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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.

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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?

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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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