 Solution

Multiply the supplied Feynman gauge propagators before reducing the numerator. Their two factors $-i$ and the source's Clifford algebra implyFor example, contracting the numerator before the propagator factors gives ${\gamma }^{\nu }[-(/p+/l)+m]{\gamma }_{\nu }=-(d-2)(/p+/l)-dm$; the extra minus from $(-i{)}^2$ reverses both signs.

Use a Feynman parameter $x$ on the massive denominator and shift $q=l+xp$. The combined denominator isThe term odd in $q$ integrates to zero in dimensional regularization. The pole can be evaluated away from on-shell infrared singularities; its local ultraviolet coefficient is independent of $D$. With the source's master integral,Terms proportional to $d-4$ multiply the pole into a finite term, so they do not change this result. The counterterm kernel from 1ai is $K_{\mathrm{ct}}=-\delta Z_2/p-\delta m_L$. Cancelling $-{\Sigma }_{\mathrm{UV}}+K_{\mathrm{ct}}$ therefore givesThe distinction between mass definitions matters. Since $Z_2{m}_0=m+\delta m_L$,Equivalently the mass-dependent part of the counterterm density is $+e^{2}m\bar{\psi }\psi /(2{\pi }^2ϵ)$, because its definition includes the explicit minus sign in $-\delta m_L\bar{\psi }\psi$. In the alternative convention $d=4-2\epsilon$, replace every $1/ϵ$ here by $1/(2\epsilon )$. These are ultraviolet minimal-subtraction poles, not the finite on-shell mass correction. The Ward identity also fixes $\delta Z_1=\delta Z_2$; the photon coefficient $\delta Z_3$ requires its own two-point function and is not obtained from this graph.