I'm supposed to evaluate $\int_0^{2\pi}\frac1{A-\cos\theta}\,\mathrm{d}\theta$
Using a contour of a unit circle, $z=\mathrm{e}^{\mathrm{i}\theta}$.
This is the same as:
$$2\mathrm{i}\oint\frac{1}{z^2-Az + 1}\,\mathrm{d}z.$$
The roots are $z=A\pm\sqrt{A^2-1} $. For the unit circle to enclose both roots, $|A|<1$.
However, the sum of the residue gives zero! The residue at smaller root $= -\frac1{2\sqrt{A^2-1}}$, while the residue at larger root $= \frac1{2\sqrt{A^2-1}}$.
I have a hunch that they should be the same in order for me to use residue theorem.
Best Answer
I did the calculation and I got a different answer. In particular, I get that
$\int_{0}^{2\pi} \frac {1}{A - \cos(\theta)} d\theta = 2i\oint \frac{1}{1-2Az + z^2} dz$. Notice that factor of $2$ in the denominator of the second integral.