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Delta function of a function with multiple zeros

  1. Aug 27, 2008 #1
    Hi everyone,

    I was wondering how to deal with delta functions of functions that have double zeros.

    For instance, how does one compute an integral of the form

    [tex]\int_{-\infty}^{\infty}dx g(x)\delta(x^2)[/tex]

    where g(x) is a well behaved continuous everywhere function?

    In general how does one find

    [tex]\int_{-\infty}^{\infty}dx g(x)\delta(f(x))[/tex]

    where f(x) has a finite number of multiple zeros along with some simple zeros. I know that

    [tex]\delta(f(x)) = \sum_{i=1}^{N}\frac{\delta(x-x_{i})}{|f'(x_{i})|}[/tex]

    where [itex]x_{i}[/itex]'s (for i = 1 to N) are simple zeros of [itex]f(x)[/itex] and it is known that [itex]f(x)[/itex] has no zeros of multiplicitiy > 1.

    but this is of course not valid here. Using this, however I could write

    [tex]\delta(x^2-a^2) = \frac{1}{2|a|}\left(\delta(x-a) + \delta(x+a)\right)[/tex]

    But the limit of this as [itex]a \rightarrow 0[/itex] tends to infinity.

    Any ideas?

    Thanks in advance.
    Cheers.
     
  2. jcsd
  3. Aug 27, 2008 #2

    haushofer

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    Science Advisor

  4. Aug 28, 2008 #3
    Thank you haushofer.
     
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