1
$\begingroup$

I have this integral

$\int_{0}^{2\pi}\frac{1}{1+\cos^2x}\,dx$

I have two double order poles at $z^2=-3\pm2\sqrt2$ I'm having trouble taking the limit and finding the residue of this double order pole $\lim_{z\to-3+2\sqrt{2}}\frac{d}{dz}[(z-(-3+2\sqrt2)^2\frac{z}{(z-(-3+2\sqrt2)(z-(-3-2\sqrt2)}]$ any help would be greatly appreciated it

  • 0
    Ok, from here we should have begun.2012-11-25

1 Answers 1

1

You make the change of variable $z=e^{ix}$. Then $ dx=\frac{1}{i}\frac{dz}{z}, $ $ \frac{1}{1+\cos^2x}=\frac{1}{1+(z+z^{-1})^2}=\frac{4\,z^2}{z^4+6\,z^2+1}, $ and $ \int_0^{2\pi}\frac{dx}{1+\cos^2x}=\frac{1}{i}\int_{|z|=1}\frac{4\,z}{z^4+6\,z^2+1}\,dz. $ To apply the residue theorem you need the poles inside the circle $\{|z|=1\}$, that is, the solutions of $ z^4+6\,z^2+1=0,\quad |z|<1. $ Solving for $z^2$ gives $ z^2=-3\pm2\,\sqrt2. $

There are no double poles. You are interested only on the poles in the unit disk. Since $ |-3-2\,\sqrt2|>1\text{ and }|-3+2\,\sqrt2|<1, $ you have to consider only $ z^2=2\,\sqrt2-3\ . $ This gives you two simple poles at $ z=\pm\sqrt{2\,\sqrt2-3\,}\ . $

  • 1
    I have added details to the answer.2012-11-25