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A Dirac Delta and Residue Calculus

  1. May 7, 2017 #1
    I'm an undergraduate student, so I understand that it may be difficult to provide an answer that I can understand, but I have experience using both the Dirac delta function and residue calculus in a classroom setting, so I'm at least familiar with how they're applied.

    Whether you're integrating along a closed loop around a singularity in the complex plane or you're integrating on a closed sphere in 3D space about a Dirac delta (like you might do in E&M), the value of the integral depends entirely on the point where the singularity/delta is located. Do these similarities betray a connection between the Dirac delta and residue calculus?
     
  2. jcsd
  3. May 8, 2017 #2

    stevendaryl

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    Well, I would say that there is a closer connection between residues and the Heaviside function [itex]H(x)[/itex] defined as follows:

    [itex]H(x) = 0[/itex] if [itex]x < 0[/itex]
    [itex]H(x) = 1[/itex] if [itex]x > 0[/itex]

    Then an integral representation of [itex]H(x)[/itex] is:

    [itex]H(x) = lim_{\epsilon \rightarrow 0} \frac{1}{2\pi i} \int_{-\infty}^{+\infty} \frac{1}{\tau -i \epsilon} e^{ix\tau} d\tau[/itex]

    which you can prove using residues.

    [itex]H(x)[/itex] is related to the delta-function by formally taking the derivative of this integral representation with respect to [itex]x[/itex]:

    [itex]\delta(x) \equiv \frac{dH}{dx} = lim_{\epsilon \rightarrow 0} \frac{1}{2\pi i} \int_{-\infty}^{+\infty} \frac{i \tau}{\tau -i \epsilon} e^{ix\tau} d\tau[/itex]
    [itex]= \frac{1}{2\pi} \int_{-\infty}^{+\infty} e^{ix\tau} d\tau[/itex]
     
  4. May 8, 2017 #3
    It is hard to encounter adequate understanding of the distributions concept in physical textbooks.
    Try
    KöSaku Yosida. Functional Analysis. Sixth Edition. Springer-Verlag. Berlin Heidelberg New York 1980
     
  5. May 8, 2017 #4
    it just remains to explain what this divergent integral means and in which sense the limit ##\epsilon\to 0## is understood
     
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