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I am trying to solve the following integral in spherical 3D coordinates $$\mathcal{I}(\theta,\phi)= \int_0^{2\pi}d\phi_{\Delta}\int_{\delta}^{\Delta/2}\sin\theta_{\Delta}d\theta_{\Delta} \frac{1}{[1-\sin\theta_{\Delta}\sin\theta\cos(\phi_{\Delta}-\phi)-\cos\theta_{\Delta}\cos\theta]}$$ where $(\theta_{\Delta},\phi_{\Delta})$ are the spherical angular coordinates that I wish to integrate with respect to, whereas $(\theta,\phi)$ are simply some free angular spherical coordinates (of some other vector).

I would like to learn some hints on computing it. I have tried switching the integration order, writing integrand as another integral, namely $$\frac{1}{[1-\sin\theta_{\Delta}\sin\theta\cos(\phi_{\Delta}-\phi)-\cos\theta_{\Delta}\cos\theta]}= \int_{-\infty}^0d\alpha \exp\Big(\alpha [1-\sin\theta_{\Delta}\sin\theta\cos(\phi_{\Delta}-\phi)-\cos\theta_{\Delta}\cos\theta]\Big)$$ But nothing seems to work. Can someone provide me with a hint? Also, what happens in the $\delta\rightarrow0$ limit?

Thanks

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  • $\begingroup$ What's $\Delta$? $\endgroup$
    – joriki
    Mar 13 at 13:41
  • $\begingroup$ The upper limit for the angular spherical coordinates $(\theta_{\Delta},\phi_{\Delta})$ $\endgroup$
    – schris38
    Mar 13 at 13:43
  • $\begingroup$ That's not what you wrote in the question – there you integrate over $\phi_\Delta$ without restrictions, and the upper limit for $\theta_\Delta$ is $\Delta/2$, not $\Delta$. $\endgroup$
    – joriki
    Mar 13 at 13:48
  • $\begingroup$ Indeed, you are right. So, $\Delta$ double the upper limit. $\endgroup$
    – schris38
    Mar 13 at 14:52

1 Answer 1

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Note that the denominator is $1$ minus the scalar product of the two unit vectors corresponding to $(\theta,\phi)$ and $(\theta_\Delta,\phi_\Delta)$. If you were to integrate over the entire solid angle, the result wouldn’t depend on $(\theta,\phi)$ (since the scalar product is invariant under rotations). So in that case you could choose a convenient value for $(\theta,\phi)$ to evaluate the integral, for instance $\theta=\phi=0$. That would lead to

$$ \int_0^{2\pi}\mathrm d\phi_\Delta\int_0^\pi\mathrm d\theta_\Delta\frac{\sin\theta_\Delta}{1-\cos\theta_\Delta}\;. $$

The denominator is quadratic in $\theta_\Delta$, while the numerator is only linear, so this integral diverges at $\theta_\Delta=0$.

Whether this singularity occurs in your truncated version of the integral depends on whether $(\theta,\phi)$ lies within the region of integration. The integral is only defined for values of $(\theta,\phi)$ outside this region. The limit $\delta\to0$ doesn’t have any special significance, since the singularity doesn’t occur at $\theta_\Delta=0$ but at $(\theta_\Delta,\phi_\Delta)=(\theta,\phi)$.

As regards evaluating this integral in closed form, I doubt that that’s possible, but you never know.

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  • $\begingroup$ Hi @joriki and thanks for the reply. I do not have to integrate over the entire solid angle, but just by a small part of it. Hence, I can not exploit symmetry. Furthermore, I believe that $(\theta_{\Delta},\phi_{\Delta})$ can also take the values $(\theta,\phi)$, as later on I have to integrate $\theta$ and $\phi$ over the entire solid angle. So, I guess that would mean that my integral is not defined... Thanks for making me understand that. If you have any additional comments, please do let me know... $\endgroup$
    – schris38
    Mar 13 at 15:56
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    $\begingroup$ @schris38: Yes, I realize that you can't exploit symmetry; I just used it to illustrate the singularity at $(\theta_\Delta,\phi_\Delta)=(\theta,\phi)$. That you also integrate over $\theta$ and $\phi$, while relevant, doesn't change the fact that the integral is not well-defined – even if you regard it as a four-dimensional integral without specifying an integration order, it still diverges. An analogue would be the integral $\int_{-1}^1\mathrm dx\int_{-1}^1\mathrm dy\frac1{|x-y|}$, which diverges no matter whether you regard it as one iterated integral or the other or as two-dimensional. $\endgroup$
    – joriki
    Mar 13 at 16:23
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    $\begingroup$ thank you so much for your help $\endgroup$
    – schris38
    Mar 13 at 16:25

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