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Representation of Dirac-delta function

  1. Aug 19, 2010 #1
    1. The problem statement, all variables and given/known data

    Show that [tex] \delta_\epsilon(x) = \frac{\epsilon}{\pi (x^2 + \epsilon^2)} [/tex] is a representation of the Dirac delta function.

    2. Relevant equations

    I already know how the function can satisfy the first two requirements of being a dirac delta function, namely

    [tex] \lim_{\substack{\epsilon \to 0}} \delta_\epsilon (x) = 0 [/tex],
    [tex] \int_{-\infty}^{\infty} \delta_\epsilon (x) dx = 1 [/tex].

    The third requirement is

    [tex] \lim_{\substack{\epsilon \to 0}} \int_{-\infty}^{\infty} f(x) \delta_\epsilon (x) dx = f(0) [/tex].

    3. The attempt at a solution

    Following the previous examples of the book, it seems that the solution should be like this..

    [tex] \frac{\epsilon}{\pi} \int_{-\infty}^{\infty} \frac{f(x)}{x^2 + \epsilon^2} dx
    = \frac{\epsilon}{\pi} \, f(c\epsilon) \int_{-\infty}^{\infty} \frac{1}{x^2 + \epsilon^2} dx
    = \frac{\epsilon}{\pi} \frac{\pi}{\epsilon} f(c\epsilon) = f(c\epsilon)
    [/tex]

    where c varies from negative infinity to positive infinity. Now as we take the limit of epsilon to zero, we get f(0) which is the answer. But I don't get how the 2nd step (the one where the f(ce) came out of the integral) was done using the mean value theorem. I've looked up the integral forms of the mean value theorem but I still couldn't figure out how the 2nd line is done.
     
  2. jcsd
  3. Aug 19, 2010 #2

    hunt_mat

    User Avatar
    Homework Helper

    How about applying the Cauchy residue theorem? This should yield a function of [tex]\varepsilon[/tex] which you can take the limit without a problem.
     
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