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Prove that derivative of the theta function is the dirac delta function

  1. Jul 21, 2011 #1
    let θ(x-x') be the function such that θ = 1 when x-x' > 0 and θ = 0 when x-x' < 0. Show that d/dx θ(x-x') = δ(x - x').

    it is easy to show that d/dx θ(x-x') is 0 everywhere except at x = x'. To show that d/dx θ(x-x') is the dirac delta function i also need to show that the integral over the real line of d/dx θ(x-x') = 1.

    this is what i tried: ∫ d/dx θ(x-x') dx' = d/dx ∫ θ(x-x') dx' = d/dx ∫ 1 dx' but now i am a little stuck and not sure if this is the way to go or not. my plan was to show that the integral equals x so that d/dx (x) = 1 like we wanted. but it seems like the bottom limit of integration will be x and i'll end up getting -1 instead of 1. can anyone give me some advice on this problem? thanks.
     
  2. jcsd
  3. Jul 22, 2011 #2

    Hurkyl

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    This is one of those problems that tends to be really easy if you think of them in a rigorous fashion rather than a heuristic "intuitive" fashion.

    What foundations are you using for the delta function and derivatives and such? When working with tempered distributions, the derivative of a distribution F is the (unique) distribution which satisfies, for all test functions [itex]\varphi[/itex],
    [tex]\int_{-\infty}^{+\infty} F'(x) \varphi(x) \, dx = -\int_{-\infty}^{+\infty} F(x) \varphi'(x) \, dx[/tex]​
     
  4. Jul 22, 2011 #3
    Just think about what is [itex]f(x) = \int_{-\infty}^{x} \delta(t) dt [/itex] for each [itex]x \ne 0[/itex]. You should obvious separate it into positive and negative [itex]x[/itex] case. Then the fundamental theorem of calculus applies.
     
  5. Jul 26, 2011 #4
    ah that's a similar idea to what i did. thanks for your answer! so would this be a more rigorous way to do this problem? i know that taking the derivative operator out of the integral isn't being very careful but the physics book seemed to do that all the time so i thought it would be "ok" since i wan't doing mathematics here.
     
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