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It is said that: $$F[\frac{df(x)}{dx}] = i\omega F(\omega)$$. This expression depends on the initial definition of Fourier transform, yes? if I define Fourier transform as: $$F(\omega)=\frac{1}{\sqrt{2\pi}}\int f(x)e^{i\omega x}dx$$ Then tha above expression will be $$F[\frac{df(x)}{dx}]=-i\omega F(\omega)$$ right?????

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  • $\begingroup$ And there are other similar definitions, all of which change Fourier transform results slightly. See the list here, for example, and notice the different columns, for various FT definitions. $\endgroup$ – Chester May 15 '16 at 11:08
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$$f(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F(\omega) \, \mathrm e^{-i \omega x} \,\mathrm d\omega$$ so we have $$f'(x) = \frac{\mathrm d}{\mathrm dx}\!\left( \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F(\omega) \,\mathrm e^{-i \omega x} \,\mathrm d\omega \right)= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} (-i \omega) \, F(\omega) \,\mathrm e^{-i \omega x} \,\mathrm d\omega$$

Hence the Fourier transform of $f'(x)$ is $\mathcal F\{f'(x)\}= -i \omega \, F(\omega)$.

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  • $\begingroup$ That means my comment about the initial definition is right yes?? $\endgroup$ – Mr. an May 13 '16 at 19:02
  • $\begingroup$ yes, it is right. $\endgroup$ – alexjo May 13 '16 at 19:22

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