So I've been given a reduced form of the neutron continuity equation that does not account for geometry:
$$\Sigma_s(E)\phi(E)=\int_0^\infty\Sigma_s(E^{'}→E)\phi(E^{'})dE^{'}+S(E),$$
for which the normalization condition is
$$\int_0^\infty f_s(E^{'}→E)dE=1.$$
Restricting focus to only elastic scattering, I get
$$f_s(E^{'}→E)dE=\begin{cases}\frac{dE}{(1-\alpha)E^{'}} & E\le E^{'}\le\frac{E}{\alpha}\\0 & \text{otherwise}\end{cases},$$
where $\alpha$ is the scattering parameter:
$$\alpha=\left(\frac{A-1}{A+1}\right)^2,$$
where $A$ is the atomic mass of the nuclei.
How do I check this normalization condition? It's been awhile, I don't even know where to start.
It sounds like you have to validate that $$ 1 = \int_0^\infty f_s(E' \to E)dE = \int_{E \le E' \le E/\alpha} \frac{dE}{(1-\alpha)E'} $$
To finish this, what is the relationship between $E$ and $E'$ and $\alpha$?