Prove a delta function identity

[Yaelcita]

Hi,

I'm stuck with the last proof I need to do

Homework Statement

I need to prove that f(x)delta(g(x)) = f(x) delta (x-x0)/abs(g'(x))
By delta I mean the Dirac delta function here. (I'm new to this forum, so i don't know how to write it all nicely like so many of you do!)

Homework Equations

The Attempt at a Solution

First of all, of course, I put all of that in an integral. And what I've done so far is just replace g(x) = u so that dx=du/g'(x) It looks promising because I have the g'(x) dividing everything, and if I could somehow use the identity that says that delta(kx) = delta/abs(k) I could almost have it but it seems to me that that identity only works for k a constant and g'(x) is not. Also, I have no idea what to do with my f(x). Can I express it as f(g^-1 (x))? It seems just very weird to me. So, basically I'm stuck there. Any help would be great!

 

[lippyka]

To prove this, we will use the definition of the Dirac delta function, which states that for any continuous function f(x):\int_{-\infty}^{\infty} f(x) \delta(g(x)) \ dx = f(x_0)where x_0 is the only point where g(x) = 0. We can rewrite the integral as:\int_{-\infty}^{\infty} f(x) \delta(g(x)) \ dx = \int_{-\infty}^{\infty} f(x) \frac{\delta(x-x_0)}{|g'(x)|} \ dxUsing the substitution u = g(x), we have \int_{-\infty}^{\infty} f(x) \frac{\delta(x-x_0)}{|g'(x)|} \ dx = \int_{-\infty}^{\infty} f(x) \frac{\delta(u)}{|g'(x)|} \frac{du}{g'(x)} By the definition of the Dirac delta function, we have \int_{-\infty}^{\infty} f(x) \frac{\delta(u)}{|g'(x)|} \frac{du}{g'(x)} = f(x_0)Thus, f(x) \delta(g(x)) = f(x) \frac{\delta(x-x_0)}{|g'(x)|} as required.