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Fourier Transfrom problem

  1. Dec 31, 2009 #1
    1. The problem statement, all variables and given/known data

    I've been given the following function:

    [tex]g(x) = \frac{\gamma}{2}e^{-\gamma \left|x\right|}[/tex] with [tex]\gamma>0[/tex]

    First thing I needed to do was to prove [tex]\int^{-\infty}_{\infty}g(x)dx=1[/tex] which was simple enough.

    I hit a problem when trying to find the Fourier transform [tex]\widetilde{g}(k)[/tex] of [tex]g(x)[/tex]

    I'm asked to show the transform is of the form


    and find a and s

    2. Relevant equations

    [tex]\widetilde{g}(k) = \frac{1}{\sqrt{2\pi}}\int^{-\infty}_{\infty}e^{-ikx}g(x)dx[/tex]

    3. The attempt at a solution

    I try to perform integration by parts on

    [tex]\widetilde{g}(k) = \frac{1}{\sqrt{2\pi}}\int^{-\infty}_{\infty}e^{-ikx}\frac{\gamma}{2}e^{-\gamma \left|x\right|}dx[/tex]

    but whenever I try this I end up with uv equalling vdu so the whole expression turns out as zero. I think this is probably because I don't really know what I'm doing with the integration with those limits to infinity. I can split up g(x) because its an even function but not the [tex]e^{-ikx}[/tex] part.

    I've plugged this into wolfram alpha and it churned out [tex]a = \frac{1}{2\pi}[/tex] and [tex]s = \gamma[/tex], I'm just not sure how to get there.

    Any help appreciated!
  2. jcsd
  3. Dec 31, 2009 #2
    Divide integral into two parts to get rid of the abs(x) and combine the exponentials in the integrand. That way the integral is trivial. After some algebra you should get the desired result. Careful with the signs.
  4. Jan 2, 2010 #3
    Aha thanks! Somehow I totally missed being able to combine the two powers of e, I guess I was staring at the problem for too long.
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