= Solution
{c}
In the right panel of part <2/a>, $p$ enters the <fermion> line, $p'=p+q$ leaves it, and a virtual <photon> of <momentum> $k$ bridges the external <photon> vertex. Write $\Gamma^\mu=\gamma^\mu+\Lambda^\mu$ and $F_1=1+F_1^{(1)}+\cdots$. The one-loop <amplitude> is
$$
\begin{aligned}
-ie\Lambda^\mu={}&\int_k(-ie\gamma^\alpha)
\frac{i(\not p'-\not k+m)}{(p'-k)^2-m^2+i0}
(-ie\gamma^\mu)\frac{i(\not p-\not k+m)}{(p-k)^2-m^2+i0}
(-ie\gamma^\beta)\frac{-ig_{\alpha\beta}}{k^2+i0},\\
&\boxed{\Lambda^\mu=-ie^2\int_k
\frac{\gamma^\alpha(\not p'-\not k+m)\gamma^\mu(\not p-\not k+m)\gamma_\alpha}
{[k^2+i0][(p'-k)^2-m^2+i0][(p-k)^2-m^2+i0]}.}
\end{aligned}
$$
This <Feynman integral> has a logarithmic <superficial degree of divergence>. Its <ultraviolet pole> is independent of the external <momenta> and $m$, because every such numerator insertion or denominator correction lowers the large-$k$ degree. Preserve an auxiliary positive <mass> $M$ while extracting the <ultraviolet pole>; setting everything to zero would create a <scaleless integral> and conceal the cancellation of <ultraviolet> and <infrared divergences>.
The leading numerator is $\gamma^\alpha\not k\gamma^\mu\not k\gamma_\alpha$. Symmetric integration replaces $k_\rho k_\sigma$ by $g_{\rho\sigma}k^2/d$. Two contractions of <gamma matrices> then give
$$
\gamma^\alpha\not k\gamma^\mu\not k\gamma_\alpha
\ \longmapsto\ \frac{(d-2)^2}{d}k^2\gamma^\mu,
\qquad
\operatorname{tr}\left[\gamma^\alpha\not k\gamma^\mu\not k\gamma_\alpha\gamma_\mu\right]
=4(d-2)^2k^2.
$$
For the <auxiliary-mass extraction of ultraviolet poles>,
$$
J_M=\int_k\frac{k^2}{(k^2-M^2+i0)^3}
=\int_k\left[\frac1{(k^2-M^2+i0)^2}+\frac{M^2}{(k^2-M^2+i0)^3}\right],
\qquad J_M\big|_{\rm UV}=\frac{i}{16\pi^2\epsilon}.
$$
The second term is <ultraviolet> convergent. Thus
$$
\frac1{4d}\operatorname{tr}(\Lambda^\mu\gamma_\mu)\big|_{\rm UV}
=-ie^2\frac{(d-2)^2}{d}J_M\big|_{\rm UV},
\qquad
\boxed{F_1^{(1)}(q^2)\big|_{\rm UV}=\frac{e^2}{16\pi^2\epsilon}.}
$$
The <ultraviolet pole of the QED vertex> is therefore $\Lambda^\mu|_{\rm UV}=e^2\gamma^\mu/(16\pi^2\epsilon)$. The Pauli structure has zero <trace> against $\gamma_\mu$ and no <ultraviolet pole> at this order.
\b[Form-factor qualification.] The two <electron electromagnetic vertex form factors> in the source describe matrix elements between external <on shell> <Dirac spinors>. A general <off shell> vertex has more structures, so the unprojected <trace> is not a projector for the complete finite on-shell $F_1$. It does correctly isolate the divergent local $\gamma^\mu$ coefficient, which is all that is required here.
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