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Short options course of the University of Oxford

Past exam of the short options course of the University of Oxford

2017

With this convention, the <Fourier inversion theorem> states that if $f,\widehat f\in L^1(\mathbb R)$, then
$$
f(x)=\frac1{2\pi}\int_{\mathbb R}\widehat f(s)e^{isx}\,ds
$$
almost everywhere and at every continuity point of $f$. Apply this to the continuous function in part (ii):
$$
e^{-a|x|}=\frac a\pi\int_{\mathbb R}\frac{e^{isx}}{a^2+s^2}\,ds.
$$
The denominator is even. Replacing $x$ by $-t$ and renaming the integration variable gives
$$
\boxed{\mathcal F\!\left(\frac1{a^2+x^2}\right)(t)=\frac\pi a e^{-a|t|}.}
$$
Both integrable members of this transform pair satisfy the inversion hypotheses.

Solution

Past exam of the short options course of the University of Oxford / 2017 / 5 / a / iii / Solution

Past exam of the short options course of the University of Oxford / 2017 / 5 / a / iii / Solution

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