Complex integration with residues $\int_0^{2\pi}\frac{d\phi}{(p+q\cos\phi)^2}=\frac{2\pi p}{(p^2-q^2)^{3/2}}$

Question

I am trying to show that $\int_0^{2\pi}\frac{d\phi}{(p+q\cos\phi)^2}=\frac{2\pi p}{(p^2-q^2)^{3/2}}$ with $p>q>0$.

I set $C:z=e^{i\phi}\rightarrow dz=ie^{i\phi}=izd\phi$. Then I have that $\cos(\phi)=\frac{1}{2}(z+\frac{1}{z})$.

Then I have $\oint_C=\frac{dz}{iz(p+q[\frac{1}{2}(z+\frac{1}{z})]^2}=\oint_C=\frac{dz}{iz(p+q(\frac{z}{2}+\frac{1}{2z}))^2}$ I know that I need to find what roots in denominator are equal to 0 in the denominator are in C in order to use residues. I am having a hard time figuring out how to do this. I know I can move the i to be -i in the numerator. Then I have $\oint_C=\frac{-idz}{z(p+q[\frac{1}{2}(z+\frac{1}{z})]^2}$. Then I have z=0 and two other roots, that I can't figure out... https://www.wolframalpha.com/input/?i=(p%2Bq(z%2F2%2B2%2Fz))%5E2%3D0

Any help would be much appreciated. Thanks.

Answer 1

Residues method:

After the usual sine/cosine substation you will get the integral

$$\int \frac{4z\ dz}{i(2pz + qz^2 + q)^2}$$

Thence your poles are at

$$z_{1, 2} = \frac{-p \pm \sqrt{p^2 - q^2}}{q}$$

By residues methods, you can easily obtain the wanted solution:

$$\boxed{\frac{2 \pi }{\sqrt{\frac{p-q}{p+q}} (p+q)}}$$

Answer 2

Another approach is to instead compute (without loss of generality, taking $q=1$, if not normalizing)

$$\int_0^{2\pi}\frac{d\phi}{p+\cos\phi}$$

by the same method as in Von Neumann's answer, and then differentiate the answer with respect to $p$. This is a bit less bothersome than the original problem.