 Solution

After modified minimal subtraction, the surviving logarithm in the given fermion self-energy is $\log [{\mu }^2/((1-x)(m^2-p^{2}x))]$. For a one-loop mass shift, set $p^2=m^2$ and let $/p$ act as $m$ on an on-shell spinor inside that correction; changing these arguments by the mass shift contributes only at order $e^4$. ThusThe needed integrals areFor the second, put $u=1-x$ and use ${\int }_0^1\log udu=-1$ and ${\int }_0^{1}u\log udu=-1/4$. ConsequentlyThe pole mass condition $m_{\mathrm{phys}}-m+{\Sigma }_R=0$ gives $m_{\mathrm{phys}}=m[1+e^2(5+3\log ({\mu }^2/m^2))/(16{\pi }^2)]+O(e^4)$. Invert this relation and replace $m$ by $m_{\mathrm{phys}}$ inside the already one-loop term to obtainThe negative sign in the running-mass conversion follows from the explicitly chosen self-energy convention. Subtracting poles alone, instead of the overbarred combination, would leave additional $\log 4\pi -{\gamma }_E$ terms.