 Schwarzschild horizon-to-singularity proper time for fall from infinity

A radial observer released from rest at infinity has $dr/d\tau =-\sqrt{2M/r}$. Integration from $r=2M$ to $r=0$ gives $\Delta \tau ={\int }_0^{2M}\sqrt{r/(2M)}dr=4M/3$ in geometric units. In ordinary units it is $4GM/(3c^3)$. The endpoint is the curvature singularity, not a regular spacetime centre; this value belongs to that specified energy, not every possible infalling worldline.