The Fourier transform of a function $f(x)$ is given by
$$ \mathcal{F}(f)(w) = \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^{\infty} f(x) e^{-i w x} dx $$
Then, there is a problem that asks to prove that if $f$ is differentiable and the integrals for $\mathcal{F}(f)(w)$ and $\mathcal{F}(f')(w) $ converge, then
$$ \mathcal{F}(f')(w) = \frac{1}{i w} \mathcal{F}(f)(w) $$
But I get that $\mathcal{F}(f')(w) = iw \mathcal{F}(f)(w) $. Is this a typo?
It is a typo. Dividing by $iw$ corresponds to the Fourier transform of $f(x)$ integrated. For example, in control engineering an integrator block is labeled $1/s$ where $s$ is the complex frequency variable of the Laplace transform, which becomes the Fourier transform for $s = iw$. Is it a first edition book?
https://en.wikipedia.org/wiki/Fourier_transform#Functional_relationships
https://en.wikipedia.org/wiki/Laplace_transform#Properties_and_theorems
