 Solution

For fixed $g>0$, write $A=\hslash T/(2m)$. The bound contribution to the quantum propagator of an attractive delta potential remains $\kappa e^{i\hslash {\kappa }^{2}T/(2m)}$. For its continuum, the stationary phase is at $k=0$, where $k^2/(k^2+{\kappa }^2)=k^2/{\kappa }^2+O(k^4)$. The Gaussian oscillatory integral and its derivative giveConsequentlyThe continuum decays as $T^{-3/2}$, but the full kernel has a persistent oscillating bound state contribution and no ordinary complex limit. This expansion requires $\hslash {\kappa }^{2}T/(2m)\gg 1$; it is not uniform in the zero-coupling limit, where the bound state disappears and the free kernel instead decays as $T^{-1/2}$.