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Evaluate $$\int_{0}^{2\pi}\frac{d \theta}{1+2p\cos \theta+p^2}$$ where $0<|p|<1$

$$\int_{0}^{2\pi}\frac{d \theta}{1+2p\cos \theta+p^2}=\frac{1}{i}\int_{|z|=1}\frac{d z}{z(1+pz+\frac{p}{z}+p^2)}=\frac{1}{i}\int_{|z|=1}\frac{d z}{z+pz^2+p+zp^2}=\\ =\frac{1}{i}\int_{|z|=1}\frac{d z}{pz^2+(p^2+1)z+p}$$

$pz^2+(p^2+1)z+p=0$ Then $z_1=0,z_2=-p$

$Res(f,0)=lim_{z\to 0}\frac{z}{z(z+p)}=\frac{1}{p}$

$Res(f,-p)=lim_{z\to -p}\frac{(z+p)}{z(z+p)}=-\frac{1}{p}$

So it is $\frac{1}{i}2\pi i (\frac{1}{p}-\frac{1}{p})=0$

But the answer is $\frac{2\pi }{1-p^2}$

Where did it went wrong?

newhere
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1 Answers1

4

Your integrand has poles at $-p$ and $-1/p$. The point $0$ is not a pole, so there is no need to incorporate the residue there. Your calculation should be as follows:

We first note that when $0 < |p| <1$ then the pole at $-1/p$ lies outside the unit circle, so it will not affect the integral.

The residue at $-p$ is $$\lim_{z \to -p}\frac{(z+p)}{p(z+p)(z+1/p)}= \frac{1}{1-p^2}$$

Thus the desired integral is $2\pi i\cdot\frac{1}{i}\frac{1}{1-p^2} = \frac{2\pi}{1-p^2}$ by residue theorem

Brevan Ellefsen
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