 Solution

For the momentum-space propagator in a uniform force, the stated Fourier convention makes $x$ act on momentum wavefunctions as $i\hslash {\partial }_p$. HenceFor $T>0$, the characteristic has $p(s)=p^{\prime }-\alpha (s-t^{\prime })$. Along it the amplitude acquires quantum phase $-{\int }_{t^{\prime }}^{t}p(s{)}^{2}ds/(2m\hslash )$. momentum is translated without a Jacobian, givingThe cubic expression is evaluated on the delta-function support. There it equals $-i[p^{\prime 2}T-\alpha p^{\prime }{T}^2+{\alpha }^2{T}^3/3]/(2m\hslash )$, making both its sign and its continuous $\alpha \to 0$ limit clear. The jump at $T=0$ is the identity momentum kernel.