Using the Duality of the Fourier Transform

Question

How would I use the duality of the Fourier transform to find the inverse Fourier transform of

$$ X(\omega) = 2\pi e^{a\omega} u(-\omega)? $$

Answer 1

Hints.

1. $F^2(g)(x)=g(-x)$
2. $F^{-1}(q)(x)$, with $q(\omega):=e^{a\omega}g(\omega)$ is equal to $$F^{-1}(q)(x)=\frac{1}{\sqrt{2\pi}}\int e^{a\omega}g(\omega) e^{i\omega x}d\omega=F^{-1}(g)(x-ia). $$

Then

$$F^{-1}(X)(x)=2\pi F^{-1}(r)(x), $$

where the function $r$ is given by $r(\omega):=e^{a\omega}F^2(u)(\omega)$; then we can write

$$F^{-1}(X)(x)=2\pi F^{-1}(F^2(u))(x-ia)=2\pi F(u)(x-ia). $$