Problem on integrating dirac delta function

[tan90ds]

Hi there,
I am trying to integrate this: http://imm.io/oqKi
I should get the second line from the integral, but I can't show it.
This should somehow relate to the Heaviside step function, or I am completely wrong.
Any ideas?

Sorry about the url, I fixed it.

 

[mathman]

What is "imm.io/oqKi "?

 

[arildno]

Anyhow, whatever "imm.io/oqKi " is:

The Dirac delta "function" isn't a function in the usual sense, so can't integrate it in the usual sense, either.

 

[HallsofIvy]

The "Dirac delta funtion" is not really a function, as arildno says. It is a "generalized function" or "distribution"- a linear operator on functions.

By definition
$$\int_a^b f(x)\delta(x)dx= f(0)$$
if a< 0< b, 0 otherwise.

That means that
$$\int_a^b \delta(g(x))f(x) dx= f(c)$$
if g(c)= 0 for some c between a and b.

 

[tan90ds]

The "Dirac delta funtion" is not really a function, as arildno says. It is a "generalized function" or "distribution"- a linear operator on functions.

By definition
$$\int_a^b f(x)\delta(x)dx= f(0)$$
if a< 0< b, 0 otherwise.

That means that
$$\int_a^b \delta(g(x))f(x) dx= f(c)<br /> if g(c)= 0 for some c between a and b.$$

[tex] <br /> This makes sense for why they put $$(x-x_0)\cos(\theta)+(y-y_0)\sin(\theta)=0$$ after the second line.<br /> <br /> Actually I am in the middle of proving the simple backprojection of the Radon transform of a dot can be viewed as a two dimensional convolution of $$\frac{1}{\sqrt{x^2+y^2}}$$ and the original function. I used the Dirac Delta to formulate the dot, so this is just for the convenience of prove.<br /> <br /> The Dirac Delta also has this property:$$\int\delta(\alpha x)dx = \frac{1}{|\alpha|}$$, I think this might help.[/tex]