Cross section normalization in $2 \to 2$ scattering

Consider Compton scattering $$p_1 + p_2 \to p_3 + p_4$$ in the laboratory frame. According to Quantum Field Theory and the Standard Model by Schwartz, the relation between the differential cross section and the normalized differential probability associated with the scattering process can be derived from Equation $(5.6)$ $$ d \sigma = \frac{V}{T} \frac{1}{|\textbf{v}_1 - \textbf{v}_2|} dP . \tag{5.6} $$ For Compton scattering in the lab frame $|\textbf{v}_1 - \textbf{v}_2| = |c \cdot \hat{\textbf{z}} - \textbf{0}| = c = 1$ in natural units, so this becomes $$ d \sigma = \frac{V}{T} dP . $$ Adapting Schwartz's Equation $(5.7)$ for the normalized differential probability in the case of $2$ outgoing particles gives $$ d P = \frac{|\langle f | S | i \rangle|}{\langle f | f \rangle \langle i | i \rangle} \frac{V d^3 \textbf{p}_3}{(2\pi)^3} \frac{V d^3 \textbf{p}_4}{(2\pi)^3} . \tag{5.7} $$ Schwartz uses this momentum-space measure rather than the Lorentz-invariant phase space measure used in quantum field theory. He also explicitly states that this distribution is normalised. I assume this must mean that $$ \int d P = \int \frac{|\langle f | S | i \rangle|}{\langle f | f \rangle \langle i | i \rangle} \frac{V d^3 \textbf{p}_3}{(2\pi)^3} \frac{V d^3 \textbf{p}_4}{(2\pi)^3} = 1 . $$ However, applying this to the definition of the differential cross section would imply that $$ \int d \sigma \stackrel{?}{=} \int \frac{V}{T} dP = \frac{V}{T}. $$ We know that this is not the case. In fact, $$ \int d \sigma = \sigma_{\text{tot}} $$ gives the total cross section $\sigma_{\text{tot}}$ which, in the case of Compton scattering, is a function of the incoming photon's energy. What is the source of this seeming contradiction and what did I get wrong?