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?
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 $\psi=\pi$.