<s>[INST] Is the rectangular function a convolution of $L^1$ functions? Do there exist functions $f,g$ in $L^1(\mathbf{R})$ such that the convolution $f \star g$ is (almost everywhere) equal to the indicator function of the interval $[0,1]$ ?
 [/INST] Following the idea of Christian Remling, I have two explicit functions that have the  property that $f*g = \mathbf{1}_{[-1,1]}$:
\begin{eqnarray}
f(x) &=& \frac{K_\frac{1}{4}\left(|x|\right)}{2^{1/4}\pi\Gamma\left(\frac{1}{4}\right)|x|^{1/4}}\\\\
g(x) &=& \frac{\mathrm{sgn}(1-x)}{\sqrt{|1-x|}}\,_1F_2\left(\begin{matrix}-\frac{1}{4} \\ \frac{1}{4},  \frac{3}{4}\end{matrix}; \frac{(1-x)^2}{4}\right) +\frac{\mathrm{sgn}(1+x)}{\sqrt{|1+x|}}\,_1F_2\left(\begin{matrix}-\frac{1}{4} \\ \frac{1}{4},  \frac{3}{4}\end{matrix}; \frac{(1+x)^2}{4}\right)\\ &\,&+ \frac{\Gamma\left(\frac{5}{4}\right)(1-x)}{\sqrt{2}\Gamma\left(\frac{7}{4}\right)}\,_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},  \frac{7}{4}\end{matrix}; \frac{(1-x)^2}{4}\right)+ \frac{\Gamma\left(\frac{5}{4}\right)(1+x)}{\sqrt{2}\Gamma\left(\frac{7}{4}\right)}\,_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},  \frac{7}{4}\end{matrix}; \frac{(1+x)^2}{4}\right)
\end{eqnarray}
Where $K_\frac{1}{4}$ is a modified Bessel function and $\,_1F_2$ is a hypergeometric function. While these aren't the simplest functions, and $g$ is kind of a nightmare, they have the nice property that
\begin{eqnarray}
\mathcal{F}[f](k) &=& \frac{1}{\sqrt{2\pi}(1+k^2)^{1/4}} \\
\mathcal{F}[g](k) &=& 2\sqrt{2\pi}(1+k^2)^{1/4}\frac{\sin(k)}{k},
\end{eqnarray}
so that $f*g = \mathcal{F}^{-1}[\mathcal{F}[f]\cdot\mathcal{F}[g]] = \mathcal{F}^{-1}\left[2\sin(k)/k\right] = \mathbf{1}_{[-1,1]}$.
The convolution integral isn't doable analytically, but it can be evaluated numerically. Doing this in Mathematica gives

where the blue is $f$, the yellow is $g$, and the green is $f*g$. Despite their divergences, both $f$ and $g$ are in $L^1$, as the divergences scale as $|x|^{-1/2}$.
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