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Tricky integration (maybe delta function)

  1. Mar 8, 2016 #1
    1. The problem statement, all variables and given/known data
    I need to integrate

    ##\frac{A}{2a\sqrt{2\pi}} \int_{-\infty}^{\infty} \frac{e^{ik(x+x')}}{(b^2+k^2)}dk##

    I have tried substitution and integration by parts and that hasn't worked. I can see that part of it is the delta function, but I don't really know how to use that fact! I think the delta function is

    ##\delta(x+x') = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{ik(x+x')}dk##

    Which is almost exactly what I have. If it helps, this is from trying to reverse Fourier transform a function. How can I integrate this?

    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Mar 8, 2016 #2

    vela

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    Try a contour integration in the complex plane.
     
  4. Mar 8, 2016 #3
    That sounds well beyond anything we've been taught! I've never heard of contour integration, in the complex plane or otherwise... Is that the only method?
     
  5. Mar 8, 2016 #4
    Maybe I have the integral wrong, let me double check.
     
  6. Mar 8, 2016 #5

    Ray Vickson

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    Let ##x+x'= y##, so you want to evaluate
    [tex] I(y) = \int_{-\infty}^{\infty} \frac{\exp(iky)}{k^2+b^2} \, dk. [/tex]
    Write ##\exp(iky) = \cos(ky) + i \sin(ky)##, and note that in the integral the denominator is even in ##k##, while ##\sin(ky)## is odd in ##k##. Therefore, without any need for computation we conclude that the imaginary part = 0 because the "##\sin##" term integrates to zero for any fixed ##y##. Thus, we have
    [tex] I(y) = 2 \int_0^{\infty} \frac{\cos(ky)}{k^2+b^2} \, dk.[/tex]

    Assuming we can legitimately differentiate wrt ##y## under the integral sign, we have
    [tex] \begin{array}{rcl}\frac{d^2}{dy^2} I(y) &= &2 \int_0^{\infty} \frac{-k^2 \cos(ky)}{k^2+b^2} \, dk \\
    &=& -2 \int_0^{\infty} \frac{(k^2 + b^2 - b^2)\cos(ky)}{k^2 + b^2} \\
    &=& - 2 \int_0^{\infty} \cos(ky) \, dy + 2 b^2 \int_0^{\infty} \frac{\cos(ky)}{k^2+b^2} \, dk \\
    &=& - 2 \int_0^{\infty} \cos(ky) \, dy + b^2 I(y)
    \end{array}
    [/tex]
    Thus, you have a DE ##I''(y) - b^2 I(y) = f(y)##, where ##f(y) = -2 \int_0^{\infty} \cos(ky) \, dk##. Do you recognize ##f(y)## from somewhere?

    Of course, to get the full ##I(y)## you need two boundary conditions on ##I(y)##, such as values of ##I(0)## and ##I'(0)##, and both of these can be obtained from results you have seen already (or can derive on your own).

    There remains a serious issue, however: we just carelessly differentiated under the integral sign and further split the resulting integral into two separate integrations. Those manipulations need justification, but for now I will leave it.
     
  7. Mar 8, 2016 #6

    vela

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    Are you allowed to just look the transform in a table?
     
  8. Mar 9, 2016 #7
    Brilliant, thank you!
     
  9. Mar 9, 2016 #8
    Maybe! I don't think we've been taught the right techniques to actually evaluate it. I'll ask my lecturer about the tables. Thanks for helping! :)
     
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