Derivatives of fourier transform

  • Thread starter mnb96
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  • #1
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Main Question or Discussion Point

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) ?
 

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  • #2
710
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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:

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