Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2020 / 1 / vii / Solution

Only the term proportional to $\alpha$ has an ultraviolet pole. Its $k^2$ cancels the massless denominator, leaving the standard Feynman integralIn the present dimensional regularization convention, $\Gamma (ϵ/2)=2/ϵ+O(1)$ and $\alpha =1+O(ϵ)$. ThereforeThe counterterm contribution is $-i{e}_D{\delta }_1{\gamma }^{\mu }$. Cancellation of the ultraviolet pole fixeswhere the displayed expression uses the minimal subtraction scheme; a different renormalization scheme may add finite $O(e^2)$ terms. In the frequently used convention $D=4-2\epsilon$, the same result is $Z_1=1-e^2/(16{\pi }^2\epsilon )+O(e^4)$. The sign agrees with $Z_1=Z_2$ from the Ward identity for QED renormalization constants in Feynman gauge.