Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / 4 / ii

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- Solution ii

Solution

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) .

Hence the Fourier transform of a polynomial multiple is

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.

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