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Contour integral of dz/(z-w)^n

  1. Feb 23, 2012 #1
    If C is a simple closed contour such that w lies interior to C, and n > 1, then
    [tex]\int_{C} \frac{dz}{(z-w)^n} = 0.[/tex] I'm confused because the function [itex]f(z) = (z-w)^{-n}[/itex] has a pole at w, so it isn't holomorphic, but the integral is still zero. The Cauchy-Goursat Theorem says that if f is holomorphic over a simply connected domain D and if C is a simple closed contour in D then the contour integral of f over C is zero. Does this mean there exist functions which are not holomorphic but behave like holomorphic functions under contour integration? (Silly question, I know)

    I also want to know how to prove that this integral is zero. Here is my attempt.

    Let C be as above. Since the (Jordan) interior of C is open, there exists a small disk of radius r > 0 around w such that its boundary, which I will parametrize by [itex]\gamma(t) = w + re^{it}[/itex] for [itex]t\in [0, 2\pi][/itex], is contained entirely interior to C. Then
    [tex]\int_{C} \frac{dz}{(z-w)^n} = \int_0^{2\pi} \frac{1}{(\gamma(t) - w)^n}\gamma'(t) dt = \int_0^{2\pi} \frac{r i e^{it}}{r^n e^{int}} dt = \int_0^{2\pi} r^{1-n} i e^{i(1-n)t} dt = \frac{r^{1-n}}{1-n}(e^{2\pi (1-n)i} - 1) = 0.[/tex](I'm using the complex analysis notes at http://math.fullerton.edu/mathews/c2003/CauchyGoursatMod.html [Broken])
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 23, 2012 #2

    HallsofIvy

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    I suggest you go back and look at that again. As long as f is holomorphic, the integral of [itex]\int f(z)/(z- w)^n dz[/itex] around a simple closed curve containing w is 0 for all n except -1.

    First, the integral around any simple closed contour containing w is the same as the integral around a circle containing w. In particular, on a circle with center at w, of radius R, we can write [itex]z= w+ e^{it}[/itex] with t going from 0 to [itex]2\pi[/itex].

    What do you get if you use that contour?
     
  4. Feb 23, 2012 #3
    HallsofIvy: I did exactly that when I parametrized my circle with [itex]\gamma(t)[/itex]. Is my work correct?
     
  5. Feb 23, 2012 #4
    You have some misunderstanding with all that. The Cauchy-Goursat theorem states that if a function is analytic everywhere interior to and on a simple closed curve, than the closed contour of the function is zero. Then 1(z-w)^n is not analytic everywhere in the integration disc so you can't rely on Cauchy-Goursat to conclude it's zero.

    If it's not analytic everywhere in the disc then the integral may or may not be zero. This is due to the existence or not of a single-valued antiderivative not over the entire region of the disc but only over the integration path.
    [tex]\oint \frac{1}{z-w} dz[/tex]
    is not zero because it does not have a single-valued antiderivative over the integration path; it's antiderivative is log(z) which is multi-valued and does not come back to it's starting point when you traverse it over a closed contour around the origin. However,
    [tex]\oint \frac{1}{(z-w)^n}dz,\quad n\ne 1[/tex]
    does have single-valued antiderivatives and even though that antiderivative is not defined at z=w, it doesn't have to be. It only needs to be defined, single-valued, and analytic over the integration path and if so, then the integral is zero. Of course it has to be zero because by the Fundamental Theorem of Calculus,

    [tex]\oint f(z)dz=F(z)\biggr|_{a}^{a}[/tex]

    and if the antiderivative F is single valued, and analytic over the contour then F(a)-F(a)=0.

    So given an integral:

    [tex]\oint f(z)dz[/tex]

    that you know has all kinds of singular points inside the contour, then if the function and it's antiderivative along the path of integration are single-valued and analytic, then the integral is zero simply because when you evaluate the antiderivative at the end points, since the contour loops back to the starting point, the integral is zero.

    So you said you wanted to prove your integral is zero. Then is the function and it's antiderivative single-valued and analytic along the entire path of integration?

    . . . ugh, I'm pretty sure that's all correct. Weigh-in guys if something's not.
     
    Last edited: Feb 23, 2012
  6. Feb 23, 2012 #5

    lavinia

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    As Jackmell rightly says, you can apply the fundamental theorem of calculus because the integrand has a single valued anti-derivative in the region around the pole.

    If you view the pole as the source of an incompressible, irrotational fluid flow then its strength is zero. Pictorially flow lines emerging from the pole loop back and return to it.
    The only exception is when n=1 in which case you get a source/sink whose strength is the residue at the pole.

    You calculation is correct up to the last equality. you need to evaluate the anti-derivative at 0 and 2pi.
     
  7. Feb 25, 2012 #6
    Thanks everyone!
     
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