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Another Residue Problem

  1. Oct 19, 2014 #1
    1. The problem statement, all variables and given/known data

    Show that ##\int _{ 0 }^{ \infty }{ \frac{x^2dx}{(x^2+9)(x^2+4)^2} } =\frac{\pi}{200}##.

    2. Relevant equations

    ##Res=\frac{1}{n!}\frac{d^n}{dz^n}[f(z)(z-z_0)^{n+1}]## where the order of the pole is ##n+1##.

    3. The attempt at a solution

    Integreading of a semicircle contour one finds poles at ##2i## and ##3i##. The second residue ##3i## was correct (I checked with mathematica). I used the ##Res=\frac{g(z)}{h^{\prime}(z)}## because it is first order, and I can just include the squared term on bottom into ##g(z)##. The first residue ##2i## is not as easy, and I need to use the equation above. ##Res=\frac{1}{1!}\frac{d}{dz}[\frac{z^2(z^2+4)^2}{(z^2+4)^2(z^2+9)}]##. This simplifies to ##\frac{d}{dz}[\frac{z^2}{z^2+9}]##. Which in turn simplifies to ##\frac{2z}{z^2+9}-\frac{2z^3}{(z^2+9)^2}##. Evaluating at ##2i## yields ##\frac{36i}{25}##. This is wrong. Mathematica yields: ##-\frac{13i}{200}##. Calculating this with the other term I got ##\frac{1}{2}2\pi i (\frac{3i}{50}-\frac{13i}{200})=\frac{\pi}{200}##. So the Mathematica term is correct, but my ##\frac{36i}{25}## is wrong. I am not sure what I did wrong here.

    Any suggestions?

    Thanks,.
    Chris
     
  2. jcsd
  3. Oct 19, 2014 #2

    mfb

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    2016 Award

    Staff: Mentor

    ##(z-z_0)^2## is not what you used in your numerator at ##Res=\frac{1}{1!}\frac{d}{dz}[\frac{z^2(z^2+4)^2}{(z^2+4)^2(z^2+9)}]##
     
  4. Oct 19, 2014 #3
    It should be cancelling the term that goes to zero in the denominator, no?

    Chris
     
  5. Oct 19, 2014 #4

    RUber

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    Homework Helper

    Is ##z_0=2i##? Or is it the negative?
    I don't think you can do both at the same time.
     
  6. Oct 19, 2014 #5
    Good point.

    Chris
     
  7. Oct 19, 2014 #6

    mfb

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    2016 Award

    Staff: Mentor

    Exactly.
    x^2 + 4 = (x+2i)(x-2i). Those are independent terms.
     
  8. Oct 19, 2014 #7
    Thanks, I fixed it and it worked finally.

    Chris
     
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