 Parton on-shell delta identity

In the leading massless collinear parton approximation, $P^2=0$ and $q^2=-Q^2$ imply $(\xi P+q{)}^2=2\xi P\cdot q-Q^2$. With Bjorken x $x=Q^2/(2P\cdot q)$ and $P\cdot q>0$, the Dirac delta change-of-variable formula yields the displayed identity. The positive-energy branch of the final parton also gives $\delta (E_q+\xi E_P-E_k)/E_k=2\delta ((E_q+\xi E_P{)}^2-E_k^2)$ locally on physical support. Keeping target mass instead introduces ${\xi }^2{M}^2$ and changes the root and Jacobian.