= Solution
{c}
The middle panel in part <2/a> has an internal <photon> of <momentum> $k$ joining two points on a <fermion> line; the internal <fermion> has <momentum> $p-k$. It is a <one-particle-irreducible Feynman diagram> with no <closed fermion loop sign>. In <Feynman gauge>,
$$
-i\Sigma(p)=\int_k(-ie\gamma^\alpha)
\frac{i(\not p-\not k+m)}{(p-k)^2-m^2+i0}
(-ie\gamma^\beta)\frac{-ig_{\alpha\beta}}{k^2+i0},
$$
or
$$
\boxed{\Sigma(p)=-ie^2\int_k
\frac{(2-d)(\not p-\not k)+dm}{[k^2+i0][(p-k)^2-m^2+i0]}.}
$$
The decomposition in the source gives $\operatorname{tr}(\not p\Sigma)=4p^2S(p^2)$. In particular,
$$
S(p^2)=-\frac{ie^2(d-2)}{p^2}\int_k
\frac{p\cdot k-p^2}{[k^2+i0][(p-k)^2-m^2+i0]}.
$$
For a direct <ultraviolet divergence> extraction, use a <Feynman parameter> $x$ multiplying the massive denominator, shift $r=k-xp$, and define $\Delta=xm^2-x(1-x)p^2$. Then $\not p-\not k=(1-x)\not p-\not r$ and
$$
\int_0^1dx\int_r\frac1{(r^2-\Delta+i0)^2}
\bigg|_{\rm UV}=\frac{i}{16\pi^2\epsilon}\int_0^1dx.
$$
Odd $r$ terms integrate to zero. Taking the <trace> explicitly supplies the weight $(2-d)(1-x)$, whose integral tends to $-1$. Hence
$$
\operatorname{tr}(\not p\Sigma)\big|_{\rm UV}
=-\frac{4e^2p^2}{16\pi^2\epsilon},\qquad
\boxed{S(p^2)\big|_{\rm UV}=-\frac{e^2}{16\pi^2\epsilon}.}
$$
As a check on the same <fermion self-energy> calculation,
$$
\Sigma(p)\big|_{\rm UV}=\frac{e^2}{16\pi^2\epsilon}(-\not p+4m).
$$
This is the <ultraviolet pole of QED fermion self-energy>. It is safe to extract it at generic external <momentum>; an <on shell> or massless limit must distinguish any <infrared divergence> from this <ultraviolet divergence>.
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