$I = \int^{2\pi}_0 \dfrac{d\theta}{2 – \cos \theta}$
This is straight from a book I'm reading, which suggests to convert $\cos\theta$ into $0.5(z+1/z)$ and then solve the integral on the unit circle. This is what I don't understand. The two singularities of this function are at $2\pm \sqrt 3$ and so the unit circle only encircles one of the singularities. The rest of the calculations I understand, but I just don't understand how you can decide to calculate this on the unit circle and not a circle of a different radius? My only idea is that changing the radius of the circle on which the contour integral is evaluated will shift the singularities appropriately, is this the case?
As an aside, is there a difference between the term "singularity" and "pole" in contour integration?
Best Answer
In a mixed real/complex-analytic way we can notice that: $$ I = \int_{0}^{2\pi}\frac{d\theta}{2-\cos\theta}=2\int_{0}^{\pi}\frac{d\theta}{2-\cos\theta}=8\int_{0}^{\pi/2}\frac{d\theta}{4-\cos^2\theta}$$ and by replacing $\theta$ with $\arctan t$ we get: $$ I = 8\int_{0}^{+\infty}\frac{dt}{3+4t^2} = 4\int_{\mathbb{R}}\frac{dt}{3+4t^2}=2\int_{\mathbb{R}}\frac{dz}{3+z^2}$$ and the last integral can be computed through the residue of the integrand function in the simple pole $z=i\sqrt{3}$ (aside: not every singularity is a simple pole. Multiple poles and essential singularities may occur, too), leading to: $$ I = \frac{2\pi}{\sqrt{3}}.$$