 Solution

ID: past-exam-of-the-short-options-course-of-the-university-of-oxford/2017/5/a/iii/solution

With this convention, the Fourier inversion theorem states that if

f, f \in L^{1} (R)

, then

f (x) = \frac{1}{2 π} \int_{R} f (s) e^{i s x} d s

f

. Apply this to the continuous function in part (ii):

e^{- a ∣ x ∣} = \frac{a}{π} \int_{R} \frac{e ^{i s x}}{a ^{2} + s ^{2}} d s .

The denominator is even. Replacing

x

- t

and renaming the integration variable gives

F (\frac{1}{a ^{2} + x ^{2}}) (t) = \frac{π}{a} e^{- a ∣ t ∣} .

Both integrable members of this transform pair satisfy the inversion hypotheses.

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