How to solve integration elegantly using contour integration

Question

Solve this integral analytically and prove the answer:

$$ \int_0^{\pi/2}\frac{d\psi}{1-\cos\theta \cdot \cos\psi} = \frac{\pi - \theta}{sin \theta}, \quad \theta \in (0,2\pi) $$ To see the pic

I have already proved it numerically using Mathematica. But still can't prove it analytically.

Please do not post images of equations (which cannot be searched). Instead, take the time and care to write the equations (in MathJax). — David G. Stork, Apr 15 '20 at 17:50

score -1 · Accepted Answer · answered Apr 15 '20 at 17:53

-1

This is a standard integral:

Let $\cos \theta = a$ (and $-1\leq a \leq 1$):

$$\int\limits_{\psi = 0}^{\pi/2} \frac{1}{1 - a \cos \psi}\ d\psi =\frac{2 \tan ^{-1}\left(\frac{a+1}{\sqrt{1-a^2}}\right)}{\sqrt{1-a^2}}$$

answered Apr 15 '20 at 17:53

David G. Stork

Hi David, Thank you for answering! Do you know how to solve this using contour integration on complex plane? And the analytical answer is ought to be $$\frac{\pi - \theta}{sin(\theta)}$$ – Shuofeng Zhang Apr 15 '20 at 17:57
You have a simple pole at $\cos \psi = 1/a$. – David G. Stork Apr 15 '20 at 17:59

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