- #1
bob900
- 40
- 0
The Dirac delta "function" is often given as :
δ(x) = ∞ | x [itex]=[/itex] 0
δ(x) = 0 | x [itex]\neq[/itex] 0
and ∫δ(x)f(x)dx = f(0).
What about δ(cx)? By u=cx substitution into above integral is, ∫δ(cx)f(x)dx = ∫δ(u)f(u/c)du = 1/c f(0).
But intuitively, the graph of δ(cx) is the same as the graph of δ(x)! At x=0, cx=0 so δ(cx)=∞, and everywhere else cx[itex]\neq[/itex]0 so δ(cx)=0.
So how can it be that ∫δ(x)f(x)dx = 1/c ∫δ(cx)f(x)dx ?
δ(x) = ∞ | x [itex]=[/itex] 0
δ(x) = 0 | x [itex]\neq[/itex] 0
and ∫δ(x)f(x)dx = f(0).
What about δ(cx)? By u=cx substitution into above integral is, ∫δ(cx)f(x)dx = ∫δ(u)f(u/c)du = 1/c f(0).
But intuitively, the graph of δ(cx) is the same as the graph of δ(x)! At x=0, cx=0 so δ(cx)=∞, and everywhere else cx[itex]\neq[/itex]0 so δ(cx)=0.
So how can it be that ∫δ(x)f(x)dx = 1/c ∫δ(cx)f(x)dx ?