(1) Let f(x_0)=0, f'(x_0) \= 0. I'm trying to prove the property

delta(f(x)) = (1/|f'(x_0)| ) delta (x-x_0)

by observing

INT(-oo,oo) delta(f) df = INT(-oo,oo) delta(x-x_0) dx.

The left-hand side contributes to the integral only when f(x)=0, i.e., when
x=x_0, and the right-hand side also only contributes when x=x_0. But how can
we then take everything out of the integral and assert

delta(f(x)) = (1/|f'(x_0)| ) delta (x-x_0) ?

(2) How do you show that INT(0,oo) f(x)delta(x) dx = (1/2) f(0) ?

I know that INT(-oo,oo) f(x) delta(x) dx = f(0), so I'd like to show that

INT(-oo,0) f(x) delta(x) dx = INT(0,oo) f(x) delta(x) dx.

But, the best I can do as of yet is to obtain

INT(-oo,0) f(x) delta(x) dx = - INT(oo,0) f(-x) delta(-x) dx = INT(0,oo)
f(-x) delta(-x) dx = INT(0,oo) f(-x) delta(x) dx,

using first a change of variable and second that delta(x) is an even
function. Unless I messed up on the change of variable, though, I don't see
how to take this the next step and assert that it = INT(0,oo) f(x) delta(x)
dx without some special condition of f (i.e. f is even).

Suggestions please! Thx.

On Dec 3 burger wrote in sci.math and sci.physics:

(1) Let f(x_0)=0, f'(x_0) \= 0. I'm trying to prove the property

delta(f(x)) = (1/|f'(x_0)| ) delta (x-x_0)

by observing

INT(-oo,oo) delta(f) df = INT(-oo,oo) delta(x-x_0) dx.

The left-hand side contributes to the integral only when f(x)=0, i.e., when
x=x_0, and the right-hand side also only contributes when x=x_0. But how can
we then take everything out of the integral and assert

delta(f(x)) = (1/|f'(x_0)| ) delta (x-x_0) ?


It's probably best to view this as a definition of delta(f), motivated
by the need for 'integration by substitution' to work.

(2) How do you show that INT(0,oo) f(x)delta(x) dx = (1/2) f(0) ?

I know that INT(-oo,oo) f(x) delta(x) dx = f(0), so I'd like to show that

INT(-oo,0) f(x) delta(x) dx = INT(0,oo) f(x) delta(x) dx.

But, the best I can do as of yet is to obtain

INT(-oo,0) f(x) delta(x) dx = - INT(oo,0) f(-x) delta(-x) dx = INT(0,oo)
f(-x) delta(-x) dx = INT(0,oo) f(-x) delta(x) dx,

using first a change of variable and second that delta(x) is an even
function. Unless I messed up on the change of variable, though, I don't see
how to take this the next step and assert that it = INT(0,oo) f(x) delta(x)
dx without some special condition of f (i.e. f is even).

For any f, f(x)delta(x) is even: f(-x) = f(x) = 0 for all non-zero x.

--
P.A.C. Smith
replying by email: s/NOSPAM//

"The vast majority of Iraqis want to live in a peaceful, free world.
And we will find these people and we will bring them to justice."
- George W. Bush (Washington DC, Oct 27 2003)

Today Phil Smith wrote in sci.math and sci.physics:

On Dec 3 burger wrote in sci.math and sci.physics:

(1) Let f(x_0)=0, f'(x_0) \= 0. I'm trying to prove the property

delta(f(x)) = (1/|f'(x_0)| ) delta (x-x_0)

by observing

INT(-oo,oo) delta(f) df = INT(-oo,oo) delta(x-x_0) dx.

The left-hand side contributes to the integral only when f(x)=0, i.e., when
x=x_0, and the right-hand side also only contributes when x=x_0. But how can
we then take everything out of the integral and assert

delta(f(x)) = (1/|f'(x_0)| ) delta (x-x_0) ?


It's probably best to view this as a definition of delta(f), motivated
by the need for 'integration by substitution' to work.

(2) How do you show that INT(0,oo) f(x)delta(x) dx = (1/2) f(0) ?

I know that INT(-oo,oo) f(x) delta(x) dx = f(0), so I'd like to show that

INT(-oo,0) f(x) delta(x) dx = INT(0,oo) f(x) delta(x) dx.

But, the best I can do as of yet is to obtain

INT(-oo,0) f(x) delta(x) dx = - INT(oo,0) f(-x) delta(-x) dx = INT(0,oo)
f(-x) delta(-x) dx = INT(0,oo) f(-x) delta(x) dx,

using first a change of variable and second that delta(x) is an even
function. Unless I messed up on the change of variable, though, I don't see
how to take this the next step and assert that it = INT(0,oo) f(x) delta(x)
dx without some special condition of f (i.e. f is even).

For any f, f(x)delta(x) is even: f(-x) = f(x) = 0 for all non-zero x.

This should read, "f(-x)delta(-x) = f(x)delta(x) = 0"

--
P.A.C. Smith
replying by email: s/NOSPAM//

"The vast majority of Iraqis want to live in a peaceful, free world.
And we will find these people and we will bring them to justice."
- George W. Bush (Washington DC, Oct 27 2003)

"burger" wrote in message ...
(1) Let f(x_0)=0, f'(x_0) \= 0. I'm trying to prove the property

delta(f(x)) = (1/|f'(x_0)| ) delta (x-x_0)

Hum.

y = f(x)

dy = f'(x) dx

Let the inverse of f(x) exist and be adequately well behaved.
Let the inverse of f(x) be called g.

x = g(y)

dx = dy / f'(g(y))

Then

int F(x) delta(f(x)) dx

= int F(g(y)) delta(y) / f'(g(y)) dy

= F(g(0)) / f'(g(0))

Now f(x_0) = 0. So g(0) = x_0.

So int F(x) delta(f(x)) dx = F(x_0) / f'(x_0)

So delta(f(x)) = delta(x - x_0) / f'(x)

Socks