 Solution

A phase-space path integral is defined as a limit of finite-dimensional integrals, not by assigning a classical derivative to every path. Choose $t_1=0<t_2<\cdots <t_{N+1}=T$, let $\Delta t_i=t_{i+1}-t_i$, and set $q_{N+1}=q_f$. On slice $i$ the precise prescription isThe remaining Hamiltonian term must have a compatible operator-ordering prescription. For example, evaluate $H$ at $(p_i,(q_{i+1}+q_i)/2)$ for midpoint/Weyl ordering. A prepoint prescription $H(p_i,q_i)$ defines a corresponding ordering instead. This choice matters for a general mixed $H(p,q)$; the separable kinetic-plus-potential Hamiltonian in the next part admits the usual Trotter prescription.

This is time slicing of a phase-space path integral. Integrate the intermediate $q_i$ and the slice momenta and only then take ${\max }_i\Delta t_i\to 0$. Typical paths of the Euclidean path integral need not be differentiable; the finite difference is the meaning of the printed $\dot{q}$. For a fixed-endpoint kernel the initial coordinate is fixed as well, whereas propagation of a wavefunction includes an integral over that initial coordinate.