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Dirac delta function proof

  1. Apr 6, 2009 #1
    1. The problem statement
    Show that:
    [tex]\int_{-\infty}^{\infty} f(x) \delta^{(n)}(x-a) dx = (-1)^n f^{(n)}(a)[/tex]

    3. The attempt at a solution
    I am trying to understand how to prove:
    [tex]\int_{-\infty}^{\infty} f(x) \delta '(x) dx =- f'(x)[/tex]
    I know that we need to use integration by parts, but I'm looking for a more detailed explanation of how to use integration by parts (what is u and what is dv?). I think if I understand this, then I will be able to apply this to the problem above.
    Last edited: Apr 6, 2009
  2. jcsd
  3. Apr 6, 2009 #2


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    u is f(x), dv is [itex]\delta'(x) dx[/itex].
  4. Apr 6, 2009 #3
    I think you left out a minus sign. There's only one thing there than can be 'u' and only one thing that can be 'dv'.
  5. Apr 6, 2009 #4
    thank you, i did leave out a minus sign. I will correct that above. Also, I think I was trying to make things more complicated than they actually are. But why does the first term when doing integration by parts go to zero?

    [tex]\boxed{[f(x)\delta(x)]_{-\infty}^{\infty}} - \int_{-\infty}^{\infty} f'(x)\delta(x) dx[/tex]
  6. Apr 6, 2009 #5
    I thought it was a definition of delta-function derivative... could someone tell me if i'm wrong?
  7. Apr 6, 2009 #6
    All that is necessary for the first term to go to zero is that delta goes to zero faster than f goes to infinity. How fast does delta go to zero?
  8. Apr 6, 2009 #7
    delta tends to zero very quickly when x does not equal a (which is zero in this case) by definition of the generalized function . If that's right, then I think I get it. Thank you for your help.
  9. Apr 6, 2009 #8
    Yes, that is correct. delta of x thuds to zero immediately x leaves 0, long before x gets to infinity.
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