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Derivative of unit step function

  1. Nov 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Show that δ(x-x') = d/dx Θ(x-x')

    2. Relevant equations
    ∫ f(x') δ(x-x') dx' = f(x)
    Θ(x-x') vanishes if x-x' is negative and 1 if x-x' is positive
    3. The attempt at a solution
    I saw a relation of the δ function but I don't know why is it like that.

    Integral of δ(x-x') from -∞ to x is 1 if x>x' and 0 if x<x'

    I'm not sure how to start. Any suggestions?
  2. jcsd
  3. Nov 15, 2015 #2


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    The statement is really only true for ##x\neq x'##. Try to prove it first for ##x<x'## and then for ##x>x'##.

    Neither side has a value at ##x=x'##. But what you can prove that can substitute for that (if one sets up the required structure based on treating them as distributions rather than functions) is that, for ##a<x'<b:\ \int_a^b\delta(x-x')dx=\Theta(b)-\Theta(a)##.

    If you can prove those three things, I think that's as close as one can get without getting deep into distribution theory.
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