I have to face an integral of the form
$$\int_0^{2\pi}d\phi\int_0^{\pi}\sin\theta d\theta
\frac{(\sin\theta\cos\phi+\sin\psi)}{(1-\sin\psi\sin\theta\cos\phi-\cos\psi\cos\theta)}$$
where the angle $\psi$ is constant and it is considered to be between $0$ and $\pi$. Is there any way to solve this? Maybe try some sort of shift of the variables of integration? If not, can I know for sure that it diverges, probably at $(\theta,\phi)=(\psi,0)$, and how it diverges (i.e. logarithmically, linearly, etc)?
What about the limits $\psi\rightarrow0$ or $\psi\rightarrow\pi$?
Thanks.
EDIT: What if I know that $\psi$ is a small, but non-zero angle? Am I allowed to Taylor expand in $\psi$? Then the integral will be convergent at all orders, right? Does this contradict the traditional method of simply trying to evaluate the integral?
Best Answer
As you note, the denominator has a singularity at $(\theta,\phi)=(\psi,0)$. At this point, the numerator is just $2\sin^2\psi$, so unless $ \psi=0$ or $\pi$, there’s nothing to cancel the singularity. To better understand the nature of the singularity, consider the integral
$$ \int_0^{2\pi}d\phi\int_0^{\pi}\sin\theta d\theta \frac1{1-\sin\psi\sin\theta\cos\phi-\cos\psi\cos\theta}\;, $$
which has the same type of singularity. The denominator is $1$ minus the scalar product of the unit vectors corresponding to $(\theta,\phi)$ and $(\psi,0)$, and since the scalar product is invariant under rotations and the integral is over the entire solid angle, the value of the integral doesn’t depend on the value of $(\psi,0)$. So we can use $(0,0)$ instead, yielding
$$ \int_0^{2\pi}d\phi\int_0^{\pi}\sin\theta d\theta \frac1{1-\cos\theta}=2\pi\left[\log(1-\cos\theta)\right]_0^\pi\;. $$
Thus, there’s a logarithmic divergence at $\theta=0$, and thus a logarithmic divergence at $(\psi,0)$ in the original integral.
For $\psi=0$, the original integral is
$$ \int_0^{2\pi}d\phi\int_0^{\pi}\sin\theta d\theta \frac{\sin\theta\cos\phi}{1-\cos\theta}\;. $$
Now both the numerator and the denominator are quadratic in $\theta$ at $\theta=0$, so there’s no singularity, and then the integral over $\cos\theta$ yields $0$. The same happens when $\psi=\pi$.