# Contour integral of $\int_0^{2\pi} \frac{1}{A - cos \theta} d\theta$

I'm supposed to evaluate $\int_0^{2\pi} \frac{1}{A - cos \theta} d\theta$

Using a contour of a unit circle, $z=e^{i\theta}$.

This is the same as:

$$2i \oint \frac{1}{z^2 - Az + 1 } dz$$

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!

Residue at smaller root $= -\frac{1}{2\sqrt{A^2-1}}$ Residue at larger root $= \frac{1}{2\sqrt{A^2-1}}$

I have a hunch that they should be the same in order for me to use residue theorem.

• How do you know $\int_0^{2\pi}...$ is same as $2\pi\oint ...$ – Lion Jun 14 '14 at 0:19
• @Lion the OP does not write $2\pi$, they write $2i$, which comes from the fact that $\cos \theta=\frac{e^{i\theta}+e^{-i\theta}}{2}$ and making the proper substitution into $d\theta$. – Hayden Jun 14 '14 at 0:25
• @Hayden OK, I get it. Thank you :) – Lion Jun 14 '14 at 0:37
• In my opinion the answer should be $2\pi i/\sqrt{1-A^2}$. – user202464 Dec 21 '14 at 2:54

$\int_{0}^{2\pi} \frac {1}{A - \cos(\theta)} d\theta = 2i\oint \frac{1}{1-2Az + z^2} dz$. Notice that factor of $2$ in the denominator of the second integral.
Firstly, if $|A|>1$, it is obvious that all the deduction is valid. Although there are two poles that $A\pm\sqrt{A^2-1}$, only $A-\sqrt{A^2-1}$ in the unit circle.
What is more, in your assumption $|A|<1$ (the same situation as $|A|\leq1$), it can easy to check the OP is an improper integral and it is divergence (Consider the easy example that $\int\frac{1}{1-\cos(\theta)}d\theta=\frac{-1}{\tan(\theta/2)}$). Maybe it is seems that residue method is also valid but I think it just the Cauchy principle value.
• For the case that $|A|>1$, the answer is right. But for $|A|\leq1$, it is invalid because the improper integral is divergence. – Lion Jun 14 '14 at 13:20