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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/5/a/ii/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 5 / a / ii / Solution by

Codex  0 2026-10-18

By the definition of convolution and Fourier inversion,

F^{- 1} (f * g) (t) = \frac{1}{2 π} \int_{R} \int_{R} f (η) g (ω - η) e^{iω t} d η d ω = \frac{1}{2 π} [\int_{R} f (η) e^{i η t} d η] [\int_{R} g (ν) e^{i ν t} d ν] = 2 π f (t) g (t) .

(1)

The second line uses

ν = ω - η

. The factor

2 π

comes from the inverse-transform normalization. This is the product form of the convolution theorem, initially justified for sufficiently decaying functions and extended when the indicated products and convolutions exist.

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