Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Contour integration problem

  1. May 14, 2009 #1
    http://ocw.mit.edu/NR/rdonlyres/Mathematics/18-04Fall-2003/FACBFB25-64E5-4AA8-8868-F623EDA94CE8/0/assignment3.pdf [Broken]

    This link on problem 4 is troubling me. What I want to do is evaluate it, but I first need to find the arguments on the segments above and below the cut. To do this I started off by factoring x^-2/3 out of the integrand, to get x^-2/3(1-x)^-1/3. After this, I restrict the argument of x in between pi and -pi, and for 1-x between 0 and 2pi. After doing out the algebra we find that the cut line is exactly between 0 and 3 on the real axis. Defining them in their respective local polar coordinates, we get on the top to be -2pi/3 and on the bottom zero for each argument. I know this is where I messed up because after factoring out the exp(-2pi/3) from the segment on the top, and calling the contour [1-exp(-2pi/3)]I and applying the residue theorem, i get the answer is imaginary! This has to be wrong, but why is it that my arguments are wrong, Any help would be GREATLY appreciated, so if you know anything about this stuff, then please help!
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 14, 2009 #2
  4. May 14, 2009 #3
    Why did I get it wrong, though, I can see why their example is right, but what is the correct way to do it if you do it my way, i.e. the x^-2/3 is restricted to pi-negative pi, and the (1-x)^-1/3 is the principle branch i.e zero to 2pi?
  5. May 14, 2009 #4
    I'm afraid that would take me a while to figure out. Actually it would take me a while to go through the one in AoPS as well but I may do so but probably not in time to help you. Sorry.
    Last edited: May 14, 2009
  6. May 14, 2009 #5
    I find the correct answer using your conventions for the branches. You get 2 pi i*(-residue at infinity) divided by [1-exp(-2 pi i/3)]

    The residue at infinity depends on your choice of the branches, it is exp(-pi i/3) (apart from a possible minus sign). Together with the factor i that makes a sin in the denominator.
  7. May 14, 2009 #6
    thank you so much count iblis! and you too squidsoft! It all makes sense now, I used the -1 for my res at infinity.

    Thanks again!
  8. May 14, 2009 #7
    wait a minute, i get the right answer, but it is negative, where did i go wrong now?
  9. May 14, 2009 #8
    Is the contour around the two branch points moving in the clockwise or the anti-clockwise direction?
  10. May 14, 2009 #9
    i chose anticlockwise
  11. May 14, 2009 #10
    nevermind i got the problem.
  12. May 14, 2009 #11
    Then the contour integral will be minus 2 pi i times the residue at infinity. But note that, by definition, the residue at infinity has itself a minus sign in its definition: residue at infinity of f(z) at infinity is the residue at zero of -1/z^2 f(1/z).

    The reason for the two minus signs is as follows. If you have a contour integral I and a contour integral IR over a contour with radius R for large R, then we have:

    IR - I1 = 2 pi i sum of residues inbetween the two contours ----->

    I1 = -2 pi i sum of residues inbetween contours + IR

    Then if we take the limit of R to infinity, we can write this as:

    I1 = -2 pi i sum of residues outside the contour in I1

    if we include the so-called residue at infinity which then must be defined as:

    2 pi i residue at infinity = - Lim R to infinity of IR

    Then the limit of IR for R to infinity can be obtained by considering the conformal transormation Z = 1/W. You then get the contour integral of

    f(1/W) 1/W^2 dW

    Note that the minus sign in the dZ = -1/W^2 dW compensats for the fact that the contour would be traversed in the opposite direction.

    So, the end result is that you must take 2 pi i times the residue at zero of 1/z^2 f(1/z)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook