- #1
Yaelcita
- 14
- 0
Hi,
I'm stuck with the last proof I need to do
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!)
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!
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!