omega,E',E)$. It has a decomposition ([@duclous], [@lorence], [@boman], [@hensel]) \[i-e-e1\] \_[22]{}(x,’,,E’,E) =\^p\_[22]{}(x,’,,E’,E)+\^s\_[22]{}(x,’,,E’,E). where $\sigma^p_{22}(x,\omega',\omega,E',E)$ is corresponding to the (new) primary electrons and $\sigma^s_{22}(x,\omega',\omega,E',E)$ is corresponding to the secondary electrons. In this scattering process the spins have been averaged out, and the two electrons completely lose their identity. Therefore, categorizing the electrons as “primary” and “secondary” is simply done by assigning the electron exiting the scattering event with the highest energy to be the primary one. The scattering cross section for primary electron $\sigma^p_{22}(x,\omega',\omega,E',E)$ has an expression $$\begin{gathered}
\sigma^p_{22}(x,\omega',\omega,E',E)
=
\sigma_{0}(x){{(E'+1)^2}\over{E'(E'+2)}}\Big({1\over{E^2}}
+{1\over{(E'-E)^2}}+{1\over{(E'+1)^2}} \\
-{{2E'+1}\over{(E'+1)^2E(E'-E)}}\Big)\chi_{22,p}(E',E)\delta(\omega'\cdot\omega-\mu_{22,p}(E',E)),\end{gathered}$$ where $\sigma_{0}(x)$ depends on the background material, and $$& \mu_{22,p}(E',E):=\sqrt{{E(E'+2)}\over{E'(E+2)}} \\
& \chi_{22,p}(E',E):=\chi_{{\mathbb{R}}_+}(E-E_0)\chi_{{\mathbb{R}}_+}(E-{E'\over 2})\chi_{{\mathbb{R}}_+}(E'-E),$$ while the cross section for the secondary electron $\sigma^s_{22}(x,\omega',\omega,E',E)$ is $$\begin{gathered}
\sigma^s_{22}(x,\omega',\omega,E',E)
=
\sigma_0(x){{(E'+1)^2}\over{E'(E'+2)}}\Big({1\over{E^2}}
+{1\over{(E'-E)^2}}+{1\over{(E'+1)^2}}\\
-{{2E'+1}\over{(E'+1)^2E(E'-E)}}\Big)\chi_{22,s}(E',E)\delta(\omega'\cdot\omega-\mu_{22,s}(E',E)),\end{gathered}$$ where $$& \mu_{22,s}(E',E):=\mu_{22,p}(E',E-E), \\
& \chi_{22,s}(E',E):=\chi_{{\mathbb{R}}_+}({{E'}\over 2}-E)\chi_{{\mathbb{R}}_+}(E-E_0).$$ Since $\sigma^s_{22}(x,\omega',\omega,E',E)=0$ for $E'\leq 2E$ the singularities at $E'=E$ do not cause any problems for the secondary electrons.

Write $$\chi_{22}(E',E)
:={}&
\chi_{22,p}(E',E)+\chi_{22,s}(E',E), \\[2mm]
\mu_{22}(E',E)
:={}&
\begin{cases} \mu_{22,