# Showing that $\int_0^\pi\frac{\cos n\theta}{\cos\theta-\cos\theta_0}d\theta=\pi\frac{\sin n\theta_0}{\sin\theta_0}$

I am reading Debnath & Bhatta "Integral Transforms and Their Applications, 3rd". They cited one example from Zayed "Handbook of Function and Generalized Function Transformations" and stated an integral (Eq.(9.5.45)), for a non-negative integer n, $$\int_0^\pi \frac{\cos(n \theta)}{\cos(\theta)-\cos(\theta_0)}d\theta=\pi \frac{\sin(n \theta_0)}{\sin(\theta_0)}$$ It turns out many books on Hilbert transform use this relation for Airfoil Design example, e.g., Prederick W.King, Chapter 11.14 "Hilbert Transform-V1".

Interestingly, I remember the following one from Paul J. Nahin, Eq.(2.3.8) of "Inside Interesting Integrals" $$\int_0^\pi \frac{\cos(n \theta)-\cos(n \theta_0)}{\cos(\theta)-\cos(\theta_0)}d\theta=\pi \frac{\sin(n \theta_0)}{\sin(\theta_0)}.$$ You can find the proof in that book.

So, if both integrals are correct, then we should have $$\int_0^\pi \frac{1}{\cos(\theta)-\cos(\theta_0)}d\theta=0,$$ which I cannot see why. Mathmatica gives an pure imaginary result here. How shall I interpret these and how can I prove the first integral?

Are those integrals even well-defined? Let $$\theta_0$$ be such that $$\cos(\theta_0)=1/2$$. For instance let $$\theta_0=\frac{\pi}{3}$$. Take $$n=1$$. Now $$\int_0^\pi\frac{\cos(n\theta_0)}{\cos\theta-\cos(\theta_0)}\;d\theta=\int_0^\pi\frac{1/2}{\cos\theta-1/2}\;d\theta.$$ This integral is actually an improper one, as $$\pi/3$$ is a singularity. And it does not converge.
Similarly, $$\int_0^\pi\frac{\cos(n\theta)}{\cos(\theta)-\cos(n\theta_0)}\;d\theta=\int_0^\pi\frac{\cos(\theta)}{\cos(\theta)-1/2}\;d\theta$$ fails to converge.
• But it converges to $\pi$ as a principal value integral. That is, $\lim_{\epsilon\to0^+} \int_0^{\pi/3-\epsilon}+\int_{\pi/3+\epsilon}^\pi$. – Jean-Claude Arbaut Dec 3 '18 at 0:58
• Hey, I see your concern regarding $\theta_0$. If you first let n=0,1,...into the integral and then there is no singularity at all. – gouwangzhangdong Dec 3 '18 at 1:11