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

$$\lim_{M\rightarrow\infty}M\sigma\tau=2.84$$

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 $M$

[*] for the large-$M$ limit I replace the sum in the defition of $F_M(t)$ by an integral and solve for the half-time,

$$\int_0^M\frac{2(M-x)}{M^2}e^{-(\sigma \tau x)^2}\,dx=\frac{1}{2}$$

this gives $M\sigma\tau=2.84092\cdots$.

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

$$\lim_{M\rightarrow\infty}M\sigma\tau=2.84$$

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 $M$

[*] for the large-$M$ limit I replace the sum in the defition of $F_M(t)$ by an integral and solve for the half-time,

$$\int_0^M\frac{2(M-x)}{M^2}e^{-(\sigma \tau x)^2}\,dx=\frac{1}{2}$$

this gives $M\sigma\tau=2.84092\cdots$.

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 $M$

[*] for the large-$M$ limit I replace the sum in the defition of $F_M(t)$ by an integral and solve for the half-time,

$$\int_0^M\frac{2(M-x)}{M^2}e^{-(\sigma \tau x)^2}\,dx=\frac{1}{2}$$

this gives $M\sigma\tau=2.84092\cdots$.

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

$$\lim_{M\rightarrow\infty}M\sigma\tau=2.84$$

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 $M$

[*] for the large-$M$ limit I replace the sum in the defition of $F_M(t)$ by an integral and solve for the half-time,

$$\int_0^M\frac{2(M-x)}{M^2}e^{-(\sigma \tau x)^2}\,dx=\frac{1}{2}$$

this gives $M\sigma\tau=2.84092\cdots$.

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 $M$

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 $M$

[*] for the large-$M$ limit I replace the sum in the defition of $F_M(t)$ by an integral and solve for the half-time,

$$\int_0^M\frac{2(M-x)}{M^2}e^{-(\sigma \tau x)^2}\,dx=\frac{1}{2}$$

this gives $M\sigma\tau=2.84092\cdots$.