Bremsstrahlung and infrared divergences in Peskin & Schroeder 6.5 At the very end of chapter 6 in Peskin and Schroeder (page 208), there is a calculation of the expected number of soft photons radiated in a hard electron scattering process. The text claims that the cross-section for the process of producing $n$ soft photons is given by the cross-section for producing no soft photons in the process (or more precisely, for producing no photons with energy greater than the detector threshold energy), times the additional factor
$$ \frac{1}{n!} \left[ \frac{\alpha}{\pi} f_\mathrm{IR}(q^2) \log\left(\frac{E_+^2}{E_-^2}\right)\right]^n  \times \exp\left[ - \frac{\alpha}{\pi} f_\mathrm{IR}(q^2) \log \left(\frac{E_+^2}{E_-^2}\right)\right] \,,$$
where the soft photons all have energy between $E_-$ and $E_+$. The first factor in this expression make sense to me, since the factor in square brackets is simply the contribution to the cross-section of producing one real soft photon, as in equation (6.82) on the previous page (which gives the contribution to the cross-section of emitting one real photon with energy between the infrared cutoff $\mu$ and the detector threshold energy $E_\ell$).
What I don't understand is where the exponential factor comes from. The cross-section for the hard process with no detectable soft photons (6.84) contains similar exponential factors, but these arise from summing over a) all possible number of virtual photon exchanges, and b) all possible number of undetectable real photon emissions. Since we are explicitly asking for $n$ real photons to be produced, I cannot see where such an exponential could arise from. Thank you.
 A: The expression is just a Poisson distribution for getting $n$ events, with the rate parameter being $$\lambda = \frac{\alpha}{\pi} f_\mathrm{IR}(q^2) \log\left(\frac{E_+^2}{E_-^2}\right)$$
A: This is too old question but I'd like give a possible answer.
The exponential factor comes from eq.(6.84) but the authors made a mistake of rewriting the argument of ${\rm log}$ in the exponential factor.
The correct statement of the probability is the following.
Given that the electron scattered $p \rightarrow p'$. This fives the tree-level probability to $1$. Then, assume $A:$ the $n$ soft photon with its energy $E_{-}$ to $E_{+}$ was emitted. This gives the factor $\frac{1}{n!}\left[\frac{\alpha}{\pi} f_{\rm IR}(q^2) \log\left(\frac{E_+^2}{E_-^2}\right) \right]$.
However, there are a possibility that $B:$ the arbitrary number of virtual photons and real photons with $E<E_l$ with $E_l$ being the lower threshold of the detector were produced in the process. Hence the correct event is $A \cap B$, so the result is
\begin{align}
{\rm Prob}(n\gamma {\rm ~with~} E_- < E < E_+) = \frac{1}{n!}\left[\frac{\alpha}{\pi} f_{\rm IR}(q^2) \log\left(\frac{E_+^2}{E_-^2}\right) \right] \exp\left[-\frac{\alpha}{\pi}f_{\rm IR}(q^2)\log\left(\frac{E_+^2}{E_-^2}\right)\right].
\end{align}
Thus, the result is not just the Poisson distribution but the distribution times an exponential factor as
\begin{align}
{\rm Prob}(n\gamma {\rm ~with~} E_- < E < E_+) = P(n) \exp\left[-\frac{\alpha}{\pi}f_{\rm IR}(q^2) \left(\log\left(\frac{-q^2}{E_+^2}\right) + \log\left(\frac{E_-^2}{\mu^2}\right)   \right)\right].
\end{align}
Indeed, if we take $-q^2 \rightarrow \infty$ or $\mu^2 \rightarrow 0$ the exponential factor asymptotes to $1$ and reproduces the Poisson distribution and hence the classical result (6.19). I think it is more natural that the result matches with (6.19) in the certain limit.
