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Homework Help: 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


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    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.
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