= Solution
For integrable $f,g$, define their <convolution> by $(f*g)(x)=\int_{\mathbb R}f(y)g(x-y)dy$. Absolute integrability of the double integral permits <Fubini's theorem>. With $s=x-y$,
$$
\begin{aligned}
\mathcal F[f*g](k)&=\int_{\mathbb R}\!\int_{\mathbb R}e^{-ikx}f(y)g(x-y)dy\,dx\\
&=\int_{\mathbb R}f(y)e^{-iky}dy\int_{\mathbb R}g(s)e^{-iks}ds=\widetilde f(k)\widetilde g(k).
\end{aligned}
$$
The <convolution theorem>, together with uniqueness of the <Fourier transform>, therefore identifies the inverse transform of that product with $f*g$. With the stated convention the inverse carries $1/(2\pi)$, so there is no extra factor in the convolution itself.
Now take $f=e^{-|x|}$. For $x\geq0$, split the convolution at $y=0$ and $y=x$:
$$
\begin{aligned}
h(x)&=\int_{-\infty}^0e^{2y-x}dy+\int_0^xe^{-x}dy+\int_x^\infty e^{x-2y}dy\\
&=\frac12e^{-x}+xe^{-x}+\frac12e^{-x}=(1+x)e^{-x}.
\end{aligned}
$$
The convolution of two even functions is even, so the <self-convolution of a bilateral exponential> gives
$$
\boxed{h(x)=(1+|x|)e^{-|x|}=\begin{cases}(1-x)e^x,&x<0,\\1,&x=0,\\(1+x)e^{-x},&x>0.\end{cases}}
$$
In particular the centre value is one; both pieces extend continuously there.
Splitting the Fourier integral at zero gives
$$
\widetilde f(k)=\int_0^\infty e^{-(1+ik)x}dx+\int_0^\infty e^{-(1-ik)x}dx=\frac2{1+k^2}.
$$
Hence $\widetilde h(k)=4/(1+k^2)^2$. The <Fourier inversion theorem> yields
$$
\boxed{h(x)=\frac2\pi\int_{-\infty}^{\infty}\frac{e^{ikx}}{(1+k^2)^2}dk.}
$$
The integral is absolutely convergent. At $x=0$ it gives $\int_{\mathbb R}(1+k^2)^{-2}dk=\pi/2$, providing a normalization check.
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