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Homework Help: Proof of Cauchy integral formula

  1. Oct 19, 2012 #1
    1. The problem statement, all variables and given/known data
    For an quiz for a diff eq class, I need to prove the Cauchy integral formula.
    The assignment says prove the formula for analytic functions. Is the proof significantly different when the function is not analytic?

    2. Relevant equations
    Basically, a proof I found online says that I take the integral of a larger contour C that encloses the contour C_r which is the contour that wraps around the cirlce of radius r around z_0. By the deformation principle, if r is sufficiently small, then the integral of f(z) around C and C_r are equal. I've tried searching for 'deformation principle' but none was of much help.

    3. Questions

    What is the deformation principle? I don't know much about contour integrals, so if someone can relate it to just regular integrals, that'd be great.

    http://www2.latech.edu/~schroder/slides/comp_var/14_Cauchy_Integral_Formula.pdf [Broken] Pg 43

    What is max C(r, z_0) supposed to mean? I think the this proof uses the same principles as in the first one, but doesn't explicitly use the delta-epsilon limit.

    Thank you
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 19, 2012 #2


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    Homework Helper

    Cauchy's formula for smooth functions is

    [tex]f(a) = \frac{1}{2\pi i}\int_{\partial D} \frac{f(z) dz}{z-a} + \frac{1}{2\pi i}\iint_D \frac{\partial f(z)}{\partial \bar{z}} \frac{dz\wedge d\bar{z}}{z-a}[/tex]

    The proof is not that much different than for analytic functions other than for analytic functions the second integral is zero.

    The deformation principle is that path between the limits of integration does not matter for analytic functions. For real integration this is also true (at least if the function is non pathological and the path does over emphasize a region).
    For example if integrating from 1 to 4 is written I(1,4) we have
    I(1,4)=I(1,2)+I(2,4)=I(1,3)+I(3,2)+I(2,4) and so on
    For complex integrals there is more freedom because for each segment the path can be curved and twisted in endless ways.

    max C(r, z_0) is the maximum modulus
    the largest absolute value the function can obtain on the circle
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