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Homework Help: Definite integral evaluation (Inverse Fourier Sine Transform)

  1. May 19, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm working on a long problem and have come to the final step. The answer seems so simple, but I can't quite get to it. I need to evaluate this integral:

    [tex]\int_0^{\infty}\ \left(e^{-k^2 t}/k\right)\sin(kx)\ dk[/tex]

    2. Relevant equations

    Mathematica gives the result as [itex]\frac{\pi}{2}Erf[x/2\sqrt{t}][/itex].

    3. The attempt at a solution

    The answer makes me think that there is some kind of convolution theorem at work here (even though there isn't a direct one for the Sine Transform), since [itex]\int_0^{\infty}\ \sin(kx)/k\ dk = \pi/2[/itex]. I can't for the life of me get that error function to come out, though. I've tried playing some with the convolution theorem dealing with the Cosine Transform, but that doesn't seem to get me anywhere. I tried converting to exponential form, since that seems like the logical way to get at the error function, but I'm left with the following integral:

    [tex]\Im\left[\int_0^{\infty}\ e^{-t(k-(ix/2t))^2}/k\ dk\right][/tex]

    The trouble is, if you do the typical thing of making a u substitution, you can't get rid of the k in the denominator, and I don't know how you'd separate the imaginary component. Any tips? Thank you so much!
  2. jcsd
  3. May 20, 2012 #2
    I think that last integral may be the right way to do it, since

    [tex]\int_0^{\infty}\ e^{-k}/\sqrt{k}\ dk \propto Erf(\sqrt{k})[/tex]
  4. May 20, 2012 #3


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    I haven't tried working the problem out completely, so this is just a suggestion. Let
    $$f(x,t) = \int_0^\infty \frac{e^{-k^2 t}}{k}\sin kx\,dk.$$ If you differentiate with respect to x, you'll get
    $$\frac{\partial f}{\partial x} = \int_0^\infty e^{-k^2 t}\cos kx\,dk,$$ which doesn't have the pesky k in the denominator. Perhaps it would be easier to solve for ##\partial_x f## and integrate the result.
  5. May 21, 2012 #4
    That worked beautifully, vela. Thank you!
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