Derivative of unit step function

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1. Nov 15, 2015

shinobi20

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. Nov 15, 2015

andrewkirk

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.