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Dirac delta function

  1. Oct 5, 2006 #1
    Suppose that we take the delta function [tex]\delta(x)[/tex] and a function f(x). We know that

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

    However, does the following have any meaning?

    [tex]\int_{-\infty}^{\infty} f(x)\delta(x-a)\delta(x-b)dx,[/tex]

    for some constants [tex]-\infty<a,b<\infty[/tex].
     
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  3. Oct 5, 2006 #2

    HallsofIvy

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    Remember that [itex]\delta[/itex] is not a true function. It is a "distribution" or "generalized function". The first integral is defined for distributions but the second isn't. (If we were to force a meaning on it, the only reasonable value it could have would be 0.)

    You could do that in more than one dimension:
    [tex]\int_{x=-\infty}^{\infty}\int_{y=-\infty}^{\infty}f(x,y)\delta(x-a)\delta(y-b)dydx= f(a,b)[/itex]
     
  4. Oct 5, 2006 #3
    Here's what I was thinking about it. The integral

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

    integrates a function f(x) times a distribution [tex]\delta[/tex], so the whole thing is an integral of a generalized function, dependent on x. Suppose now that I define a generalized function [tex]g(x)[/tex] according to

    [tex]g(x) \equiv f(x)\delta(x-a).[/tex]

    Then the second integral becomes

    [tex]\int_{-\infty}^\infty f(x)\delta(x-a)\delta(x-b)\,dx = \int_{-\infty}^\infty g(x) \delta(x-b).[/tex]

    My instinct is to then conclude that

    [tex]\int_{-\infty}^\infty g(x)\delta(x-b)\, dx = g(b) = f(b)\delta(b-a)[/tex]

    Thus, the second integral is itself something which has meaning only if it appears within an integral. Yet you don't think that this is correct?
     
    Last edited: Oct 5, 2006
  5. Oct 5, 2006 #4

    daniel_i_l

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    the function g(x) is basiclly a constant (f(a)) times the dirac delta function. so in your second integral you want the integral of the product of two different DD functions times a constant, that would equal 0
     
  6. Oct 5, 2006 #5
    I'm afraid I don't see how you conclude that g(x) is a constant. Going with the definition I gave above,

    [tex]g(x)\equiv f(x)\delta(x-a)[/tex]

    for some function f(x). The only relationship between g(x) and f(a) is

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

    i.e., the integral of g(x) is equal to f(a).
     
  7. Oct 5, 2006 #6

    daniel_i_l

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    I didn't say that it was a constant, i said that is was equal to a constant times the DD function. what i meant to say that was in this case you could look at g(x) as being g(x) = f(a)*DD(x-a) when you're doing the integral. so you're right, it doesn't really equal f(a)*DD(x-a) but when you integrate you get the same resault when you interchange the two.
    and anyway f(b)DD(b) = 0.
     
    Last edited: Oct 5, 2006
  8. Oct 5, 2006 #7
    - Shoehorn according to "Schwartz theorem2 you can't multiply 2 distribution..this is really "nasty" since avoiding this (if we could) we would have solved divergences in the form [tex] \int_{0}^{\infty}dx x^{n} [/tex] using "Casual perturbation theory2 (see wikipedia) although i believe that for big n:

    [tex] \delta (x-a) \delta (x-b) =n^{2}\pi e^{-n^{2}((x-a)^2 +(x-b)^2)}

    [/tex] for n tending to [tex] \infty [/tex]
     
  9. Oct 5, 2006 #8

    matt grime

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    How do you conclude something that is not even a function is actually a lebesgue square measurable function?
     
  10. Oct 5, 2006 #9
    Oops, sorry. I was editing a post earlier and pasted that in by accident. Please ignore it - it's obviously not true. :-)
     
  11. Oct 5, 2006 #10

    Hurkyl

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    No.

    However, if a (or b) was a variable, then this expression could be considered a distribution. Convolving it with h(a) gives:

    [tex]
    \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}
    f(x) h(a) \delta(x-a) \delta(x-b) \, da \, dx
    = f(b) h(b)
    [/tex]
     
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