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Delta Function Identity

  1. Jan 11, 2010 #1
    I know I haven't entered the formulae with the proper syntax, but I'm extremely exhausted at the time of posting, so please just read it and give advice, forgiving me this once for not using proper form (it's basically in latex code format).

    1. The problem statement, all variables and given/known data

    Show [itex]f(x)\frac{d}{dx}\delta(x) = f(0)\frac{d}{dx}\delta(x)-f'(0)\delta(x)[/itex]

    2. Relevant equations
    [tex]\delta(x)\intf(x)dx = \int f(x)\delta(x)dx = f(0)[/tex]
    [tex](f',\phi):=(f,-\phi ')[/tex]


    3. The attempt at a solution
    [tex]f(0)\delta'(x) - f'(0)\delta(x)[/tex]
    [tex]=\delta'(x)\intf(x)\delta(x)dx-\delta(x)\intf'(x)\delta(x)dx[/tex]
    [tex]=...[/tex]
     
    Last edited by a moderator: Jan 11, 2010
  2. jcsd
  3. Jan 11, 2010 #2

    HallsofIvy

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    I would use "integration by parts". To integrate
    [tex]\int (f(x) d(\delta(x))/dx) dx[/tex]
    let u= f(x) and [itex]dv= d(\delta(x))/dx dx[/itex] so that du= f'(x) dx and [itex]v= \delta(x)[/itex].=

    Then [tex]\int (f(x) d(\delta(x))/dx) dx= f'(x)\delta(x)- \int f'(x)\delta(x) dx[/tex]
    [tex]= f'(x)\delta(x)- f'(0)[/itex].

    So [itex]d(\delta(x))/dx[/itex] is the distribution that maps f(x) to [itex]f'(x)\delta(x)- f'(0)[/itex].
     
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