phi \cos\theta+\sin\phi \sin\theta \cos\psi = 1+{\sin^2{\eta}\over 2rR}$$ but I don't think this approach is fruitful.

Even a simple hint about variable substitution will help me a lot. Thank you.


With AlexR's help, I did one integration with respect to $\psi$, so this is the new integral with respect to $\phi$ $$\int_{0}^{2\pi}{\cos \phi \sin \phi K({2Rr\sin \theta \sin\phi \over R^2+r^2 -2Rr\cos(\phi-\theta)})\over\sqrt{R^2+r^2 -2Rr\cos(\phi-\theta)}}d\phi$$ where $K$ is the complete elliptic integral of the first kind.

Now I'm terrified with the presence of special function in the integrand. Can this integral even be done?

  • 1
    $\begingroup$ If you use Fubini to swap the integration order, you only have $a+ b\cos \psi$ under the $\sqrt\cdot$ in the inner integral with constants $a,b$ independent of $\psi$... $\endgroup$ – AlexR Nov 21 '13 at 11:52
  • $\begingroup$ I gave it to Mathematica five minutes ago and it's still thinking, so I'm suspecting it may not have a closed form. $\endgroup$ – Neal Nov 21 '13 at 12:22
  • $\begingroup$ Was still thinking when I closed it because it's time to go in to work. $\endgroup$ – Neal Nov 21 '13 at 12:45
  • $\begingroup$ What is the original physics problem? $\endgroup$ – zy_ Nov 21 '13 at 15:38
  • $\begingroup$ math.stackexchange.com/questions/575947/… I asked new question with the context. $\endgroup$ – generic properties Nov 22 '13 at 10:45

By spherical symmetry, the integral should be independent of $\theta$, so set$\theta=0$ which gives $\pi$ times an integral which vanishes by symmetry. Hence the value is 0.

| cite | improve this answer | |
  • $\begingroup$ Thank you for your answer, but I'm not sure that there's really spherical symmetry in this problem. I mean, in this integral, there's two special axes, namely, the z-axis($\phi=0$) and the '$\phi = \theta$'axis. So, I think there's no symmetry left at all. $\endgroup$ – generic properties Nov 23 '13 at 15:15

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.