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Homework Help: Evaluating an integral

  1. Feb 27, 2007 #1
    1. The problem statement, all variables and given/known data
    2. Relevant equations

    find in the limit [tex]r\rightarrow\infty[/tex]

    the solution (or rather a hint) given by the book:
    "The integrand, considered as a complex function of p, has brunch cuts on the imaginary axis starting at [tex]\pm im[/tex].


    To evaluate the integral we push the contour up to wrap around the upper branch cut. Defining [tex]\rho = - ip[/tex], we obtain

    [tex]\frac{1}{4(\pi)^2r}\int^\infty_{m}\frac{\rho\exp(-\rho r)d\rho}{\sqrt{\rho^2-m^2}}[/tex]

    in the limit, tends to


    3. The attempt at a solution

    I can't find any theorem in complex analysis that permits a "push" of the contour shown in the figure, so I try the contour shown below:


    but when I take limit R goes to infinity, the maximum modulus integral bound around the semicircle doesn't go to zero. so I'm stuck. Expert pls help me.
    Last edited: Feb 27, 2007
  2. jcsd
  3. Feb 27, 2007 #2
    BTW: Which book?

    As for contours away from the real and imaginary axes, I would consider integrating with \rho = i*m - \delta +Re^{i\theta }
    from \theta = \theta_0 to \pi/2 - \theta_0 and then \pi/2 + \theta_0 to \pi - \theta (or opposite orientation ?)
    and have \delta and \theta_0 vanish appropriately as R-> \infty.

    The idea would be to have the imaginary part in the exponential be positive so that you get exponential decay in the semi circular part of the contour for large R. The rest maybe would converge to the desired integrals, but I did not check this all the way. However, a change in sign in the imaginary part of the exp function in the integrand occurs as arg \rho goes from \pi/2-\theta_0 \pi/2+\theta_0 which I suppose makes the contributions add (and not cancel) along the imaginary axis in the limit R-> \infinity.

    (I didn't give equations of contours exactly like the pictures - well I'm only trying to help.....)

  4. Feb 27, 2007 #3
    The book is Peskin & Schroeder, Introduction to Quantum Field Theory
    the problem is somewhere in p.g. 1~30

    Let me spend some time figure out what you are saying.
  5. Feb 27, 2007 #4
    I am thinking of contour that's almost a semi-circle, except for a thin wedge at a small angle that goes slightly below i*m and the larger arc not quite \pi radians across and not quite lying on the x-axis.

    On second thought you may get away with
    \theta = 0 to \pi/2 - \theta_0 and then \pi/2 + \theta_0 to \pi
    (Maybe part of contour can exactly on x-axis - Am I overzealous in trying to keep the argument of the exponent in having a strictly **positivie imaginary part**?).

    Just making the U shape in your picture more like a skinny V shape with vertex slightly under the singularity and making the contour not quite lying on the x-axis.
    Last edited: Feb 27, 2007
  6. Mar 1, 2007 #5
    I bet you mean this:

    along the big "semicircle", let [tex]p = R\exp(i\theta)[/tex], integrate from [tex]\theta_0[/tex] to [tex]\pi/2-\theta_0[/tex] then [tex]\pi/2+\theta_0[/tex] to [tex]\pi - \theta_0[/tex]

    Let [tex]\theta_0\rightarrow 0, R\rightarrow\infty[/tex] later.

    along the small "circle", let [tex]p = im + z[/tex] where [tex]z = \delta\exp(i\omega)[/tex], integrate from [tex]\pi/2 - \theta_1[/tex] to [tex]\pi/2 + \theta_1 - 2\pi[/tex].

    Let [tex]\delta\rightarrow 0, \theta_1\rightarrow 0[/tex] later?

    :wink: Is that what you mean? I check (loosely) that the maximum modulus bound tends to zero
  7. Mar 1, 2007 #6
    Nice picture. I was thinking of a V with a small angle instead of a "small circle" going under the singularity, but I think you got the picture - literally.
    Good luck.
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