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Question on Dirac Delta function

  1. Aug 6, 2013 #1
    I want to proof (1)##\delta(x)=\delta(-x)## and (2) ## \delta(kx)=\frac{1}{|k|}\delta(x)##

    (1) let ##u=-x\Rightarrow\;du=-dx##
    [tex]\int_{-\infty}^{\infty}f(x)\delta(x)dx=(0)[/tex]
    but
    [tex]\int_{-\infty}^{\infty}f(x)\delta(-x)dx=-\int_{-\infty}^{\infty}f(-u)\delta(u)du=-f(0)[/tex]

    I cannot proof (1) is equal as I don't know if ##f(u)=-f(-u)##. Please help.




    (2)Let ##u=kx\Rightarrow\;\frac{du}{k}=dx##. The sign of the limit reverse depends on the polarity of k.
    [tex]\int_{-\infty}^{\infty}f(x)\delta(kx)dx=\frac{1}{k}\int_{-\infty}^{\infty}f(\frac{u}{k})\delta(u)du=^+_-\frac{1}{k}f(0)[/tex]
    Since sign is +ve if k is positive. Sign is -ve if k is -ve. This means the result is always ##\frac{1}{|k|}f(0)##. Am I correct?

    Thanks
     
  2. jcsd
  3. Aug 6, 2013 #2

    micromass

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    The bounds of the integral change too, so you would have

    [tex]\int_{-\infty}^{\infty}f(x)\delta(-x)dx=-\int_{\infty}^{-\infty}f(-u)\delta(u)du[/tex]

    intead of what you had. Does that change anything?
     
  4. Aug 6, 2013 #3
    Got it. How about (2)

    Thanks
     
  5. Aug 6, 2013 #4

    micromass

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    You should switch the bounds in (2) too, if ##k## is negative. Aside from that, it's alright.
     
  6. Aug 7, 2013 #5
    Thanks
     
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