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Math Help - 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 \int_{0}^{2\pi}\frac{1}{(a+\cos\theta)^2} d\theta=\frac{2\pi a}{(a^2-1)^{3/2}} for a>1

    How do I start?
    I know this function has a pole in a=-cos \theta
    I know the function can be rewritten as: \frac{2}{2a^2+4a\cos(\theta)+\cos(2\theta)+1} or as \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 \int_{0}^{2\pi}\frac{1}{(a+\cos\theta)^2} d\theta=\frac{2\pi a}{(a^2-1)^{3/2}} for a>1

    How do I start?
    I know this function has a pole in a=-cos \theta
    I know the function can be rewritten as: \frac{2}{2a^2+4a\cos(\theta)+\cos(2\theta)+1} or as \frac{1}{(a+\frac{e^{i\theta}+e^{-i\theta}}{2})^2}
    That last suggestion is the best one. Write the integral as \int_0^{2\pi}\frac{1}{\bigl(a+\frac12(e^{i\theta}+  e^{-i\theta})\bigr)^2}d\theta. Then make the substitution 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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  3. #3
    Super Member Random Variable's Avatar
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    Re: A (probably complex) integral

    If the function is  f(\sin x) or f(\cos x ) and the limits are  (0, 2 \pi) or  (-\pi , \pi) , the contour will be a circle.
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  4. #4
    CSM
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    Re: A (probably complex) integral

    z=e^{i\theta},   \,\,\,dz=ie^{i\theta}d\theta=izd\theta,   \,\,\,d\theta=\frac{dz}{iz}

    \int_C \frac{1}{(a+\frac{1}{2}(z+\frac{1}{z}))^2} \frac{dz}{iz}
    With C the unit circle.
    =\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 (5a^2+4z+2+4\frac{a}{z}+\frac{a}{z^2})=0?
    Last edited by CSM; June 15th 2011 at 02:57 PM.
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  5. #5
    CSM
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    Re: A (probably complex) integral

    I rewrote it as \oint_{|z|=1}\frac{-4iz}{(2az+z^2+1)^2}dz which only has a pole in the unit circle and for a>1 in z=\sqrt(a^2-1)-a:=q.
    This is a second order pole (?)
    So I have to calculate \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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