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Dirac Delta Scaling

  1. Jan 21, 2010 #1
    Using the defining property of the dirac delta function,
    [tex]\int{dx f(x) \delta(x-c)}[/tex]

    show that
    [tex]\delta(ax)=\frac{1}{|a|}\delta(x)[/tex]

    I think all I need to do is make the right u substitutions and it will come out right, but I can't think of how to make the substitutions...after a long time working on this problem.
     
  2. jcsd
  3. Jan 21, 2010 #2

    Dick

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    You mean delta(ax)=delta(x)/|a|. Not c. Just substitute u=ax. Compare delta(x) with delta(u). Don't forget du=a*dx. And show what you are actually doing this time. The integral of f(x-c)*delta(x)*dx equals f(c). The integral of f(u-c)*delta(u)*du must also equal f(c). What's the relation between the two deltas?
     
    Last edited: Jan 21, 2010
  4. Jan 21, 2010 #3
    Thanks Dick, I kind of get the right answer, but it's just not the way I was thinking about it, so starting with the defining property we have
    [tex]\int{dx f(x) \delta(x-c)}[/tex]

    I make the u substitution u=ax, and du=a dx
    [tex]\frac{1}{a}\int{du f(u/a) \delta(u/a-c)}[/tex]

    So that right there tells me that [tex]\delta(ax)=\frac{1}{a}\delta(x)[/tex], but the problem states that it should be [tex]\delta(ax)=\frac{1}{|a|}\delta(x)[/tex]

    So I'm not sure where the absolute value comes in.
     
  5. Jan 21, 2010 #4

    Dick

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    If a is negative, then the change of variable makes integration limits reverse from -infinity to +infinity to +infinity to -infinity. To turn it back into a proper delta function, you have to turn them back. That introduces an extra sign. See what I'm saying?
     
  6. Jan 21, 2010 #5
    yes that makes sense. Thanks!
     
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