General-service infinite-server queue ($M/G/\infty$)

Poisson arrivals of rate $\lambda$ receive independent service times distributed as $S$, with infinitely many servers so service starts on arrival. Starting empty, the occupancy at time $t$ is Poisson with mean $\lambda {\int }_0^t\mathbb{P}(S>u)du$, by independent thinning of the marked arrivals. If $\mathbb{E}S<\infty$, the equilibrium count is Poisson with mean $\lambda \mathbb{E}S$. Occupancy need not be a Markov process for a general service law; remaining service times carry additional state information.

Stop new arrivals at time $t$ in an initially empty general-service infinite-server queue. The number remaining $v$ time units later is Poisson with mean $\lambda {\int }_v^{t+v}\mathbb{P}(S>u)du$. Thus the probability that clearance has not occurred is one minus the exponential of the negative of that mean. Letting $t\to \infty$ gives the displayed equilibrium formula, since ${\int }_v^{\infty }\mathbb{P}(S>u)du=\mathbb{E}(S-v{)}_{+}$. The clearance time is zero when no customers remain at closing.