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Derivatives of fourier transform

  1. Mar 24, 2009 #1
    Can anyone explain me how to prove the following identity?

    [tex]\frac{\partial \hat{f}}{\partial x}(0,0) = \int \int x^2f(x,y)dxdy[/tex]

    where [tex]\hat{f}[/tex] denotes the Fourier Transform of f(x,y) ?
     
  2. jcsd
  3. Mar 25, 2009 #2
    Well,
    I figured it out by myself. It was actually not hard, but unfortunately I noticed that in my first post the statement of the problem was wrong. I will show the solution for one dimension: the 2-dimensional case is trivial.

    [tex]\frac{d}{d\omega}F\{f\} = \int_{-\infty}^{\infty}\frac{\partial}{\partial \omega}f(x)e^{-i\omega x}dx = -i\int_{-\infty}^{\infty}xf(x)e^{-i\omega x}dx[/tex]

    The first step was Leibniz integral rule; now, iterating this process n-times yields:

    [tex]\frac{d^n}{d\omega^n}F\{f\} = (-i)^{n}F\{x^{n}f\}[/tex]

    We have just to set [tex]n=2[/tex] and [tex]\omega=0[/tex] in order to obtain:

    [tex]\frac{d^2}{d\omega^2}F\{f\} = -\int_{-\infty}^{\infty}x^{2}f(x)dx[/tex]

    That was it: the second derivative (with opposite sign) of the Fourier transform of f at 0, is equivalent to the second moment of f.
     
    Last edited: Mar 25, 2009
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