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

1. Jul 21, 2011

### 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.

2. Jul 22, 2011

### Hurkyl

Staff Emeritus
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 $\varphi$,
$$\int_{-\infty}^{+\infty} F'(x) \varphi(x) \, dx = -\int_{-\infty}^{+\infty} F(x) \varphi'(x) \, dx$$​

3. Jul 22, 2011

### mathfeel

Just think about what is $f(x) = \int_{-\infty}^{x} \delta(t) dt$ for each $x \ne 0$. You should obvious separate it into positive and negative $x$ case. Then the fundamental theorem of calculus applies.

4. Jul 26, 2011

### 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.