 Half-line Schrodinger boundary-trace elimination

Orient the upper-quadrant contour from $i\infty$ down to zero and then to $+\infty$. The reflected half-line transform has zero inverse contribution for positive $x$. Adding $c$ times this contribution, and deforming boundary integrands through the second quadrant where $|e^{ikx-i{k}^2(t-\tau )}|\le e^{-x\mathrm{Im}k}$, gives boundary coefficients $(1-c)k{\tilde{g}}_0-i(1+c){\tilde{g}}_1$. Thus $c=-1$ removes the unknown Neumann trace for a prescribed Dirichlet value, while $c=1$ removes the unknown Dirichlet trace for a prescribed Neumann value. The initial reflected term has coefficient $c$ and cannot be discarded after the time-dependent spectral decomposition.