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Contour Integral

  1. Feb 11, 2010 #1

    kreil

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    1. The problem statement, all variables and given/known data
    Calculate the following integral along three different circular contours,

    [tex]\int_{C_j}\frac{dz}{z(3z-1)^2(z+2)}[/tex]

    where
    [tex]C_1:0<r_1<1/3[/tex]
    [tex]C_2:1/3<r_2<2[/tex]
    [tex]C_3: r_3>2[/tex]

    3. The attempt at a solution

    The function has singularities at z=0, z=1/3 and z=-2. Thus all three contours enclose singularities and Cauchy's integral theorem doesn't hold (none of the integrals are immediately zero).

    Along each circular contour,

    [tex]z=re^{i \theta}\implies dz=ire^{i \theta}d \theta[/tex]

    Am I going to need to use partial fractions for this? What is the best way to get started?
     
    Last edited: Feb 11, 2010
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  3. Feb 11, 2010 #2

    vela

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    Perhaps calculate the residue for each pole.
     
  4. Feb 11, 2010 #3

    kreil

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    we havent covered that..i think i need to it a different (probably harder) way
     
  5. Feb 11, 2010 #4
    Partial fractions might be the easiest way. You could also expand the function into its Maclaurin series and apply Cauchy's integral formula to the individual terms (it will apply to most terms, and you can do the exceptions by hand).
     
    Last edited: Feb 11, 2010
  6. Feb 11, 2010 #5

    kreil

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    After expanding I got the following:

    [tex]\frac{1}{z(3z-1)^2(z+2)}=\frac{1}{2z}+\frac{9}{7(3z-1)^2}-\frac{98}{z+2}[/tex]

    When I put this back into the integral and substitute for z and dz (defined in OP) I can't get the wolfram integrator to integrate the second term:

    [tex]\frac{9ir}{7}\int_0^{2 \pi}\frac{e^{i \theta}}{(3re^{i \theta}-1)^2}d \theta[/tex]

    did I do something incorrectly, or does this integral need further manipulation?
     
  7. Feb 12, 2010 #6
    Do you know that the Fundamental Theorem of Calculus holds true for complex-valued functions? How would you usually evaluate

    [tex]\int_a^b \frac{f'(x)}{(3 f(x)-1)^2}dx[/tex]
     
    Last edited: Feb 12, 2010
  8. Feb 12, 2010 #7

    kreil

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    u = f(x)
    du = f'(x) dx

    doesnt this essentially go backwards though? I already subbed in [itex]z=re^{i \theta}[/itex]
     
  9. Feb 12, 2010 #8
    Not exactly. The initial substitution reduced your integral from an integral of a complex variable to an integral in terms of a real variable. The second substitution computes an antiderivative for the function of a real variable. Your integrand assumes complex values, but it is a function of a real variable.
     
  10. Feb 12, 2010 #9

    kreil

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    This will give the last two integrals equal limits (since [itex]e^{2i \pi}=e^0=1[/itex]) so am I correct to say that the second and third terms drop off and the total integral is [itex]i \pi[/itex]? Also, this would mean that all three integrals are equal to this (despite the fact that they enclose differing numbers of singularities)...?
    Just to be clear the whole expression is:

    [tex]\frac{i}{2}\int_0^{2\pi}d \theta+\frac{9ir}{7}\int_0^{2 \pi}\frac{e^{i \theta}}{(3re^{i \theta}-1)^2}d \theta-98\int_0^{2 \pi}\frac{ire^{i \theta}}{re^{i \theta}+2}d \theta[/tex]
     
    Last edited: Feb 12, 2010
  11. Feb 12, 2010 #10
    If r < 1/3, the total integral is i*Pi indeed. You should get a different answer for r > 2 though.

    What happens to the third fraction when r > 2?
     
  12. Feb 12, 2010 #11

    vela

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    No. You have to be careful. When you integrate the last term, you'll get log(z+2), and you'll have to worry about a branch cut.
     
  13. Feb 12, 2010 #12

    kreil

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    when you say r>2 do you mean r--> inf?

    also what about 1/3 < r < 2?
     
  14. Feb 12, 2010 #13

    kreil

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    but when you use the u-sub the last two integrals are essentially

    [tex]\int_a^af(x)dx=0[/tex]....right?
     
  15. Feb 12, 2010 #14
    I mean any r>2, not necessarily the limit as r -> inf (though that wouldn't make a difference).

    Try this: what does Euler's integral theorem let you conclude about the difference between the integral of 1/(z+2) over the circle |z| = 3 vs the circle |z+2|=1?
     
  16. Feb 12, 2010 #15

    kreil

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    sorry im not sure what euler's integral theorem is (we havent covered it yet)

    if i was to let r--> inf though, the last term would tend to 1
     
  17. Feb 12, 2010 #16
    Sorry, I meant Cauchy's integral theorem. My bad!
     
  18. Feb 12, 2010 #17
    The problem with the complex log is that it is not continuous: you will jump from [itex]2 \pi[/tex] to 0 at some point. That often causes trouble.
     
  19. Feb 12, 2010 #18

    kreil

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    Haha ok. In that case I believe the integrals would be equal? (they enclose the same singularity so one could deform the contours to be equal without crossing a singularity or changing the value of the integral)
     
  20. Feb 12, 2010 #19
    Yes, they are equal. And the integral over |z+2|=1 is a bit easier to compute.
     
  21. Feb 12, 2010 #20

    kreil

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    Deforming the z>2 contour in that way in this problem would cross a singularity (z=1/3) wouldn't it?

    Also, the integral expression I have produced is general and not part a-c specific. I'm confused as to how the last two terms are not zero after the u-sub, since they have equal limits....should i be integrating wrt u and then subbing back in [itex]re^{i \theta}[/itex]? Why do the last two terms drop for r<1/3 but not for parts (b) and (c)?
     
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