 Solution

Lorentz-covariant tensor integration replaces $q_{\mu }{q}_{\nu }$ by $g_{\mu \nu }{q}^2/d$. The numerator consequently contributes $(d-1)k^2{q}^2/d$. The supplied master integral with powers $a=1,b=3$ givesInserting the corrected parameter integral givesThe pole uses $\Gamma (ϵ/2)=2/ϵ+O(1)$, $D^{-ϵ/2}=1+O(ϵ)$ and ${\int }_0^1(1-x)dx=1/2$. ThusThere is no momentum-independent pole at $\lambda =0$ in this transverse gauge after removing the photon tadpole. The two-point counterterm insertion is $-i(\delta Z_{\varphi }{k}^2+\delta M^2)$, so minimal subtraction requiresThese coefficients cancel the actual one-loop insertion derived from the action. Applying the misprinted weight in part e would give a wave-function pole four times too small. In a $d=4-2\epsilon$ convention the same answer is $\delta Z_{\varphi }=3e^2/(16{\pi }^2\epsilon )$. Finally, $\delta M^2=0$ does not mean that the bare mass parameter is unchanged: $Z_{\varphi }{m}_0^2=m^2$ gives $m_0^2-m^2=-m^2\delta Z_{\varphi }+O(e^4)$. Photon, charge and quartic counterterms require their respective Green functions and cannot be determined from this scalar two-point function alone.