Summing unresolved coherent emissions removes the lower cutoff

Define ${\lambda }_u=C{\int }_{\epsilon }^{\mathrm{\Delta }}d\nu \mathrm{/}\nu$ and ${\lambda }_r=C{\int }_{\mathrm{\Delta }}^{\mathrm{\Lambda }}d\nu \mathrm{/}\nu$. By [F2], ${\lambda }_u=C\log (\mathrm{\Delta }\mathrm{/}\epsilon )$, ${\lambda }_r=C\log (\mathrm{\Lambda }\mathrm{/}\mathrm{\Delta })$, and ${\lambda }_u+{\lambda }_r=C\log (\mathrm{\Lambda }\mathrm{/}\epsilon )$. The endpoint $\nu =\mathrm{\Delta }$ has zero one-particle measure, so assigning it to the resolved interval does not change either mean.

Independence in [F1] gives $\mathbb{P}(N_u=n,N_r=0)=\exp (-{\lambda }_u){\lambda }_u^n\mathrm{/}n!\exp (-{\lambda }_r)$ for every $n\in {\mathbb{N}}_0$. This also covers $C=0$: at $n=0$, the convention $0^0=1$ gives probability one, and at positive $n$ the probability is zero. In particular the case $n=0$ is the exclusive no-emission probability.

Summing the joint probabilities over $n$ and using the everywhere convergent exponential series [F3] gives ${\sum }_{n=0}^{\mathrm{\infty }}\mathbb{P}(N_u=n,N_r=0)=\exp (-{\lambda }_u-{\lambda }_r){\sum }_{n=0}^{\mathrm{\infty }}\frac{{\lambda }_u^n}{n!}=\exp (-{\lambda }_r)\mathrm{.}$ The terms are nonnegative and the series is absolutely convergent, so this sum is the probability of the disjoint union of the events $\{N_u=n,N_r=0\}$.

Substituting ${\lambda }_r=C\log (\mathrm{\Lambda }\mathrm{/}\mathrm{\Delta })$ from step 1.1 yields $\mathbb{P}(N_r=0)=\exp (-C\log (\mathrm{\Lambda }\mathrm{/}\mathrm{\Delta }))$, with no remaining $\epsilon$. The conclusion is a model deduction from the finite-band Poisson law, not a statement about virtual amplitudes in QED.