Limit of Lorentz distributions

  • Thread starter Niles
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  • #1
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Main Question or Discussion Point

Hi guys

Can one prove the identity

[tex]

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

[/tex]

or is it just intuitively clear (by looking at a graph)?
 

Answers and Replies

  • #2
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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
Mute
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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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