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Limit of Lorentz distributions

  1. Mar 15, 2010 #1
    Hi guys

    Can one prove the identity


    \frac{\epsilon}{x^2 + \epsilon^2} \underset{\epsilon\to 0^+}{\to} \pi \delta(x)


    or is it just intuitively clear (by looking at a graph)?
  2. jcsd
  3. Mar 15, 2010 #2
    First show that [tex]\int_{-\infty}^\infty\frac{\epsilon}{x^2+\epsilon^2}dx=\pi[/tex] for any [tex]\epsilon>0[/tex]. Then show that for any fixed [tex]x\ne0[/tex], [tex]\lim_{\epsilon\to0^+}\frac{\epsilon}{x^2+\epsilon^2}=0[/tex].
  4. Mar 15, 2010 #3
  5. Mar 15, 2010 #4


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    Homework Helper

    Alternatively, consider

    [tex]\lim_{\epsilon \rightarrow 0^+} \int_{-\infty}^{\infty} dx~\frac{\epsilon}{x^2 + \epsilon^2} f(x)[/tex]
    and consider the change of variables [itex]y = x/\epsilon[/itex], and show that the result is [itex]\pi f(0)[/itex]. It is in this sense that

    [tex]\frac{\epsilon}{x^2 + \epsilon^2} \underset{\epsilon\to 0^+}{\to} \pi \delta(x).[/tex]
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