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 and inverse transform

  1. Oct 30, 2007 #1
    1. The problem statement, all variables and given/known data

    Let f(x) be an integrable complex-valued function on [tex]\mathbb{R}[/tex]. We define the Fourier transform [tex]\phi=\mathcal{F}f[/tex] by
    [tex]\[\phi(t)=\int_{\infty}^{\infty} e^{ixt} f(x) dx.\][/tex]

    Show that if [tex]f[/tex] is continuous and if [tex]$\phi$[/tex] is integrable, then
    [tex]\[f(x)=\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-ixt} \phi(t) dt.\][/tex]


    3. The attempt at a solution

    So far I have
    [tex]\[\int_{-\infty}^{\infty} e^{-ixt} \phi(t) dt = \lim_{a \to \infty} \int_{-a}^{a} e^{-ixt} \phi(t) dt = \lim_{b \to \infty} \frac{1}{b} \int_0^b da \int_{-a}^a e^{-ixt} \phi(t) dt = \lim_{b \to \infty} \frac{1}{b} \int_0^b da \int_{-\infty}^\infty \frac{2 \sin((y-x)a)}{y-x} f(y) dy.\][/tex]

    What to do next?
     
    Last edited: Oct 30, 2007
  2. jcsd
  3. Oct 31, 2007 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    It's not a matter of "what to do next?" but "Why in the world are you doing that?". For one thing, there is no integral from 0 to [itex]\infty[/itex] in what you had before, where are you getting that? And even when you have two integrals, you still only have one variable: dt. You want to show that, for any continuous f (in which case [itex]\phi[/itex] must be integrable)
    [tex]\frac{1}{2\pi}\int_{-\infty}^\infty e^{-ixt} \int_{-\infty}^\infty e^{iut}f(u)dudt= f(x)[/tex]
     
  4. Nov 17, 2009 #3
    What your coarse probably wants from you is to use the following,

    [tex]\delta(t - a) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i \omega (t -a) } d\omega. [/tex]
    If you want a full but rather simple proof, for a graduate student with a mathematical focus, of this identity I can send it to you.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Fourier transform and inverse transform
Loading...