= Solution
Use the <massless collinear parton approximation> in a high-energy frame: $P^2=0$, $k=\xi P$, and $k'=\xi P+q$ with $k'^2=0$. This neglects target-mass corrections to the <parton model>; it does not literally set a stationary massive target to a massless particle in the earlier flux formula.
For a <quark> of dimensionless charge $Q_f$, the electromagnetic <vector current> <matrix element> is $Q_f\bar u(k')\gamma^\mu u(k)$. The <spin average> and <gamma-matrix trace> give
$$
\frac12\sum_{\rm spins}J^\mu J^{\nu *}=2Q_f^2\left(k^\mu k'^\nu+k^\nu k'^\mu-g^{\mu\nu}k\cdot k'\right).
$$
Integrating the three-momentum <Dirac delta function> in the <parton> <hadronic tensor> leaves
$$
\widetilde W^{\mu\nu}=\frac{\delta(q^0+\xi E_P-E_{k'})}{2\xi E_{k'}}\left(k^\mu k'^\nu+k^\nu k'^\mu-g^{\mu\nu}k\cdot k'\right).
$$
Since $k\cdot k'=\xi\nu$, this is
$$
\widetilde W^{\mu\nu}=\frac{\delta(q^0+\xi E_P-E_{k'})}{E_{k'}}\left[\xi P^\mu P^\nu+\frac12(P^\mu q^\nu+q^\mu P^\nu)-\frac\nu2g^{\mu\nu}\right].
$$
For the massless <electron> momenta, $p^2=p'^2=0$ and $q=p-p'$. Substitution into the <leptonic tensor> gives
$$
q^\mu L_{\mu\nu}=4\left[-(p\cdot p')p'_\nu+(p\cdot p')p_\nu-(p\cdot p')(p_\nu-p'_\nu)\right]=0,
$$
and likewise $q^\nu L_{\mu\nu}=0$. These <Ward identities> eliminate every term with an exposed $q$ index in the contraction. Therefore
$$
\boxed{\widetilde W^{\mu\nu}\doteq\frac{\delta(q^0+\xi E_P-E_{k'})}{E_{k'}}\left[\xi P^\mu P^\nu-\frac{\nu}{2}g^{\mu\nu}\right].}
$$
Here $\doteq$ means equality \b[after contraction with the <leptonic tensor>]. The shortened tensor is not itself conserved; the omitted terms restore <current conservation> in the full <hadronic tensor>.
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