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Thread: A (probably complex) integral

  1. #1
    CSM
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    Question 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.)
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  2. #2
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    Re: A (probably complex) integral

    Quote Originally Posted by CSM View Post
    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.
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    Super Member Random Variable's Avatar
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    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.
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  4. #4
    CSM
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    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 $\displaystyle (5a^2+4z+2+4\frac{a}{z}+\frac{a}{z^2})=0$?
    Last edited by CSM; Jun 15th 2011 at 01:57 PM.
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  5. #5
    CSM
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    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...
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