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Proving a Dirac delta property

  1. Aug 30, 2015 #1
    1. The problem statement, all variables and given/known data

    Prove that [tex]\delta[a(x-x_1)]=\frac{1}{a}\delta(x-x_1)[/tex]

    2. Relevant equations

    In my attempt I have used [tex]\delta(ax)=\frac{1}{a}\delta(x)[/tex] but I'm not sure I'm allowed to use it in this proof.

    3. The attempt at a solution

    Some properties of Dirac delta function are proven using a test function. Thence I tried [tex]I=\int f(x)\delta[a(x-x_1)]=[/tex]

    In the following I tried the substitution y = x - x1, getting [tex]I=\int f(y+x_1)\delta(ay)=\frac{1}{a}\int f(y+x_1)\delta(y)=\frac{1}{a}f(x_1)=\frac{1}{a}\int f(x)\delta(x-x_1)[/tex]

    But I don't know if this proof is correct, as I have used a property similar to the one I'm trying to prove. Is it ok to do this?
     
  2. jcsd
  3. Aug 30, 2015 #2

    blue_leaf77

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    Next, do another integration with ##\delta[a(x-x_1)]## replaced by ##\frac{1}{a}\delta(x-x_1)##, with the same ##f(x)##. Then compare the results with the first integration, taking into account that ##f(x)## is an arbitrary function.
     
  4. Aug 30, 2015 #3

    vela

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    I don't think you're allowed to use that. Try the substitution ##u = a(x-x_1)##.
     
  5. Aug 31, 2015 #4

    HallsofIvy

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    What is your definition of the delta function? Exactly how you would prove this depends upon which definition you are using and I believe there are several.
     
  6. Aug 31, 2015 #5
    Hallsoflvy: I have used the definition by the integral [tex]\int f(x)\delta(x)=f(0)[/tex]
    vela: with this substitution it worked.

    Thank you all for the help
     
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