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How to find analytic continuation?

  1. Nov 24, 2011 #1
    The following is the problem from Fetter and Walecka (problem 3.7)

    If f(z) is defined to be the integration of rho(x) * (z-x)^(-1) from -infinity to +infinity. rho is in the following form
    rho(x)=gamma * ( gamma^2+x^2 )^(-1).
    Evaluate f(z) explicitly for Im(z)>0 and find its analytic continuation to Im(z)<0

    First, I assumed gamma is real and positive and used the residue theorem, then I got
    f(z)=pi * (z+i*gamma)^(-1), My question is how to find its analytic continuation to Im(z)<0.

    Another question is that I found f(z) takes a different form if I assumed gamma is pure imaginary. It doesn't make sense to me.

    Could you help me with these two questions? Thanks a lot.
     
  2. jcsd
  3. Nov 24, 2011 #2
    Lemme' see if I have this right: You have the function f(z) defined in the upper half-plane for constant gamma as:

    [tex]f(z)=\int_{-\infty}^{\infty}\frac{\gamma}{(\gamma^2+x^2)(z-x)}dx,\quad Im(z)>0[/tex]

    and when you compute that for [itex]Re(\gamma)\ne 0[/itex] you get:

    [tex]f(z)=\frac{\pi}{z+i\gamma}[/tex]

    and you want the analytic continuation of [itex]f(z)[/itex] into the lower half-plane. But the quantity [itex]\displaystyle\frac{\pi}{z+i\gamma}[/itex] is analytic everywhere in the complex plane except the point [itex]z=-i\gamma[/itex] and is equal to your function in the upper half-plane when [itex]\gamma[/itex] is not on the imaginary axis. Then isn't this expression the analytic extension of the function to the lower half-plane? Granted, it's not equal to the integral down there but that's not what you're asking for. You want the analytic continuation of the function f(z) which has an integral representation given when Im(z)>0 and at least for now, when gamma is not on the imaginary axis.

    That's what I'm goin' with. What do you think?
     
  4. Nov 24, 2011 #3

    vela

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    If the imaginary part of z is greater than 0, don't you have another pole in the upper half plane when evaluating the integral using the residue theorem?
     
  5. Nov 24, 2011 #4
    Ok, I see that now. Need to add another. Sorry about that but I think the general principle is still valid right?
     
  6. Nov 24, 2011 #5
    Thanks a lot. My question now becomes "can we call [itex]\displaystyle\frac{\pi}{z+i\gamma}[/itex] it an analytic continuation?", since there is a pole when Im(z)<0.
     
  7. Nov 24, 2011 #6
    Ok, look. I made a mistake rushing with this. Need to compute the residues correctly. I think they're only two. When I do that I get:

    [tex]f(z)=\frac{\pi}{z+i\gamma}[/tex]

    and that's analytic over the complex plane except when [itex]z=-i\gamma[/itex] but equals your function in the upper half plane so in my opinion that is the analytic continuation of the function f(z) which is representied by the integral when z is in the upper half-plane.
     
  8. Nov 24, 2011 #7
    When Im(z)<0, then the integral representation of the function f(z) no longer represents the function. With regards to analytic continuation of f(z), the integral is equal to the function only when Im(z)>0. I think we need to distinguish between the function f(z) and the integral that's attempting to represent the function when Im(z)>0.

    I think that's right anyway.

    If I may add to this cus' I'll like to get a grip on it too: the purpose of analytic continuation is to continue the function to a larger domain , not analytically continue the expression that represents the function in a restricted domain. And keep in mind analytic continuation is unique. So that if the expression above is identically equal to f(z) in the upper half-plane but is also analytic in a larger domain (in this case all of C except at that pole), then that is the only expression that can be the analytic continuation of f(z).
     
    Last edited: Nov 24, 2011
  9. Nov 24, 2011 #8

    vela

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    I think you also need to consider the case where [itex]z=i\gamma[/itex]. In that case, the integrand will have a second-order pole [itex]i\gamma[/itex].
     
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