Assume $F \in L^2([0,\infty))$, so that the Laplace-transform $L[F]$ is well-defined. Assume furthermore, that $$ y \mapsto \frac{L[F](iy)}{1+L[F](iy)} $$ is in $L^2(\mathbb{R})\cap L^1(\mathbb{R})$, so that we obtain a well-defined function $R \in L^2(\mathbb{R})$ by the $L^2$-limit $$ R(t) = \frac{1}{2\pi}\lim_{T\to\infty} \int_{-T}^{T} \frac{L[F](iy)}{1+L[F](iy)} e^{iyt}dy. $$ Is it true that then $R(t)=0$ for $t < 0$?
This argument has been implicitly used in the paper http://www.tandfonline.com/doi/abs/10.1080/00411459408203873 by Glassey and Schaeffer. But I doubt that this argument is valid in general. By the classical theory of Laplace transforms one would make sure that the function $z \mapsto 1+L[F](z)$ has no poles in the complex half-plane $\Re(z) > 0$. Otherwise, if there are poles in the right half-plane, the property $R(t) =0$ for $t<0$ cannot be assumed to hold true. Furthermore, don't you have to assume in addition that $$ z \mapsto \frac{L[F](z)}{1+L[F](z)} $$ tends to $0$ as $\Re(z) \to \infty$? The author of the recent article http://www.tandfonline.com/doi/abs/10.1080/23324309.2015.1075556 makes exactly the same mistake.