Let $a>0$ and define
$$g(\xi):=\frac{\sin a\xi}{\xi(1+\xi^{2})}$$
I want to prove that $g\in\mathscr{S}'(\mathbb{R})$ and consquently that $g\in L^{1}(\mathbb{R})$ (but this implication is trivial due to the density of $L^{1}$ in $\mathscr{S}'$).
Now, a distribution $u$ is in $\mathscr{S}'$ if and only if there exists a constant $C>0$ such that
$$|u(\phi)|\le c\|\phi\|_{\mathscr{S}(k,N)}:=C\sup_{|\alpha|\le k,|\beta|\le N,x\in\mathbb{R}}|x^{\beta}\partial^{\alpha}\phi(x)|\qquad(\phi\in C^{\infty}_{0}(\mathbb{R}))$$
If $g$ is locally integrable then $\text{test }g$ defines a distribution. But this is a circular argument as $g$ locally integrable implies that $g\in L^{p}$, so I'm not sure that I can proceed with this argument.
I don't know if it helps, but we also have that
$$\frac{1}{4}\mathscr{F}(1_{[-a,a]})\mathscr{F}(e^{-|\cdot|})(\xi)=\frac{\sin a\xi}{\xi(1+\xi^{2})}=:g(\xi).$$
The trick is explained here at the end of page 1 :
If $f$ is continuous and bounded by $C$, with $$A = \sup_x |(1+x^2) \phi(x)|$$ you get $$\sup_x |(1+x^2) \phi(x)f(x)| \le A C$$ i.e. $$|\phi(x)f(x)| \le \frac{AC}{1+x^2}$$ and $\displaystyle F(\phi) = \int_{-\infty}^\infty f(x) \phi(x) dx$ is a tempered distribution since $$\left|\int_{-\infty}^\infty f(x) \phi(x) dx\right| \le A C \int_{-\infty}^\infty \frac{1}{1+x^2}dx = \pi A C$$