1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
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. Mar 21, 2009 #1
    1. The problem statement, all variables and given/known data
    I am trying to evaluate the integral:
    [tex]
    I = \int^{\infty}_{-\infty} \frac{z e^{irz}\ }{(z-k)(z+k)} dz
    [/tex]

    The way I attepted it was to use contour integration around a semicircle in the top half of the argan diagram, with two small indents above the poles. This means that the contour doesn't contain either pole, but has contributions from the indents around them.

    I calculated the residue at each pole to be:

    [tex]
    R(k) = e^{irk}/2
    [/tex]

    [tex]
    R(-k) = e^{-irk}/2
    [/tex]

    Using the method here: http://en.wikipedia.org/wiki/Residue_(complex_analysis)#Calculating_residues

    From jordans lemma, the contribution from the semicircle tends to zero as its radius tends to infinity, and the small indents around each pole contribute [tex]- i\! \pi \! residue[/tex]. The overall intergral should be:

    [tex]
    I = i \pi (e^{irk}\ + e^{-irk})/2
    [/tex]

    However the answers I have say that the intergral is

    [tex]
    I = i \pi e^{irk}
    [/tex]

    Is there an error somewhere that I'm missing? Does it depend where you choose the contour to be?


    2. Relevant equations



    3. The attempt at a solution


    [Edited because of typos]
     
    Last edited: Mar 21, 2009
  2. jcsd
  3. Mar 21, 2009 #2
    It's impossible for the residue to have a z-dependence, so try to explain in more detail how you calculate them. Then we can better judge where the error lies.
     
  4. Mar 21, 2009 #3
    Sorry, that was a typo, there should be no z there. The residues should read:
    [tex]

    R(k) = e^{irk}/2

    [/tex]

    [tex]

    R(-k) = e^{-irk}/2

    [/tex]

    They were worked out using
    [tex]
    res(f,c) = \frac{1}{(n-1)!} \lim_{z \to c} \frac{d^{n-1}}{dz^{n-1}}\left( f(z)(z-c)^{n} \right )
    [/tex]

    Both the poles I'm looking at are first order, n = 1. The overall integral is then:

    [tex]

    R(-k) = i \pi (e^{irk} + e^{-irk})/2

    [/tex]

    Apologies for the mistake.
     
  5. Mar 21, 2009 #4
    i get the answer given.

    you have two simple poles, one at +k and one at -k. Your contour is a smemicircle in the upper half plane which INCLUDES the pole at +k but doesn't include the one at -k.

    The Cauchy Residue Theorem tells us that

    [itex]\int_{\Gamma} \frac{ze^{irz}}{(z-k)(z+k)}dz = 2 \pi i \sum_{j=1}^k res(f)(a_j)[/itex]

    where [itex]\Gamma[/itex] is the contour and we only sum over the singularities contained INSIDE [itex]\Gamma[/itex] - by this means the pole at -k doesn't affect the value of your contour integral.

    so we get [itex]\int_{\Gamma} \frac{ze^{irz}}{(z-k)(z+k)}dz = 2 \pi i \frac{e^{irk}}{2}=i \pi e^{irk}[/itex] using the value of the residue at z=+k.

    then just argue that the part of your integral that's not on the real axis goes to 0 as R goes to infinity as you've done above and then its just the contribution on the real line that is left so you conclude

    [itex]\int_{-\infty}^{\infty} \frac{xe^{irx}}{(x-k)(x+k)}dx = i \pi e^{irk}[/itex]
     
  6. Mar 21, 2009 #5
    So why would you choose the contour to go around the pole at +k rather than both poles or none of them? does this mean the integral has several possible values?

    Also, while the pole at -k isn't include in the contour, doesn't the half circle that goes around it, whose radius tends to zero have some contribution? As in "Mathematical methods for physicists", Riley Hobson Bence chapter 24?
     
