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

demonelite123

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.

Hurkyl

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]

mathfeel

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.

demonelite123

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.