Restore the dimensional-regularization factor $\mu^\epsilon$ and define
$$
N_0=2i(1-x)\not p+4m,
\qquad N_1=i(1-x)\not p+m.
$$
Using $\Gamma(\epsilon/2)=2/\epsilon-\gamma+O(\epsilon)$ gives
$$
\boxed{\Sigma(\not p)=-\frac{e^2}{16\pi^2}\int_0^1dx
\left\{
\frac{2N_0}{\epsilon}
+N_0\left[\log\frac{4\pi\mu^2}{\Delta}-\gamma\right]
-2N_1
\right\}+O(\epsilon).}
$$
The first term is the <ultraviolet divergence>. In the <modified minimal subtraction scheme>, subtraction of $2/\epsilon-\gamma+\log4\pi$ leaves
$$
\Sigma_{\overline{\rm MS}}(\not p)
=-\frac{e^2}{16\pi^2}\int_0^1dx
\left[N_0\log\frac{\mu^2}{\Delta}-2N_1\right].
$$
On the tree-level mass shell, $p^2=-m^2$ and $i\not p=-m$, so $\Delta=x^2m^2$. The <mass counterterm> is
$$
\boxed{\delta m_{\overline{\rm MS}}
=-\frac{3e^2m}{16\pi^2}
\left(\frac2\epsilon-\gamma+\log4\pi\right).}
$$
The pole condition $m_{\rm phys}=m-\Sigma_{\overline{\rm MS}}|_{i\not p=-m}+O(e^4)$ therefore gives
$$
\boxed{m_{\rm phys}=m(\mu)\left[
1+\frac{e^2}{8\pi^2}\int_0^1dx
\left((1+x)\log\frac{\mu^2}{x^2m^2}-x\right)
\right]+O(e^4).}
$$
Evaluating the elementary parameter integral yields the equivalent expression
$$
m_{\rm phys}=m(\mu)\left[1+\frac{e^2}{4\pi^2}
\left(1+\frac34\log\frac{\mu^2}{m^2}\right)\right]+O(e^4).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 304 / 3 / e / Solution

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An ultraviolet divergence comes from arbitrarily large loop momentum, equivalently from singular short-distance behavior. Locality makes its regulator dependence cancellable by local counterterms in a renormalizable theory.


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