# Residues of complex circular functions

Maths just a moment. 0 answers, 0 views

I am trying to evaluate an integral of a circular function:

$I=\int \frac{1}{\alpha\cos\theta+\beta\sin\theta+\gamma} d\theta$ over the interval $[0, 2\pi]$, where $\alpha,\beta, \gamma$ are parameters.

I have specified constraints such that the integral will always converge, i.e. $|\alpha|+|\beta|\leq|\gamma|$, and want to compute this integral in terms of its parameters.

By changing the variables to complex ones, we are left with a quadratic on the denominator, which can be solved to reveal the two simple poles.

The integral is now 2 $\oint \frac{1}{(ai+b)z^2+(2ic)z+(ai-b)} dz$, the poles occurring at $\frac{-\gamma c+/-i\sqrt{\gamma^2-\alpha^2-\beta^2}}{\alpha i+\beta}$.

I want to use $\oint f(z)= 2\pi i \Sigma Res(f,c)$, but when evaluating the residue at each singularity the integral ends up being 0, which I think is wrong somewhere.