1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dirac delta function

  1. Jan 7, 2008 #1
    1. The problem statement, all variables and given/known data

    Hi there, I'm stuck at a problem where I have (sorry i don't know how to use mathtype so I'll try my best at making this clear) the integral of a dirac delta function squared:

    int[delta(x*-x)^2] between minus infinity and infinity (x*=constant)

    I know that the function equals x*(int[delta(x)^2]) but I'm at a bit of a loss as to why :/


    3. The attempt at a solution

    I'm pretty sure the function int[delta(x-x*)f(x)dx]=f(x*) has something to do with it but I can't quite figure it out.

    If f(x)=delta(x-x*) then f(x*)=delta(x*-x*) :? can f(x) even be another delta function?

    Thanks for any replies,
    JSG
     
    Last edited: Jan 7, 2008
  2. jcsd
  3. Jan 7, 2008 #2

    mda

    User Avatar

    I am not a mathematician, but as far as I know int[delta(x)^2] is infinite, hence your integral should also be infinite.

    One way to deal with these sort of problems is to replace the delta with a finite approximant and take the limit as the function width goes to zero.
     
  4. Jan 8, 2008 #3

    pam

    User Avatar

    [tex]\delta[f(x-x*)]=\delta(x-x*)/|df/dx|_{x=x*}[/tex].
    This integral diverges. Perhaps the problem was for [tex]\delta(x^2-x*^2)[/tex].
     
  5. Jan 8, 2008 #4

    Mute

    User Avatar
    Homework Helper


    The question asked is actually for the square of the delta function, not its argument.

    The problem with the square of a delta function is that by the property

    [tex]\int_{-\infty}^{\infty} dx f(x) \delta(x - x_0) = f(x_0)[/tex]

    one would expect the answer to be the delta function at x_0 if one just blindly substitutes the delta function for f(x) there, but since the delta function is not really defined outside of an integral, that won't work. So, slightly less naively than that I would say there is no well defined answer and that integral just doesn't make sense. I believe I once found a pdf somewhere on the internet that stated that.

    However, I once asked my supervisor about this, and he gave some sort of shifty physicist explanation of it, which I think resulted a result like

    [tex]\int_{-L}^{L} dx \delta^2(x) f(x) = 2 \pi L \int_{-L}^{L} dx \delta(x) f(x)[/tex]

    where I think he used the fourier integral representation for one of the delta functions in the first integral. Now, I probably didn't recall this correctly and the RHS is probably wrong, but that was the idea, or something. I know I wasn't really convinced by it, and I think it had something to do with him using a fourier representation of the delta function using finite limits of +/- L, which I didn't like, so presummably L would be tending to infinity in the above expression. But again, I can't remember what my prof did, and it came off as shifty physics logic, so this may not help you much.
     
    Last edited: Jan 8, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Dirac delta function
  1. Dirac delta function (Replies: 8)

  2. Dirac Delta Function (Replies: 1)

Loading...