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Derivative of an integral containing a Dirac delta

  1. Oct 1, 2005 #1
    If I had a function g(x) defined by

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

    where [tex]\delta(x)[/tex] is the dirac delta function, what would dg(x)/dx be? The fundamental theorem of calculus requires that [tex]f(x) \delta(x)[/tex] needs to be a continuous and differentiable function before I can immediately say that dg(x)/dx = f(x) \delta(x), which is clearly not the case.
     
  2. jcsd
  3. Oct 1, 2005 #2

    HallsofIvy

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    [tex]g(x) = \int_{-\infty}^{\infty} f(x) \delta(x) dx[/tex]
    is not a function of x! It's derivative is 0. In fact that's true of any definite integral of any integrable function! Perhaps a little more interesting would be:
    What is the derivative of
    [tex]g(x)= \int_{-\infty}^x f(t)\delta(t)dt[/tex]
    but not a whole lot more! If x< 0, g(x)= 0 and the derivative is 0. If x> 0, g(x)= f(0) and the derivative is 0. However, g(x) is not differentiable at 0.

    (In terms of 'distributions' or 'generalized functions', which is what [itex]\delta(x)[/itex] really is, that is differentiable at 0: the derivative of g(x) is [itex]\delta(x)f(x)[/itex].)
     
  4. Oct 1, 2005 #3
    Hmm...

    The Heaviside function,

    [tex] H(x) = \int_{-\infty}^x \delta(t) \: dt [/tex]

    shows what you mean, but how is it that many people define [tex] H(0) = 1/2 [/tex]? I think I'm missing the point...
     
  5. Oct 2, 2005 #4

    lurflurf

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    That is because it is nice if
    f(x-)=f(x)=f(x+)
    but if
    f(x+)!=f(x-)
    the next best thing is if
    f(x)=(f(x-)+f(x+))/2
     
  6. Oct 2, 2005 #5

    HallsofIvy

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    The value H(0)= 1/2 is convenient but really irrelevant. H(0) is still distcontinuous at 0. Since the crucial point with distributions is their integral properties values at individual values of x are not important.
     
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