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

Complex contour integrals

  1. Feb 25, 2016 #1
    I am trying to teach myself complex analysis . There seems to be multiple ways of achieving the same thing and I am unsure on which approach to take, I am also struggling to visualise the problem...Would someone show me step by step how to solve for example

    $$\int_{\Gamma}\frac{2\lambda}{[(\lambda-\lambda_{+})(\lambda-\lambda_{-})]^{2}}\frac{d\lambda}{i}$$

    where ##\lambda_{+}, \ \ \lambda_{-}## are the roots given by
    ##\lambda_{\pm}=\frac{-i\zeta\pm ir}{q}##

    So i believe the answer should be ##\frac{x}{r^{3}}## or something like that but I can't get there. Could someone show me how to approach a problem like this... Like visualise the contour etc
     
    Last edited by a moderator: Feb 25, 2016
  2. jcsd
  3. Feb 25, 2016 #2

    RUber

    User Avatar
    Homework Helper

    Do you know how the contour ##\Gamma## is defined? Should we assume that the roots, ##\gamma_{-}## and ##\gamma_{+}## are enclosed by the contour?
    If they are then use the residue theorem to solve. It is very difficult to help you visualize a contour that is not defined.
    Cauchy's Residue Theorem tells you that
    ##\int_\Gamma f(\gamma) \, d\gamma = 2\pi i \sum_{a \in \text{poles}}\mathrm{Res}(f,a)##
    You already have the roots.
    You should see that the poles are second order in this function, so use the higher order residue formula:
    (From Wikipedia's page on Residue_(complex_analysis))
    More generally, if c is a pole of order n, then the residue of f around z = c can be found by the formula:

    fc16ae96cdbc7aea1023462e07d19753.png
    For a 2nd order pole, this is simply

    ##\mathrm{Res}(f,a)=\lim_{\gamma \to a} \frac{d}{d\gamma} (\gamma - a)^2 f(\gamma) ##

    Next, find the residues for ##\gamma_+## and ##\gamma_-##:
    ##\mathrm{Res}(f,\gamma_+)=\lim_{\gamma \to \gamma_+} \frac{d}{d\gamma}\left[ (\gamma-\gamma_+)^2\frac{2\gamma}{ i (\gamma-\gamma_+)^2 (\gamma-\gamma_-)^2}\right]##
    Simplifying and taking the derivative, you get:
    ##\mathrm{Res}(f,\gamma_+)=\lim_{\gamma \to \gamma_+}\frac{2(\gamma_- +\gamma) }{i (\gamma_- - \gamma)^3 }=\frac{2(\gamma_- +\gamma_+) }{i (\gamma_- - \gamma_+)^3 }##
    Due to the symmetry, you will get something very similar for ##\mathrm{Res}(f,\gamma_-)##.
    Then, you plug in the expression you have for ##\gamma_+ , \gamma_-## to simplify even more.
    Finally, you use the residue theorem to give you that the integral is equal to ##2\pi i ( \mathrm{Res}(f,\gamma_+)+\mathrm{Res}(f,\gamma_-))##.
    And you are done.
     
  4. Feb 25, 2016 #3
    RUber's post should be very helpful. The most important point is that you can't even begin to think about a complex contour integral without knowing what contour you are integrating around.

    If you are just beginning to learn about contour integrals, it's probably a good idea to do some "by hand" — i.e., just going back to the definitions — before graduating to the extremely useful method of residues.

    For instance, suppose you want to integrate f(z) = 1/z about the contour given by the unit circle C about the origin in ℂ. So: parametrize the contour C, by say

    z(t) = eit, 0 ≤ t ≤ 2π,​

    which turns the original integral

    C dz/z​

    into

    ∫ (1/eit) i eit dt = ∫ i dt​

    (where the integral is from t = 0 to t = 2π) , since dz = i eit dt.

    Thus we have shown

    C dz/z = 2πi,​

    "by hand".
     
  5. Feb 25, 2016 #4

    FactChecker

    User Avatar
    Science Advisor
    Gold Member

    To fill in a couple of important points regarding the comments above. Assuming that the contour is a closed curve. When applying the residue theorem, you do not need to know a lot about the contour. You do need to know how many times and in what direction it winds around each root, clockwise or counterclockwise.
     
  6. Feb 26, 2016 #5
    Ok thanks for the replies everyone....

    So the way I obtained the integral in the first place was to go from

    $$\vartheta \rightarrow \lambda,$$ by using $$\lambda = e^{i\vartheta}$$

    So i guess the contour is an integration over the unit circle like Zinq said.

    Ok thanks for your comments
     
  7. Feb 26, 2016 #6
    So further from this, my original problem was to calculate the so called Whittaker contour integral

    so we start from

    $$f=\int_{0}^{2\pi} \frac{1}{(x+izcos(\vartheta)+iysin(\vartheta))^{2}}d\vartheta$$

    This should give $$f=2\pi/r^(3)$$

    could someone show me how? Essentially i changed from $$\vartheta \rightarrow \lambda$$ where $$\lambda=e^{i\vartheta}$$ and then i tried the contour integral but I still don't see how one can get the function f above?
     
    Last edited: Feb 26, 2016
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook