Integral over angular spherical coordinates vol. 3

Question

This question is related to two older posts of mine: Integral over spherical angular coordinates vol. 2 and Integral over angular spherical coordinates. None of the above contain the answer to what I am going to ask now.

Suppose I have the integral over the spherical angular coordinates

$$\int_0^{2\pi}d\phi\int_0^{\pi}\sin\theta d\theta \bigg[\frac{1}{(1-\sin\psi\sin\theta\cos\phi-\cos\psi\cos\theta)}- \frac{1}{(1-\cos\theta)}\bigg]$$

with $0<\psi<\pi$ an arbitrary angle. How am I supposed to determine whether or not this integral diverges? I know that the first term has a single pole at $(\theta,\phi)=(\psi,0)$ and the second term has a single pole, again, at $(\theta,\phi)=(0,0)$. How can this information help me determine the behavior of the integral?

Can it be such that the divergence of the first term cancels with the divergence of the second term, leaving us with a finite result?

$\large{\textbf{EDIT:}}$ My attemtp to study whether or not the integral above is divergent is as follows:

1. Let the first integral be labelled as $\mathcal{I}_1$ and the second one be labelled as $\mathcal{I}_2$.

2. The pole for $\mathcal{I}_1$ is located at $(\theta,\phi)=(\psi,0)$, so I expand around that point. I also split the $\theta$ integral, choosing the integration limits to be $\theta\in[\psi-\frac{\Lambda_{\theta}}{2},\psi-\frac{\lambda_{\theta}}{2}]$ in the first integral and $\theta\in[\psi+\frac{\lambda_{\theta}}{2},\psi+\frac{\Lambda_{\theta}}{2}]$ in the second one. The capital $\Lambda_{\theta}$ is there such that it justifies the approximations made (so, it is small), whereas the small cutoff $\lambda_{\theta}$ is soon to be taken to zero if possible. Hence, $$\mathcal{I}_1= 2\sin\psi \int_{\lambda_{\phi}}^{\Lambda_{\phi}}d\phi\bigg[ \int_{\psi-\frac{\Lambda_{\theta}}{2}}^{\psi-\frac{\lambda_{\theta}}{2}} d\theta \frac{1}{\phi^2+(\theta-\psi)^2}+ \int_{\psi+\frac{\lambda_{\theta}}{2}}^{\psi+\frac{\Lambda_{\theta}}{2}} d\theta \frac{1}{\phi^2+(\theta-\psi)^2}\bigg]$$ which yields (after changing the order of integration $$\mathcal{I}_1= 2\sin\psi \bigg\{ \int_{\psi-\frac{\Lambda_{\theta}}{2}}^{\psi-\frac{\lambda_{\theta}}{2}} d\theta \frac{1}{(\theta-\psi)}\bigg[ \tan^{-1}\Big(\frac{\Lambda_{\phi}}{\theta-\psi}\Big)- \tan^{-1}\Big(\frac{\lambda_{\phi}}{\theta-\psi}\Big)\bigg]+ \int_{\psi+\frac{\lambda_{\theta}}{2}}^{\psi+\frac{\Lambda_{\theta}}{2}} d\theta \frac{1}{(\theta-\psi)}\bigg[ \tan^{-1}\Big(\frac{\Lambda_{\phi}}{\theta-\psi}\Big)- \tan^{-1}\Big(\frac{\lambda_{\phi}}{\theta-\psi}\Big)\bigg]\bigg\}$$ and upon expanding the inverse tangent functions and integrating wrt $\theta$, one gets $$\mathcal{I}_1= 2\sin\psi(\Lambda_{\phi}-\lambda_{\phi}) \bigg\{ \int_{\psi-\frac{\Lambda_{\theta}}{2}}^{\psi-\frac{\lambda_{\theta}}{2}} d\theta \frac{1}{(\theta-\psi)^2}+ \int_{\psi+\frac{\lambda_{\theta}}{2}}^{\psi+\frac{\Lambda_{\theta}}{2}} d\theta \frac{1}{(\theta-\psi)^2}\bigg\}$$ and finally $$\mathcal{I}_1= 2\sin\psi(\Lambda_{\phi}-\lambda_{\phi}) \bigg\{ \big(-\frac{2}{\Lambda_{\theta}}+\frac{2}{\lambda_{\theta}}\big)+ \big(\frac{2}{\lambda_{\theta}}-\frac{2}{\Lambda_{\theta}}\big) \bigg\}= \frac{8\sin\psi}{\Lambda_{\theta}\lambda_{\theta}} (\Lambda_{\phi}-\lambda_{\phi}) (\Lambda_{\theta}-\lambda_{\theta})$$ In the limit $\lambda_{\phi},\lambda_{\theta}\rightarrow0$, this diverges as $\sim\frac{8\Lambda_{\phi}\sin\psi}{\lambda_{\theta}}$

3. The pole for $\mathcal{I}_2$ is located at $(\theta,\phi)=(0,0)$, so we expand around that point to get $$\mathcal{I}_1= \int_{\lambda_{\phi}}^{\Lambda_{\phi}}d\phi \int_{\lambda_{\theta}}^{\Lambda_{\theta}} \theta d\theta \frac{2}{\theta^2}= 2(\Lambda_{\phi}-\lambda_{\phi}) \ln(\frac{\Lambda_{\theta}}{\lambda_{\theta}})$$ which yields $$\mathcal{I}_1= \frac{2}{\lambda_{\theta}}(\Lambda_{\phi}-\lambda_{\phi}) (\Lambda_{\theta}-\lambda_{\theta})+...$$ In the limit $\lambda_{\phi},\lambda_{\theta}\rightarrow0$, this diverges as $\sim\frac{2\Lambda_{\phi}\Lambda_{\theta}}{\lambda_{\theta}}$

So, I observe that one divergence is $\psi$ dependennt, whereas the other is not. Also, once is proportional to the scale $\Lambda_{\theta}$, whereas the other is not. However, both diverge with the same way as $\lambda_{\theta}$ goes to zero. Does that mean that the divergent part does not cancel out, or am I wrong somewhere??

Any help will be appreciated!

P.S.: I do not really care about the result. I just wish to understand why the integral is convergent if it is so...