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Thread: Residues and Integrals over Unit Circles

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    Last edited by KatyCar; Oct 31st 2010 at 05:38 PM.
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    Super Member girdav's Avatar
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    Hello,
    we use the parameterization g(t) =e^{it} for t\in\left[0,2\pi\right]. We can denote by R(\cos t,\sin t) the function we have to compute the integral. We have \int_0^{2\pi}R(\cos t,\sin t) dt=\int_{C(0,1)}\frac{1}{iz}R\left(\frac 12 \left(z+\frac{1}{z}\right),\frac{1}{2i}\left(z-\frac{1}{z}\right)\right) dz because g'(t) = ie^{it}.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by KatyCar View Post


    b) \int_{0}^{2\pi }  \frac{1}{1 + 8cos^2\theta}d\theta
    Notice first that by a usual substitution ( y=x-\pi, and don't forget the rule about integrals of even functions) our integral, call it I, is equal to \displaystyle 2\int_0^{\pi}\frac{d\theta}{1+8\cos^2(\theta)}. So, \displaystyle \frac{1}{2}I=\int_0^{\pi}\frac{d\theta}{1+8)\frac{  e^{2i\theta}+2+e^{-2i\theat}}{2})}=2\int_0^{\pi}\frac{d\theta}{10+8e^  {4i\theta}+16e^{2i\theta}}e^{2i\theta}=2\oint_{|z|  =1}\frac{dz}{10+8z^2+16z}. From where the rest is simple residue calculation.
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    Quote Originally Posted by KatyCar View Post
    Hello Everyone,

    I'm stumped on how to solve these problems by converting to a complex integral over the unit circle and then applying Residue theroem.

    a) \int_{0}^{2\pi } (cos^4\theta + sin^4\theta) d\theta

    b) \int_{0}^{2\pi }  \frac{1}{1 + 8cos^2\theta}d\theta

    Any help would be greatly appreciated!
    The systematic way to do these problems is to start by making the substitution z = e^{i\theta}. Then dz = ie^{i\theta}d\theta = izd\theta, so that d\theta = \dfrac{dz}{iz}. Also, as \theta goes from 0 to 2\pi, z goes round the unit circle, so that the integral becomes a contour integral round the unit circle. Finally, \cos\theta = \frac12(z+z^{-1}) and \sin\theta = \frac1{2i}(z-z^{-1}).

    For the first problem, \cos^4\theta+\sin^4\theta = \frac1{16}\bigl((z+z^{-1})^4 + (z-z^{-1})^4\bigr) = \frac18(z^4+6+z^{-4}), and so

    \displaystyle\int_0^{2\pi}(\cos^4\theta+\sin^4\the  ta)\,d\theta = \oint\tfrac18(z^4+6+z^{-4})\frac{dz}{iz} = \frac1{8i}\oint(z^3+6z^{-1}+z^{-5})\,dz}.

    The only pole of the integrand comes at the origin, and the residue is obviously 6 (being the coefficient of z^{-1}), so you should find it easy to use the residue theorem to get the value of the integral.

    Do the second problem in the same way. You should find that

    \displaystyle\int_0^{2\pi}\frac1{1+8\cos^2\theta}\  ,d\theta = \oint\frac1{1+2(z+z^{-1})^2}\frac{dz}{iz} = \ldots = \frac1i\oint\frac z{(2z^2+1)(z^2+2)}\,dz (after a bit of simplification). There are two poles inside the unit circle, at z = \pm\frac1{\sqrt2}i. Find the residue at each of them, and use the residue theorem to get the value of the integral.
    Last edited by Opalg; Oct 30th 2010 at 10:26 AM. Reason: corrected error
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