This question is similar to Convolution of $e^{-|x|} \ast e^{-|x|}$ but I'm looking to apply Fourier transform. Defining $f(x)=e^{-a|x|}$ with $a>0$, I know that $$\hat{f}(\xi) = \frac{2a}{a^2 + 4\pi^2\xi^2}.$$
I can apply the property $\widehat{(f\ast f)} = \hat{f}.\hat{f} = (\hat{f})^2$. As such, using inverse Fourier transform, I can deduce that
$$e^{-a|x|} \ast e^{-a|x|} = \int_{\mathbb{R}} (\frac{2a}{a^2 + 4\pi^2\xi^2})^2 e^{2i\pi x\xi} d\xi$$
Luckily, I covered a previous question where I established that the Fourier transform of $x\mapsto \frac{1}{1+x^2}$ is $\xi\mapsto\pi e^{-|\xi|}$, but it's still unclear to me how to apply this fact to the integral I have above.
Let $I(t,x)$ be given by the integral
$$\begin{align} I(t,x)&=\frac1{2\pi}\int_{-\infty}^\infty \frac1{t+k^2}e^{ikx }\,dk\\\\ &=\frac{e^{-\sqrt{t}|x|}}{2\sqrt{t}} \end{align}$$
Differentiating with respect to $t$ and setting $t=a^2$ reveals
$$\begin{align} \left.\left(\frac{\partial I(t,x)}{\partial t}\right)\right|_{t=a^2}&=-\frac1{2\pi}\int_{-\infty}^\infty \frac{1}{(a^2+k^2)^2}e^{ikx}\,dk\\\\ &=-\frac{1+|ax|}{4|a|^3}\,e^{-|ax|} \end{align}$$
Multiplying by $-4a^2$, we find that
$$\int_{-\infty}^\infty \left(\frac{2a}{a^2+4\pi \xi^2}\right)^2e^{i2\pi \xi x}\,d\xi=\frac{1+|ax|}{|a|}\,e^{-|ax|}$$