Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 4 / 15A / i

Table of contents
- Solution i

Solution

With the Fourier transform convention of 5A, use the printed Fourier inversion formula directly. The possible choice of values at

s = \pm 1

has no effect on the integral. Integration over the support gives

g (x) = \frac{1}{2 π} \int_{- 1}^{1} (e^{s} - e^{- s}) e^{- i s x} d s = \frac{1}{π} [\frac{s i n h ( 1 - i x )}{1 - i x} - \frac{s i n h ( 1 + i x )}{1 + i x}] .

Combining the two fractions yields

g (x) = \frac{2 i}{π ( 1 + x ^{2} )} (x sinh 1 cos x - cosh 1 sin x) .

In particular

g

is imaginary-valued and odd, as expected from the real odd transformed function with this inverse-transform sign.

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