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.

First remove the projector denominator explicitly:Let $A=(l+k{)}^2+m^2$ and $B=l^2$. The Feynman parameter for a simple and a squared propagator follows by differentiating the elementary two-denominator Feynman formula with respect to $B$ givesShift $q=l+xk$. Then $xA+(1-x)B=q^2+D$, with $D=x(1-x)k^2+x{m}^2$. The numerator is unchanged by this parallel shift:Therefore the correctly combined integral isThe printed weight $x/2$ is not compatible with this denominator and prefactor. Exchanging $x$ with $1-x$ changes the mass term in $D$ as well and cannot supply the missing normalization. A concrete counterexample is $m=0$ with nonzero spacelike external momentum and $2<d<4$: after Wick rotation, the remaining integral is nonzero and symmetric under $x\leftrightarrow 1-x$. The source's $4e^2\int (x/2)dx$ expression is then exactly one quarter of the correct expression. Thus the claimed equality cannot be imposed when computing the pole.

Apply the dimensionally regulated tadpole integral before removing the infrared regulator:At fixed regulator $0<\mathrm{Re}ϵ<2$ away from gamma-function poles, the power of $m_{\gamma }^2$ has positive real part. ConsequentlyBoth the correctly weighted tadpole and the source's doubled version vanish. This is the near-four-dimensional limit appropriate to the question; the order of limits is important. Equivalently the exactly massless tadpole is a scaleless integral and is zero in dimensional regularization. The statement is not valid for every arbitrary fixed dimension: for example, $ϵ=5/2$ gives a nonzero prefactor and growth as $m_{\gamma }^{-1/2}$. One must keep the regulator in its stated near-four-dimensional domain.

Route the photon momentum as $l$ and the internal scalar momentum as $l+k$. The transverse projector obeys $P_{\alpha \beta }{l}^{\beta }=0$, so the two cubic vertices giveMultiplying the two vertices and propagators therefore givesFor the seagull, $g^{\alpha \beta }{P}_{\alpha \beta }=d-1$. Including the photon-loop symmetry factor gives insteadThe printed seagull coefficient is twice this result. The vertex's factor two differentiates the two photon fields; closing them into one line requires the compensating symmetry factor. Direct expansion of the action gives the same check: the interaction density $-e^2{A}^2{\varphi }^{†}\varphi$ has one photon contraction, yielding coefficient $-e^2$, not $-2e^2$. The identical-photon origin of the vertex factor is also explicit in the standard scalar-electrodynamics rules.^[ref] This discrepancy does not affect the massless-photon limit in part d or the ultraviolet pole in part f.