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the decay time in your "simple case" is well approximated by the large-$M$ limit

$$\lim_{M\rightarrow\infty}M\sigma\tau=2.8409$$

here is a plot of

$$f_M(s)=\left.\frac{F_M(t)}{1-F_M(\infty)}\right|_{t=s/(M\sigma)}$$

for $M=5,10,100$, that shows the half-time $s\approx 3$ is quite accurate already for not so large values of $M$