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Dirac delta integral

  1. Jan 27, 2013 #1
    1. The problem statement, all variables and given/known data

    compute the integral:

    [tex]\int_ {-\infty}^\infty \mathrm{e}^{ikx}\delta(k^2x^2-1)\,\mathrm{d}x[/tex]


    2. Relevant equations

    none that I have

    3. The attempt at a solution

    I don't actually have any work by hand done for this because this is more complex than any dirac delta integral I have done...I assume that we must break down the x^2 in the delta function because there would be 2 roots, and thus, 2 answers? I'm not totally sure how to start this problem and can't find anything on the internet like it.

    I know the delta function (from finding the roots of x) occurs at [tex]x=-1/k[/tex] and [tex]x=1/k[/tex]. But not sure where to go from there
     
  2. jcsd
  3. Jan 27, 2013 #2

    Dick

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    Here's something on the internet that's relevant. http://en.wikipedia.org/wiki/Dirac_delta_function Skip down to the "Composition with a function" section.
     
  4. Jan 27, 2013 #3

    SammyS

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    That's very cool (in my estimation).

    Direct link to the section Dick mentioned: LINK .
     
  5. Jan 27, 2013 #4
    I ended up getting cos(k) as the final answer. Is that right?
     
  6. Jan 27, 2013 #5

    Dick

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    How did you get that? It's not what I get.
     
  7. Jan 27, 2013 #6
    thank you!

    my last step before the final answer was [tex]\frac{1}{2}(e^{ik}+e^{-ik}) = cos(k)[/tex]
     
  8. Jan 27, 2013 #7

    Dick

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    You're welcome. But the cool reference I pointed you to says the delta function is the same as the sum of δ(1/k) and δ(-1/k) each divided by |g'| evaluated at those points. What happened to all of that stuff?
     
  9. Jan 28, 2013 #8
    i started out with:

    [tex]\delta[(x+1)(x-1)] = \frac{1}{2}[\delta(x+1)+\delta(x-1)][/tex]
    [tex]\frac{1}{2} \int_{-\infty}^\infty e^{ikx}\delta(x+1)\,dx + \frac{1}{2} \int_{-\infty}^\infty e^{ikx} \delta(x-1)\,dx [/tex]

    from there i boiled it down to cos(k)

    did I mess up in the beginning?

    I just read your last post..sorry i was in the middle of typing this while you posted last. So i found this in the composition section function however as you point out it looks like I started off wrong and need the include the "k"
     
  10. Jan 28, 2013 #9
    [tex]\frac{1}{2} \int_{-\infty}^\infty e^{ikx}\delta(x+\frac{1}{k})\,dx + \frac{1}{2} \int_{-\infty}^\infty e^{ikx} \delta(x-\frac{1}{k})\,dx [/tex]

    that should be how I solve it using what you said, correct?

    and the final solution i get is [tex] \frac{1}{2}[e^{i}+e^{-i}][/tex]
     
  11. Jan 28, 2013 #10

    Dick

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    Yes, if you had δ(x^2-1) that would be fine. But you have δ(k^2*x^2-1). Makes a difference. You had the right general idea originally that the delta function should involve x=1/k and x=(-1/k). You just need to correct that a little.
     
  12. Jan 28, 2013 #11

    Dick

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    You're getting closer. Look at the |g'(x)| factor you are supposed to divide those delta functions by. g(x)=k^2*x^2-1.
     
  13. Jan 28, 2013 #12
    oooohhhhh I gotcha, many, many thanks for the clarification
     
  14. Jan 28, 2013 #13
    with your correction I got

    [tex] \frac{1}{2xk^2}[e^i+e^{-i}][/tex]
     
    Last edited: Jan 28, 2013
  15. Jan 28, 2013 #14

    Dick

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    Closer yet. But if g(x)=k^2*x^2-1, g'(x)=2*k^2*x. Putting x=1/k I get |g'(1/k)|=|2k|, not 2k^2.
     
  16. Jan 28, 2013 #15
    Oh i see how you did that plugging in 1/k for x, I didnt think to do that
     
  17. Jan 28, 2013 #16

    Dick

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    Right. There shouldn't be any x in the final answer. You should put x=1/k,-1/k into g'(x) and the absolute value is pretty important.
     
  18. Jan 28, 2013 #17
    [tex] \frac{1}{2k}[e^i+e^{-i}][/tex]

    finally all good :D ?
     
  19. Jan 28, 2013 #18
    [tex] \frac{1}{2k}[e^i+e^{-i}][/tex]

    finally all good :D ?

    sorry double posted by accident
     
  20. Jan 28, 2013 #19

    Dick

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    Absolute value. Absolute value. The bars around |g'| mean something. And you can simplify e^i+e^(-i) just like you did before.
     
  21. Jan 28, 2013 #20
    many thanks for the help and putting up with my careless errors at midnight :D
     
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