Let$$I(\epsilon)\equiv \int_0^\pi\frac{\sin \theta\; d\theta}{A \cos \theta-i \epsilon}$$ where $A>0$ is a constant
Is $\lim_{\epsilon\to 0}I(\epsilon)$ finite?
My attempt:
$$I(\epsilon)=\int_0^{\frac{\pi}{2}}\frac{\sin \theta\;d\theta}{A\cos \theta-i\epsilon}-\int_0^{\frac{\pi}{2}}\frac{\sin \theta\;d\theta}{A\cos \theta+i\epsilon}=2i\epsilon\int_{0}^{\frac{\pi}{2}}\frac{\sin \theta\;d\theta}{A^2 \cos^2 \theta+\epsilon^2}$$
From here I donot see how to conclude whether $lim_{\epsilon\to 0}I(\epsilon)$ is is finite or not.
P.S. I think we have to do some analytic continuation here, but I am not sure how to go about doing it
Note that $$2\epsilon\int_{0}^{\frac{\pi}{2}}\frac{\sin \theta\;d\theta}{A^2 \cos^2 \theta+\epsilon^2}=\left[-\frac{2}{A}\arctan\left(\frac{A\cos(\theta)}{\epsilon}\right)\right]_0^{\pi/2}= \frac{2}{A}\arctan\left(\frac{A}{\epsilon}\right)$$ and the right-hand side has a finite limit as $\epsilon\to 0$. Therefore $$\lim_{\epsilon\to 0}I(\epsilon)=\frac{2i}{A}\lim_{\epsilon\to 0}\arctan\left(\frac{A}{\epsilon}\right)=\frac{i\pi}{A}.$$