TITLE: Fourier Transform of Derivative
QUESTION [67 upvotes]: 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 $F(s)$?

REPLY [112 votes]: 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)$