The path integral in quantum mechanics involves a factor $e^{iS_{N}/\hbar}$, where
$$S_{N}\equiv \sum\limits_{n=1}^{N+1}[p_{n}(x_{n}-x_{n-1})-\epsilon H(p_{n},x_{n},t_{n})]$$
In the limit $N \rightarrow \infty$, $S_{N}$ becomes the usual action $S$ for a given path.
When the Hamiltonian vanishes, the potential energy of the system offsets the kinetic energy of the system. Therefore, in the limit $N \rightarrow \infty$ the propagator $\langle x_{b}, t_{b}|x_{a}, t_{a}\rangle$ becomes
$$S_{N} \equiv \sum\limits_{n=1}^{N+1}\epsilon\bigg[p_{n}\bigg(\frac{x_{n}-x_{n-1}}{\epsilon}\bigg)\bigg]=\int\ dt\ p\dot{x}$$
Can you justify this answer using a physical argument?
