Assuming you mean the Dirac delta function here. I think you should use different function labels in order to avoid confusion. Try this on for size. Let
Then
Let Then the integral
Does that work for you?
Suppose we have some well behaved curve y=f(x) lying in a plane. The length of a curve is given by
L = int( sqrt( 1+f'(x) ) )dx
I want to express this integral using the delta fuction as an integral over some area
A = int( delta( y-f(x) ) )dxdy.
Can this be done? If so, how to prove it?
Thanks.
It came from my head. My reasoning went like this: the Dirac delta function (technically, a distribution), under an integral sign, is a "replacer". That is, you actually define the Dirac delta distribution by its action under an integral sign:
So the Dirac delta distribution "picks out" one value of the domain, and substitutes it into the function f(x).
Now, I wanted to end up with and I wanted to use the delta distribution Hence, the function into which I wanted to replace would have been a simple
Make sense?