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Properties of the Delta Function

  1. Jul 8, 2014 #1
    1. The problem statement, all variables and given/known data

    Delta functions said to live under the integral signs, and two expressions (##D_1(x)## and ##D_2(x)##) involving delta functions are said to be equal if:

    ##\int _{ -\infty }^{ \infty }{ f(x)D_{ 1 }(x)dx } =\int _{ -\infty }^{ \infty }{ f(x)D_{ 2 }(x)dx }##

    (a) Show that:

    ##\delta (cx)=\frac{ 1 }{ |c| } \delta (x)##

    Where ##c## is a real constant. (Be sure to check the case where ##c## is negative.)

    2. Relevant equations

    Posted above.

    3. The attempt at a solution

    Let ##u=cx## ##\therefore## ##\frac{1}{c}du=dx##

    This yields:

    ##\frac{1}{c}\int_{-\infty}^{\infty}{f(u/c)\delta (u)du}=\frac{1}{|c|}\int_{-\infty}^{\infty}{f(x)\delta (x)dx}##. This works for a test case where ##c > 0##, but obviously fails when ## c < 0##. I am not sure where this absolute value came from.

    Hints please.

    Thanks,
    Chris
     
  2. jcsd
  3. Jul 8, 2014 #2

    TSny

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    Don't forget to think about the limits of integration when making a change of integration variable.
     
  4. Jul 8, 2014 #3
    Isn't still negative infinity to infinity. Oh wait, not of the sign changes, so the sign flips. I think I got it now.

    Thanks,
    Chris
     
  5. Jul 8, 2014 #4

    ChrisVer

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    Gold Member

    you can choose to try for [itex]f(x)=1[/itex]
     
  6. Jul 8, 2014 #5
    I used f(x)=x+2, and it works if I consider that using ##-c## would flip the signs of the infinities on the integral thus flipping the sign back to positive.

    Chris
     
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