= Solution
For $E<0$, inversion of the free operator gives $G_0(x,y)=m e^{-\kappa|x-y|}/(\kappa\hbar^2)$, with $\kappa=\sqrt{-2mE}/\hbar$. The bound-state equation is $\psi=-G_0U\psi$, hence
$$
\psi(x)=-\frac m{\kappa\hbar^2}\int\!\int e^{-\kappa|x-y|}V(y,z)\psi(z)\,dy\,dz.
$$
For the separable kernel $V(y,z)=-\alpha F(y)F^*(z)$, define
$$
\boxed{C=\int_{-\infty}^{\infty}F^*(z)\psi(z)\,dz.}
$$
Then
$$
\boxed{\psi(x)=\frac{\alpha mC}{\kappa\hbar^2}\int e^{-\kappa|x-y|}F(y)\,dy.}
$$
An overall <normalization> still has to be chosen after finding a permitted <energy>. A nontrivial <bound state> must have $C\ne0$: if $C=0$, the potential acts as zero and a square-integrable negative-energy free solution on the whole line cannot exist. Decay or square-integrability assumptions on $F$ are needed to make these <integrals> well defined; smoothness alone would not ensure convergence.
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