# Integration over angular spherical coordinates

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

• What's $\Delta$? Mar 13 at 13:41
• The upper limit for the angular spherical coordinates $(\theta_{\Delta},\phi_{\Delta})$ Mar 13 at 13:43
• 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 $\Delta$. Mar 13 at 13:48
• Indeed, you are right. So, $\Delta$ double the upper limit. Mar 13 at 14:52

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.

• 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... Mar 13 at 15:56
• @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. Mar 13 at 16:23
• thank you so much for your help Mar 13 at 16:25