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Homework Help: Proofs for Dirac delta function/distribution

  1. Sep 22, 2010 #1
    [SOLVED] Proofs for Dirac delta function/distribution

    1. The problem statement, all variables and given/known data

    Prove that
    [tex]
    \delta(cx)=\frac{1}{|c|}\delta(x)
    [/tex]


    2. Relevant equations

    [tex]\delta(x)[/tex] is defined as
    [tex]
    \delta(x)=\left\{\stackrel{0 for x \neq 0}{\infty for x=0}
    [/tex]

    It has the properties:
    [tex]
    \int^{\infty}_{-\infty}\delta(x)dx=1
    [/tex]
    [tex]
    \int^{\infty}_{-\infty}f(x)\delta(x-a)dx=f(a)
    [/tex]

    Two expressions [tex]D_{1}[/tex] and [tex]D_{2}[/tex] involving [tex]\delta(x)[/tex] are considered equivalent if
    [tex]
    \int^{\infty}_{-\infty}f(x)D_{1}dx = \int^{\infty}_{-\infty}f(x)D_{2}dx
    [/tex]

    3. The attempt at a solution

    Starting from

    [tex]
    \int^{\infty}_{-\infty}f(x)\delta(cx)dx = \int^{\infty}_{-\infty}f(x)\frac{1}{|c|}\delta(x)dx
    [/tex]

    I make a substitution u=cx, and du=cdx.

    [tex]
    \frac{1}{c}\int^{\infty}_{-\infty}f(u/c)\delta(u)du = \frac{1}{|c|}\int^{\infty}_{-\infty}f(x)\delta(x)dx
    [/tex]

    Substitute f(u/c)=g(u), and on right side evaluate integral.

    [tex]
    \frac{1}{c}\int^{\infty}_{-\infty}g(u)\delta(u)du = \frac{1}{|c|}f(0)
    [/tex]

    Evaluate left side.

    [tex]
    \frac{1}{c}g(0)= \frac{1}{|c|}f(0)
    [/tex]

    By x=u/c, g(0)=f(0), so


    [tex]
    \frac{1}{c}f(0)= \frac{1}{|c|}f(0)
    [/tex]


    So, I got this far, but I don't understand where the absolute value on the right side comes from.
     
    Last edited: Sep 22, 2010
  2. jcsd
  3. Sep 22, 2010 #2
    see, when c is negative, your limits on the integral after the substitution change signs, therefore, for negative c, you should have a modulus sign.
     
  4. Sep 22, 2010 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If c is negative, when you do the substitution u=cx the u-limits become +infinity to -infinity, instead of -infinity to +infinity for x.
     
  5. Sep 22, 2010 #4
    Ok, thank you both. In retrospect, that was kind of obvious :P
     
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