= Half-line Schrodinger boundary-trace elimination
{c}
{title2=$c=-1\text{ for Dirichlet},\quad c=1\text{ for Neumann}$}
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 $|e^{ikx-ik^2(t-\tau)}|\leq e^{-x\operatorname{Im}k}$, gives boundary coefficients $(1-c)k\widetilde g_0-i(1+c)\widetilde g_1$. 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.
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