Use <dimensional regularization> with $d=4-\epsilon$ and choose <Feynman gauge>. Combining the electron and photon denominators with a <Feynman parameter>, shifting the loop momentum, and discarding the odd term leaves the numerator
$$
(2-d)(1-x)\gamma^\mu p_\mu+dm.
$$
The pole of the rotationally symmetric integral consequently gives
$$
\Sigma_{\rm pole}(p)
=\frac{e^2}{8\pi^2\epsilon}
\int_0^1dx\,[-2(1-x)\gamma^\mu p_\mu+4m]
=\frac{e^2}{8\pi^2\epsilon}(-\gamma^\mu p_\mu+4m).
$$
With the inverse-propagator convention
$$
S^{-1}(p)=\gamma^\mu p_\mu-m+\delta_2\gamma^\mu p_\mu-\delta_m-\Sigma(p),
$$
the <minimal subtraction scheme> chooses the counterterm pole to equal $\Sigma_{\rm pole}$. Hence
$$
\delta_2=-\frac{e^2}{8\pi^2\epsilon},
\qquad
\delta_m=-\frac{me^2}{2\pi^2\epsilon}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 304 / 2 / a / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-304/2/a/ii/solution