1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fourier Transform / Integral

  1. Sep 22, 2005 #1
    Hi,

    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]
    is
    [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

    mathman

    User Avatar
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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

    Integrate

    [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.

    Regards,
    George
     
    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...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Fourier Transform / Integral
  1. Fourier transformation (Replies: 1)

  2. Fourier transforms (Replies: 5)

  3. Fourier transform (Replies: 1)

  4. Fourier transform (Replies: 0)

Loading...