Solving a Problem with Integrals in R3

[cyc454]

Hi! I've got a problem with an integral. Let's assume we've got something like this:

∫_R³d³x₁∫_R³d³x₂∫_R³d³x₃∫_R³d³x₄P(|x₁|)P(|x₃|)δ(x₁+x₂)δ(x₃+x₄)W(|x₁+x₂|)W(|x₃+x₄|)


x_i is a vector
The "δ" is the Dirac delta.
P(|x|_i) & W(|x_i+x_j|) are some functions
I would like to make it looks a bit simpler---I mean get rid of deltas and two integrals. How can I make it?
Thanks for help and sorry for spelling mistakes!

 

[Bryson]

What have you attempted? Do you understand the properties of the delta function?

 

[cyc454]

if the x_i=-x_j then δ ≠0.

∫_R³δ(x)d³x should be equal 1. Well, actually it should looks

like this:

∫^∞_-∞δ(x)dx=1

but it is the same I thing.. This is all I know.

 

[Bryson]

Okay, also note that \int \cdots \int f(\vec{x}) \delta(\vec{x} - \vec{x}_o) d^Nx = f(\vec{x}_o). This can allow you to fix some variables.

My next question is, are we integrating from -\infty \rightarrow \infty? If the variable being integrated is not within the bounds, we can simplify things greatly.

I must say, it has been awhile since I have done integrals of this form.

 

[cyc454]

Well, we are integrating it over the entire R³..
I don't get it. There isn't any function depending on x. there is only P and W that depend on |x| or |x_i+x_j|

PS I can't put P and W before the integrals, can I?
PPS One more thing. There is a integral:
∫d³x₁
and let's assume x₁=x₂+x₃ so the d³x₁=d³x₂+d³x₃. So after substitution
∫d³x₁=∫d³x₂+∫d³x₃? is it correct?