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Math Help - integral question

  1. #1
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    integral question

    Evaluate :

    \int_0^\infty \frac{x^2}{x^4+5x^2+4} dx

    \int_0^\infty \frac{x^2}{(x^2+1)(x^2+4)} dx

    now I would like to extended this out into the complex plane and using the Cauchy integral to find the value, but not sure on how to do so exactly and what bounds to choose.

    I figure I'll have something like:

    \int_C \frac{z^2}{(z^2+1)(z^2+4)} dx
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  2. #2
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    Do you have to use CA?.

    \int\frac{x^{2}}{x^{4}+5x^{2}+4}dx=\frac{4}{3}\int  _{0}^{\infty}\frac{1}{x^{2}+4}dx-\frac{1}{3}\int_{0}^{\infty}\frac{1}{x^{2}+1}dx

    You should get \frac{\pi}{6}

    They involve arctan
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  3. #3
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    If it must be done with CA, here is how I would look at it.

    You have:

    \int_{0}^{\infty}\frac{x^{2}}{(x^{2}+1)(x^{2}+4)}d  x

    Which is correct.

    Let's do \frac{1}{2}\int_{-\infty}^{\infty}\frac{x^{2}}{(x^{2}+1)(x^{2}+4)}dx

    Inside {\sigma}(0,R) there are simple poles at i and 2i, with

    residues \frac{-1}{6i}, \;\ \frac{-4}{(-3)(4i)}=\frac{1}{3i},

    respectively.

    \int_{-\infty}^{\infty}\frac{x^{2}}{(x^{2}+1)(x^{2}+4)}dx  =

    2{\pi}i\left[\frac{-1}{6i}+\frac{1}{3i}\right]=\frac{\pi}{3}

    Since we want 0 to infinity, multiply by 1/2 and get \frac{\pi}{6}
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    Quote Originally Posted by galactus View Post

    Inside {\sigma}(0,R) there are simple poles at i and 2i, with

    residues \frac{-1}{6i}, \;\ \frac{-4}{(-3)(4i)}=\frac{1}{3i},

    respectively.
    I'm just a little confused on how you got those residues...
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  5. #5
    Moo
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    Quote Originally Posted by lllll View Post
    I'm just a little confused on how you got those residues...
    The order of these poles is 1.
    For a pole z_0 of order 1 of a function f, find this limit :

    \lim_{z \to z_0} (z-z_0)f(z) and it'll give the residue.

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    More generally... :
    A singularity z_0 is a pole of a function f if there exists a positive integer n, such that :
    \lim_{z \to z_0} ~ (z-z_0)^n f(z)=A \neq 0, where A is a finite value. (n will be the order)

    Then there is a direct formula giving the residue :

    \text{Res}_{z=z_0} ~ f(z)=\lim_{z \to z_0} ~ \frac{1}{(n-1)!} \cdot \frac{d^{n-1}}{dz^{n-1}} \left\{(z-z_0)^n f(z)\right\}
    Last edited by Moo; December 6th 2008 at 04:53 AM.
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