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The residue theorem

  1. Jul 31, 2013 #1
    I have a doubt on this following procedure using the residue theorem:
    Initially we have
    ψ(k,t)=[itex]\frac{1}{2\pi}\int_{L_{\omega}}\frac{S(k,\omega)}{D(k,\omega)}e^{-i\omega t}d\omega[/itex]
    Then the author said using the residue theorem, we have
    ψ(k,t)=[itex]-iƩ_{j}\frac{S(k,\omega_j(k))}{\partial D/ \partial \omega (k,\omega_j(k))}e^{-i\omega_j(k) t}[/itex]
    where [itex]S(k,\omega_j(k)), D(k,\omega_j(k))[/itex] are functions relating [itex]k[/itex] and [itex]\omega[/itex].
    What kind of residue theorem is the author using?
    Thank you very much
     
    Last edited: Jul 31, 2013
  2. jcsd
  3. Jul 31, 2013 #2

    HallsofIvy

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    The "residue theorem" the author is using, the only one I know, is
    "The integral of f(z) around a closed curve is [itex]2\pi i[/itex] times the sum of the residues at the poles of f inside the curve."
     
  4. Jul 31, 2013 #3
    Hi,

    Thanks for your reply.

    I know residue theorem. I just can't relate the usage the author has displayed to the canonical form, like I found in Wiki or what you have said.

    To be more specific, could you tell me how did you get the partial derivative of D w.r.t. ω in the denominator?
     
  5. Jul 31, 2013 #4

    pasmith

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    A pole occurs where [itex]D(k,\omega) = 0[/itex], at which point [itex]\omega = \omega_j(k)[/itex]. Locally the Taylor series for D is (for fixed k):
    [tex]
    D(k,\omega) = (\omega - \omega_j(k))\frac{\partial D}{\partial \omega}(k,\omega_j(k)) + \dots[/tex]
    Thus close to [itex]\omega = \omega_j(k)[/itex] (assuming [itex]S(k,\omega_j(k)) \neq 0[/itex]) the integrand is approximately
    [tex]
    \frac{S(k,\omega_j(k))}{(\omega - \omega_j(k))\frac{\partial D}{\partial \omega}(k,\omega_j(k))}e^{-i\omega_j(k)t}
    [/tex]
    so the residue is
    [tex]
    \frac{S(k,\omega_j(k))}{\frac{\partial D}{\partial \omega}(k,\omega_j(k))}e^{-i\omega_j(k)t}.
    [/tex]

    I don't know where the minus sign in the original comes from; presumably the contour is being traversed clockwise instead of anticlockwise.
     
  6. Jul 31, 2013 #5
    Suppose the function you are integrating has a simple pole at ##z_0##. If the function being integrated can be written in the form ##\frac{p(z)}{q(z)}##, with ##q(z_0)=0## and ##q'(z_0)\neq 0##, then its residue at the point ##z_0## is given by ##\frac{p(z_0)}{q'(z_0)}##.
     
  7. Aug 1, 2013 #6
    Thanks a lot, what you said makes sense.
     
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