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Property of the dirac delta function

  1. May 4, 2012 #1
    Hello team!

    I saw the other day in a text book that the Dirac delta function of the form d(x-a) can be written as d(a-x) but the method was not explained. I was wondering if anyone know where this comes from. I've been googling but can seem to find it out. Any help would be appreciated.

    Cheers!
    Jonathan
     
  2. jcsd
  3. May 4, 2012 #2

    micromass

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    How did you define the Dirac delta function?? Did you do it by distributions?
     
  4. May 5, 2012 #3
    The formal definition preceding the statement was:
    f(a) = ∫f(x)d(x-a)dx the integral goes from inf to -inf
    I was thinking maybe you make a change of variable.
     
  5. May 5, 2012 #4
    Hi. So you would like to prove delta function is even function, I.e. delta x =delta -x.
     
  6. May 5, 2012 #5
    Yes, i guess this is the crux of the problem.
     
  7. May 5, 2012 #6
    Hi.

    ∫f(x)Δ(a-x)dx, x[-∞,+ ∞]
    =∫f(-t)Δ(a+t)dt, t[-∞,+ ∞]
    =f(a)

    Subtracting each other,
    for any f(x) ∫f(x){Δ(x-a)-Δ(a-x)}dx=0
    so Δ(x-a)=Δ(a-x).

    Regards.
     
  8. May 8, 2012 #7
    I'm also interested in this proof.

    if i start out with

    [itex]f(a)=\int_{-\infty}^{\infty}f(x)\delta(a-x) dx [1][/itex]

    and make the change of variable [itex]x\rightarrow -t[/itex]

    [itex]\Rightarrow dx\rightarrow -dt[/itex]

    then

    [itex]f(a)=-\int_{-\infty}^{\infty}f(-t)\delta(a+t)dt [2][/itex]


    i'm a bit confused how you then get one of them in the form δ(x-a)
     
  9. May 8, 2012 #8
    You forgot another part of the definition; The dirac delta function is a function that is 0 everywhere except at zero and:
    Δ(0) = infinity

    With this in mind:
    x-a = a-x when x = a
     
    Last edited: May 8, 2012
  10. May 8, 2012 #9

    Mute

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    More precisely, the dirac delta function has the property that

    [tex]\int_b^c dx f(x)\delta(x-a) = \left\{\begin{array}{c} f(a),~a \in [b,c] \\ 0,~\mbox{otherwise} \end{array}\right.[/tex]

    The delta function doesn't really have a well-defined meaning outside of an integral, so as far as we're concerned, if integrating f(x) against [itex]\delta(x-a)[/itex] or [itex]\delta(a-x)[/itex] gives you the same result, then [itex]\delta(x-a) = \delta(a-x)[/itex].
     
  11. May 8, 2012 #10
    Hi.
    [itex]f(a)=-\int_{\infty}^{-\infty}f(-t)\delta(a+t)dt [2][/itex]
    isn't it? Regards.
     
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