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Math Help - Residue theorem

  1. #1
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    Residue theorem

    usind Residue theorem prove that :

      \int \frac {d \theta}{ 1+ {sin^2 x}} = \sqrt 2 \pi  ,,, -\pi\to \pi
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    i using  {sin^2x} = \frac{1-cos2\theta}{2} but the answer not the same
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    First off, I think that you mean \int_{-\pi}^\pi\frac{d\theta}{1+\sin^2\theta}.

    I would suggest try using u=\tan\left(\frac{\theta}{2}\right). I haven't tried it out yet but it usually helps. After this you should have a normal rational function which is usually easier to integrate.
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  4. #4
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    The redisue theorem will only work when you are integrating over some closed positively oriented contour where the function you want to integrate is analytic inside and on the contour except at say points z_{1},z_{2}\dots z_{n}
    we can then write the answer in the following way
    \int f(z)dz=2\pi i \sum^{n}_{j}Res(z_{j})
    Now since the function you want to integrate \frac{1}{1+\sin^{2}\theta} is analytic everywhere the answer to your integral should be zero but there might be trick here, try decomposing the function in the following way
    \frac{1}{1+\sin^{2}\theta}=\frac{1}{(1+i\sin\theta  )(1-i\sin\theta)}
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  5. #5
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    Quote Originally Posted by flower3 View Post
    usind Residue theorem prove that :

      \int \frac {d \theta}{ 1+ {sin^2\,{\color{red}\theta}}} = \sqrt 2 \pi  ,,, -\pi\to \pi
    To do this using the residue theorem, you need to make the substitution z = e^{i\theta}. Then \sin\theta = \tfrac1{2i}(z-z^{-1}), and dz = ie^{i\theta}d\theta so that d\theta = dz/iz. The integral becomes \oint\frac1{1-\frac14(z-z^{-1})^2}\,\frac{dz}{iz} (integral round the unit circle), which you can evaluate by computing residues.
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    Quote Originally Posted by Opalg View Post
    To do this using the residue theorem, you need to make the substitution z = e^{i\theta}. Then \sin\theta = \tfrac1{2i}(z-z^{-1}), and dz = ie^{i\theta}d\theta so that d\theta = dz/iz. The integral becomes \oint\frac1{1-\frac14(z-z^{-1})^2}\,\frac{dz}{iz} (integral round the unit circle), which you can evaluate by computing residues.
    i do it but i don't get the answer please help meeee
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  7. #7
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    Quote Originally Posted by flower3 View Post
    i do it but i don't get the answer please help meeee
    First, by a bit of elementary algebra, simplify the integral to \oint\frac{4zi}{z^4-6z^2+1}\,dz. Next, identify the poles inside the unit circle by factorising the denominator. It is a quadratic in z^2, with roots z^2 = 3\pm2\sqrt2. As luck would have it, 3\pm2\sqrt2 =(\pm\sqrt2\pm1)^2, so the denominator has four roots z=\pm\sqrt2\pm1.

    Now for a quotient of the form \frac{f(z)}{g(z)}, the residue at a simple root z_0 of g(z) is given by \frac{f(z_0)}{g'(z_0)}. In this case f(z) = 4zi and g(z) = z^4-6z^2+1. So the residue at z_0 is \frac{4z_0i}{4z_0^3-12z_0} = \frac i{z_0^2-3}. The poles inside the unit circle are given by z_0 = \pm(\sqrt2-1). For both of these, z_0^2 = 3-2\sqrt2 and so the residue is \frac i{-2\sqrt2}.

    Finally, the integral is equal to 2\pi i times the sum of the two residues, namely 2\pi i\Bigl(\frac{2i}{-2\sqrt2}\Bigr) = \sqrt2\pi.
    Last edited by Opalg; August 16th 2009 at 01:24 AM.
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