Show that in the rest frame of the target proton,
$$
\nu=E-E^{\prime}, \quad y=\frac{E-E^{\prime}}{E},
$$
where $E$ and $E^{\prime}$ are the initial and final electron energies, respectively.
Evaluation of the cross section for ep $\rightarrow \mathrm{eX}$ is a straightforward repetition of the same calculation for $\mathrm{e}^{-} \mu^{-} \rightarrow \mathrm{e}^{-} \mu^{-}$(or ep $\rightarrow$ ep) scattering with the substitution of $W_{\mu \nu}$, given by (8.27), for $L_{\mu v}^{\text {muon }}$ (or $L_{\mu v}^p$ ). Using the expression (6.25) for $\left(L^e\right)^{\mu,}$ and noting $(8.25)$, we find
$$
\left(L^e\right)^{\mu \nu} W_{\mu \nu}=4 W_1\left(k \cdot k^{\prime}\right)+\frac{2 W_2}{M^2}\left[2(p \cdot k)\left(p \cdot k^{\prime}\right)-M^2 k \cdot k^{\prime}\right] .
$$
In the laboratory frame, this becomes
$$
\left(L^e\right)^{\mu \nu} W_{\mu \nu}=4 E E^{\prime}\left\{\cos ^2 \frac{\theta}{2} W_2\left(\nu, q^2\right)+\sin ^2 \frac{\theta}{2} 2 W_1\left(\nu, q^2\right)\right\},
$$
see (6.44). By including the flux factor, (4.32), and the phase space factor for the outgoing electron, (4.24), we can obtain the inclusive differential cross section for inelastic electron-proton scattering, ep $\rightarrow \mathrm{eX}$,
$$
d \sigma=\frac{1}{4\left((k \cdot p)^2-m^2 M^2\right)^{1 / 2}}\left\{\frac{e^4}{q^4}\left(L^e\right)^{\mu \nu} W_{\mu \nu} 4 \pi M\right\} \frac{d^3 k^{\prime}}{2 E^{\prime}(2 \pi)^3},
$$
where $\overline{|\mathscr{R}|^2}$ is given by the expression in the braces [recall (6.18)]. The extra factor of $4 \pi M$ arises because we have adopted the standard convention for the normalization of $W^{\mu v}$. Inserting (8.32) in (8.33) yields
$$
\left.\frac{d \sigma}{d E^{\prime} d \Omega}\right|_{\text {lab }}=\frac{\alpha^2}{4 E^2 \sin ^4 \frac{\theta}{2}}\left\{W_2\left(\nu, q^2\right) \cos ^2 \frac{\theta}{2}+2 W_1\left(\nu, q^2\right) \sin ^2 \frac{\theta}{2}\right\}
$$
where, as usual, we neglect the mass of the electron.