after a decomposition of the fraction we get : $$\mathcal{F}\left\{\frac 1 {x^2+a^2} \right\} = \frac{i}{2a} [\Xi_{➷ }(\xi) -\Xi^{➹}(\xi)]$$
where : $$\Xi^{➹}(\xi) = \int_{-\infty}^{+\infty} \frac{e^{-i\xi x}}{x -ia}\,dx $$
$$\Xi_{➷ }(\xi)= \int_{-\infty}^{+\infty} \frac{e^{-i\xi x}}{x +ia}\,dx $$
let $\Gamma_R = \gamma_R \cup[-R,R]$
then $$\lim_{R \to +\infty} \oint_{\Gamma_R} \frac{e^{-i \xi z}}{z-ia}\,dz = \Xi^{➹}(\xi)+ \lim_{R \to +\infty} \oint_{\gamma_R} \frac{e^{-i \xi z}}{z-ia} \,dz$$
on the one hand we have : $$\lim_{R \to +\infty} \oint_{\Gamma_R } \frac{e^{-i \xi z}}{z-ia}\,dz = 2\pi i \star \operatorname*{\textbf{Res}}_{ia} = 2\pi ie^{a \xi }$$
on $\gamma_R$ we have $z = Re^{i\theta},\, \theta \in [0,\pi]$
using Jensen's inequality $$\left| \oint_{\gamma_R} \frac{e^{-i \xi z}}{z-ia}\,dz\right| \leq \oint_{\gamma_R } \frac{|e^{-i \xi z}|}{|z-ia|}\,dz = \frac{1}{R}\oint_{\gamma_R } \frac{dz}{\sqrt{(\cos^2 \theta + (\sin \theta - a)^2}} \to 0 \text{ as } R \to \infty \text{, right ?}$$
so $$\Xi^{➹}(\xi) = 2\pi ie^{a \xi }$$
for the other one I just need to swap the contour upside down.
but I'm fairly certain I'm wrong when I concluded that the integral tends to $0$ because even though the $R$ was out it still somehow depended on it.
my main goal is to find the above Fourier transform, is there another way to do it ? if not how do you deal with $\lim_{R \to +\infty} \oint_{\gamma_R } \frac{e^{-i \xi z}}{z-ia}\,dz$
Edit : I just realized that : $$ \frac{1}{R}\oint_{\gamma_R } \frac{dz}{\sqrt{(\cos^2 \theta + (\sin \theta - a)^2}} = \int_{0}^{\pi} \frac{ie^{i\theta}\,d\theta}{\sqrt{(\cos^2 \theta + (\sin \theta - a)^2}}$$
so my conclusion is wrong.