# Deriving Callan-Gross from Parton Model

I want to derive the Callan-Gross relation from the parton model, not fom the Rosenbluth and Mott cross sections, but I am having some problems obtaining the textbook result. I am following M.D. Schwartz: Quantum Field Theory and the Standard Model (pp.672, 675, 678).

Starting from the hard scattering coefficient obtained from the partonic scattering amplitude for $$\gamma^\ast q_i\rightarrow q_i$$ (eq. 32.32), $$\hat{W}^{\mu\nu}(z,Q^2)=2\pi Q^2_i\delta(1-z)\left[A^{\mu\nu}+\frac{4z}{Q^2}B^{\mu\nu}\right],$$ where $$A^{\mu\nu}:=-g^{\mu\nu}+\frac{q^\mu q^\nu}{Q^2}$$, $$B^{\mu\nu}:=\left(p^\mu+\frac{pq}{Q^2}q^\mu\right)\left(p^\nu+\frac{pq}{Q^2}q^\nu\right)$$, and the convolution formula for the hardonic tensor $$W^{\mu\nu}(x,Q^2)$$ obtained from factorisation, we arrive at \begin{align*} W^{\mu\nu}(x,Q^2) &=2\pi\int^1_x\frac{d\xi}{\xi}\sum_if_i(\xi)Q^2_i\delta(1-\frac{x}{\xi})\left[A^{\mu\nu}+\frac{4x}{Q^2\xi}B^{\mu\nu}\right]\\ &=2\pi\int^1_xd\xi\sum_if_i(\xi)Q^2_i\delta(\xi-x)\left[A^{\mu\nu}+\frac{4x}{Q^2\xi}B^{\mu\nu}\right]\\ &=2\pi\sum_if_i(x)Q^2_i\left[A^{\mu\nu}+\frac{4}{Q^2}B^{\mu\nu}\right],\end{align*} such that $$W_1(x,Q^2)=2\pi\sum_if_i(x)Q^2_i=\frac{Q^2}{4}W_2(x,Q^2)$$.

Now, the textbook says that the result should be $$W_1(x,Q^2)=\frac{Q^2}{4x^2}W_2(x,Q^2)$$ (eq. 32.23, 32.24). Did I make a mistake somewhere in my calculations?