# A (probably complex) integral

• Jun 15th 2011, 11:42 AM
CSM
A (probably complex) integral
This is an exercise from Stein & Shakarchi - Complex Analysis

II.7) Prove that $\displaystyle \int_{0}^{2\pi}\frac{1}{(a+\cos\theta)^2} d\theta=\frac{2\pi a}{(a^2-1)^{3/2}}$ for $\displaystyle a>1$

How do I start?
I know this function has a pole in $\displaystyle a=-cos \theta$
I know the function can be rewritten as: $\displaystyle \frac{2}{2a^2+4a\cos(\theta)+\cos(2\theta)+1}$ or as $\displaystyle \frac{1}{(a+\frac{e^{i\theta}+e^{-i\theta}}{2})^2}$

What I do not know: what to do? Do I use the residue theorem? Do I look for a nice contour?

(Also in general: how to tackle complex integrals? Step by step.)
• Jun 15th 2011, 12:37 PM
Opalg
Re: A (probably complex) integral
Quote:

Originally Posted by CSM
This is an exercise from Stein & Shakarchi - Complex Analysis

II.7) Prove that $\displaystyle \int_{0}^{2\pi}\frac{1}{(a+\cos\theta)^2} d\theta=\frac{2\pi a}{(a^2-1)^{3/2}}$ for $\displaystyle a>1$

How do I start?
I know this function has a pole in $\displaystyle a=-cos \theta$
I know the function can be rewritten as: $\displaystyle \frac{2}{2a^2+4a\cos(\theta)+\cos(2\theta)+1}$ or as $\displaystyle \frac{1}{(a+\frac{e^{i\theta}+e^{-i\theta}}{2})^2}$

That last suggestion is the best one. Write the integral as $\displaystyle \int_0^{2\pi}\frac{1}{\bigl(a+\frac12(e^{i\theta}+ e^{-i\theta})\bigr)^2}d\theta.$ Then make the substitution $\displaystyle z = e^{i\theta},$ so that the integral becomes an integral round the unit circle and you can use the residue theorem.
• Jun 15th 2011, 01:27 PM
Random Variable
Re: A (probably complex) integral
If the function is $\displaystyle f(\sin x)$ or $\displaystyle f(\cos x )$ and the limits are $\displaystyle (0, 2 \pi)$ or $\displaystyle (-\pi , \pi)$, the contour will be a circle.
• Jun 15th 2011, 01:41 PM
CSM
Re: A (probably complex) integral
$\displaystyle z=e^{i\theta}, \,\,\,dz=ie^{i\theta}d\theta=izd\theta, \,\,\,d\theta=\frac{dz}{iz}$

$\displaystyle \int_C \frac{1}{(a+\frac{1}{2}(z+\frac{1}{z}))^2} \frac{dz}{iz}$
With $\displaystyle C$ the unit circle.
$\displaystyle =\frac{1}{4}\int_C \frac{1}{(5a^2+4z+2+4\frac{a}{z}+\frac{a}{z^2})i}d z$
So it has a pole in $\displaystyle (5a^2+4z+2+4\frac{a}{z}+\frac{a}{z^2})=0$?
• Jun 16th 2011, 03:30 AM
CSM
Re: A (probably complex) integral
I rewrote it as $\displaystyle \oint_{|z|=1}\frac{-4iz}{(2az+z^2+1)^2}dz$ which only has a pole in the unit circle and for $\displaystyle a>1$ in $\displaystyle z=\sqrt(a^2-1)-a:=q$.
This is a second order pole (?)
So I have to calculate $\displaystyle \lim_{z\rightarrow q} \frac{d}{dz} (z-q)f(z)=\lim_{z\rightarrow \sqrt(a^2-1)-a} \frac{d}{dz} (z-\sqrt(a^2-1)+a)\frac{-4iz}{(2az+z^2+1)^2}$?
I'm asking because that will be quite ugly I guess...