Given $f_a(x) =\frac{a}{\pi(x^2+a^2)} $ where a > 0 calculate $\mathscr{F}(e^{-a|x|}) $ and calculate $\mathscr{F}(f_a) $ knowing that $\mathscr{F}(\mathscr{F}(f))(x) = 2\pi f(-x) $.
What I did so far:
I managed to compute $\mathscr{F}(e^{-a|x|}) $ and got $\mathscr{F}(e^{-a|x|}) $ = $ \frac{2a}{a+\xi^2}$ , now I have no idea how to calculate $\mathscr{F}(f_a) $ using the given property and the later result.
My instinct is to use the definition of Fourier Transform, but I can't see where is the opportunity to use the property and the first result.
You have
\begin{align} \mathcal{F}(e^{-a | x |}) = \frac{2a}{a^{2} + \xi^{2}} \implies \frac{1}{2 \pi} \mathcal{F}(e^{-a | x |}) &= \frac{a}{\pi(a^{2} + \xi^{2})} = f_{a}(\xi) \tag1 \end{align}
and so applying the Fourier transform to both sides and using $\mathcal{F}(\mathcal{F}(f))(x) = 2 \pi f(-x)$, we find
\begin{align} \mathcal{F}(f_{a}(x)) &= \mathcal{F} \left( \frac{1}{2 \pi} \mathcal{F}(e^{-a | \xi |}) \right) \\ &= \frac{1}{2 \pi} \mathcal{F}(\mathcal{F}(e^{-a | \xi |})) \qquad \qquad \text{(by linearity of the integral)} \\ &= \frac{1}{2 \pi} \cdot 2 \pi e^{-a |- \xi |} \qquad \qquad \ \ \text{(using the given property)} \\ &= e^{-a | \xi |} \end{align}