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Dirac delta function in reciprocal function

  1. Oct 22, 2011 #1
    From dirac, if A=B, then [itex] \frac{A}{x}=\frac{B}{x}+c\delta(x)[/itex] (1) How this formula is derived?

    Since [itex]\frac{dlnx}{dx} = \frac{1}{x}-i\pi\delta(x)[/itex]
    We can get [itex]\frac{A}{x} = A\frac{dlnx}{dx}+Ai\pi\delta(x)[/itex]
    [itex]\frac{B}{x} = B\frac{dlnx}{dx}+Bi\pi\delta(x)[/itex]
    So if A=B, [itex] \frac{A}{x}=\frac{B}{x}.[/itex]

    Another argument is if we integrate the equation (1) from -a to a, a->[itex]\infty[/itex] and assume in a small region [itex][-\varepsilon, \varepsilon ][/itex], [itex]\int_{-\varepsilon}^{\varepsilon}\frac{1}{x}dx=0,[/itex] so we can get [itex] \int_{-a}^{a}\frac{1}{x}dx=0, but \int_{-a}^{a}c\delta(x)dx=c,[/itex] so the left side of equation (1) doesn't equal to the right side. Please correct me if I am wrong!
  2. jcsd
  3. Oct 23, 2011 #2


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    I'm not sure what you're getting at. It seems like you're talking about the identity

    [tex]\lim_{\epsilon \rightarrow 0} \frac{1}{x\pm i\epsilon} = \mathcal P \frac{1}{x} \mp i\pi \delta(x),[/tex]
    where [itex]\mathcal P[/itex] denotes a principle value.

    Beyond that, though, I'm not sure what your question is.
  4. Oct 23, 2011 #3
    Hi Mute,

    In Dirac's book, he demonstrates that if A=B, one can't infer A/x=B/x, but only A/x=B/x+cδ(x). I cannot get the latter result. And if we integrate the latter equation, it doesn't look correct.

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