  7. Mar 23, 2009 #6
    Any ideas? Is the integral Multivalued?
     
  8. Mar 23, 2009 #7
    Have you tried calculating the contribution of the small arc and see if it tends to zero? Let me know if you need some help on it.
     
  9. Mar 23, 2009 #8
    I think your result is correct.
    It seems there is a deviation from k, such that [tex] k \rightarrow k+i\eta [/tex] where [tex]\eta[/tex] is a positve infinitesimal.
    (i.e. the poles are a little bit off from the real axis)
     
    Last edited: Mar 23, 2009
  10. Mar 23, 2009 #9
    what page of chapter 24???

    also if you see the integral you want the value of goes from [itex]-\infty[/itex] to [itex]+\infty[/itex] so part of your contour is going to need to lie on the real axis. If you draw the two poles and put part of your contour on the real axis, there is no way to now make a closed contour that contains both poles in its interior, is there?

    secondly - the only reason we take a semi circle is because its relatively easy to deal with - you just use Jordan's lemma to see how the semi circular arc behaves. you could take a ridiculous contour that goes all over the complex plane and provided it includes that pole you'll get teh same answer but it's going to be basically impossible to evaluate.

    hope this helps.

    oh and the integrand isn't multivalued as far as i can tell...
     
  11. Mar 23, 2009 #10
    Thanks for the replies. The section I was referring to is page 864 RHB third edition.

    I'm treating the small indents around the poles as semicircles A, so their contribution to the integral is:


    [itex]
    \int_{A} f(z) dz
    [/itex]

    Where f(z) is the function being integrated. Expanding it as a laurent series:

    [tex]
    \sum^{\infty}_{n = -m} a_{n} \int_{A} (z - z_{0})^{n} dz
    [/tex]

    Provided the pole is simple (as in this case) then the function can be written as the sum of a analytic function and [tex] a_{-1} (z - z_{0})^{-1} [/tex]. When you carry out the integration, the analytic part vanishes and you're left with, written in polar form:

    [itex]
    \int_{A} f(z) dz = \int _{0} ^{\pi} a_{-1} \frac{i r e^{i\theta}}{re^{ir\theta}} d\theta
    [/itex]

    Which just gives:

    [itex]
    \int_{A} f(z) dz = i \pi a_{-1}
    [/itex]

    Where r is the radius of the semicircle and the extra [tex] ire^{ir\theta} [/tex] comes from the substitution of z = [tex] re^{ir\theta} [/tex].

    Then as r tends to zero the integral along the real axis becomes continuous.
     
  12. Mar 23, 2009 #11
    i didn't read your initial post very thoroughly and started treating it with poles at +-ik.

    i actually can't see any problems with what you've done.

    i hadn't seen that result on p864. i would just have used jordan's lemma on both of the small semicircular deviations as well - guess you could give that a try...
     
  13. Mar 23, 2009 #12
    Doesn't jordan's lemma only apply as the radius tends to infinity?
     
  14. Mar 24, 2009 #13
    As long as the poles(k,-k) lie exactly on the real axis and as long as we are dealing with the Cauchy principal value, ijmbarr's original result is correct.

    The poles must be a little bit off from the real axis(k -> k+i*eta).
    To be specific, the pole at k should be slightly higher than the real axis, and the one at -k should be slightly lower than it.
    This often happens when we deal with the Green's function in scattering theory.

    I think either you didn't read the book carefully enough or the book is wrong.
     
    Last edited: Mar 24, 2009
  15. Mar 24, 2009 #14
    Why do you choose the k+ to be higher and the k- pole to be lower? It seems like you could just as easily choose it the other way around and change the result?
     
  16. Mar 24, 2009 #15
    I totally agree with you. One should specify in advance whether he wants to use k+i*eta or k-i*eta.
     
  17. Mar 24, 2009 #16
    I mean how does the physics of the situation determine which you choose?
     
  18. Mar 24, 2009 #17
    for those small indentations you should be able to apply Jordan's lemma with [itex]r \rightarrow 0[/itex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Contour integration problem
Loading...