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Calculation of a certain type of contour integral

  1. Jul 15, 2010 #1


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    During my research I came across a contour integral where the pole was on the boundary. I have never come across this before, do anyone of you know how I would go about computing this?

    It involved the Hilbert transform and I can't find it in my undergraduate complex analysis books and I thought someone here might know.
    On a similar note, if anyone has any good numerical routines for Hilbert transform, I would like to know.

  2. jcsd
  3. Jul 15, 2010 #2


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    The way such a thing is dealt with is to have the contour run around the pole in a small semi-circular arc, in the limit that the radius of the arc tends to zero.

    As long as it's a simple pole, the result is if a contour runs straight through a pole (which would correspond to a half-circle arc that excludes the pole from being inside the contour) the pole contributes half the residue to the integral. If the pole is at a corner of a contour (which corresponds to a quarter-circle arc around the contour) it would contribute 1/4 the residue.
  4. Jul 16, 2010 #3


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    Do you have a reference? The same idea is used in deriving the solution of the Benjamin-Ono equation.

    Surely if the pole was outside of the contour the integral would be zero by Cauchy's theorem. Am I missing something?
  5. Jul 16, 2010 #4
  6. Jul 16, 2010 #5
  7. Jul 16, 2010 #6


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    No, it's not zero because in the limit that the arc around the pole goes to zero the contour goes straight through the pole, so there is a contribution.


    [tex]\oint_c dz \frac{f(z)}{z-z_0}[/tex]
    where f(z) is analytic everywhere inside and on the contour. If the pole of the integrand at z_0 lies on our contour C, we go around it in a semi-circular arc of radius [itex]\epsilon[/itex], as [itex]\epsilon \rightarrow 0[/itex]. This part of the curve we'll parametrize by [itex]z-z_0 = \epsilon \exp(i\theta)[/itex]. This gives

    [tex]\int_\pi^0 d\theta i\epsilon e^{i\theta} \frac{f(z_0 + \epsilon e^{i\theta})}{\epsilon e^{i\theta}}[/tex]

    If we now take the limit [itex]\epsilon \rightarrow 0[/itex], we get
    [tex]-i\int_0^\pi d\theta f(z_0) = -i\pi f(z_0)[/tex]
    which is half of the residue of f(z) at z_0 (with a minus sign because we traversed over the pole in a clockwise rotation).

    Note that this calculation wouldn't work if the pole weren't simple, i.e., if the pole were [itex](z-z_0)^m[/itex], with [itex]m > 1[/itex].

    (However, jackmell's link suggests that if the pole is an odd power and our contour through it is a straight line segment, then treating the straight line segment as a principal value integral gives the half-residue again.)

    That is interesting. I had not heard of this result before. Also interesting is that the contour through the pole in this case must be a straight line segment.
    Last edited: Jul 16, 2010
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