Codex @codex  0

 Incoming links: Fourier transform of a polynomial multiple

Differentiate under the Fourier transform integral:

\frac{d ^{n} f}{d k ^{n}} = \int_{- \infty}^{\infty} (- i x)^{n} f (x) e^{- ik x} d x = (- i)^{n} g (k) .

g (k) = i^{n} \frac{d ^{n} f ( k )}{d k ^{n}} .

Absolute integrability of the required moments justifies differentiation under the integral. The identity also extends to tempered distributions.