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Fourier Transform / Integral d

  1. Sep 22, 2005 #1

    I got an exam in calculus in a few weeks, and lots of questions coming up. Here's one of them:

    We learned that the Fourier Transform of

    [tex] f(x) = e^{-|x|} [/tex]
    [tex] \hat f(\omega) = \sqrt{\frac{2}{\pi}}\frac{1}{1+\omega^2}[/tex]

    I've got no problem with this one. Now, since [tex] \hat f(\omega)[/tex] is Lebesgue - integrable, the inverse Fourier transform exists and should be

    [tex] \hat \hat f(-x) = e^{-|x|} [/tex]

    To show this the 'hard way', I want to calculate the integral

    [tex] \sqrt{\frac{1}{2 \pi}} \int \limits_{-\infty}^\infty \frac{1}{1+t^2} e^{i \omega t} \mathrm{d} t [/tex]

    Well, I just don't know how to do this one. Partial Integration doesn't seem to work, and I can't find neither a good substition nor a clever use of Fubini's Theorem.

    Would be thankful for any help.
  2. jcsd
  3. Sep 22, 2005 #2


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    I have never seen it derived directly. However, it is easily found in a table of definite integrals.
  4. Sep 22, 2005 #3

    George Jones

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    One way to do this is by using complex contour integration and the residue theorem.


    [tex] \int \frac{1}{1+z^2} e^{i \omega z} \mathrm{d} t [/tex]

    along the closed contour that consists of the portion of the real axis from [itex]-R[/itex] to [itex]R[/itex] together with a closing counterclockwise semicircle of radius [itex]R[/itex] in the upper halfplane. Let [itex]R\rightarrow \infty[/itex]. Then:the closed contour encloses a pole at [itex]z=i[/itex]; the contibution from the semicircle goes to zero.

    Repeat for a clockwise semicircle in the lower halfplane.

    Last edited: Sep 22, 2005
  5. Sep 22, 2005 #4
    As I was just told, it seems to be done by using the residue theorem of complex analysis.

    edit: George was faster, thanks! Working it out...
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