Fourier Transform of Derivative

Consider a function f(t) with Fourier Transform F(s). So F(s) = \int_{-\infty}^{\infty} e^{-2 \pi i s t} f(t) \ dt

What is the Fourier Transform of f'(t)? Call it G(s).So G(s) = \int_{-\infty}^{\infty} e^{-2 \pi i s t} f'(t) \ dt

Would we consider \frac{d}{ds} F(s) and try and write G(s) in terms of F(s)?

Answer

A simpler way, using the anti-transform:

f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) \, e^{i \omega t} d\omega

f'(t) = \frac{d}{dt}\!\left( \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) \, e^{i \omega t} d\omega \right)= \frac{1}{2\pi} \int_{-\infty}^{\infty} i \omega \, F(\omega) \, e^{i \omega t} d\omega

Hence the Fourier transform of f'(t) is i \omega \, F(\omega)

Attribution
Source : Link , Question Author : NebulousReveal , Answer Author : leonbloy

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