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Set the external four-momenta to zero after identifying the proper diagrams. In each box or triangle, at least one endpoint of the internal photon is a cubic vertex adjacent to an external scalar of zero momentum. That vertex is proportional to $l_{\rho }$, whereas the supplied numerator obeys $l^{\rho }{P}_{\rho \sigma }(l)=0$. All seven such graphs vanish algebraically. This use of the transverse numerator is stronger than merely saying that one has chosen a generic Lorenz gauge.

The two surviving seagull vertex bubbles have vertex factors $2i{e}_D^2{g}^{\mu \rho }$ and $2i{e}_D^2{g}^{\sigma \nu }$. Combining their two propagators givesThis scalar-QED four-point vertex pole in transverse gauge calculation determines the proportionality constant as well as the required tensor structure. Keep $m>0$ while extracting the ultraviolet pole. Lorentz covariance inside the regulated integral replaces $l^{\mu }{l}^{\nu }$ by $g^{\mu \nu }{l}^2/D$, giving the factor $(D-1)g^{\mu \nu }/D$.

For its scalar integral, Feynman parameter givesThere is no infrared divergence at $l=0$ with nonzero $m$. At pole order the two photon assignments and the factor $3/4$ therefore yieldThe common engineering scale factor is included in $e_D^2$; finite logarithms are immaterial to the pole. Adding $2i{e}_D^2{\delta }_4{g}^{\mu \nu }$ cancels it whenThis is minimal subtraction scheme, up to optional finite terms. It agrees with the transverse-gauge scalar QED wave-function pole and $Z_4=Z_2$. Setting $m=0$ at the outset would instead make the zero-momentum integral scaleless and obscure its separate ultraviolet divergence.

The integral printed in the text can be evaluated without reconstructing Figure 1. Put $A=l^2-m^2+i0$ and $B=(l+k{)}^2-m^2+i0$, and let $I^{\mu }$ denote that integral without its factor $\xi$. The numerator contraction responsible for its covariant gauge dependence isWith the scalar electrodynamics three-point vertex $-ie(p+p^{\prime }{)}^{\mu }$, three such vertices, two scalar propagators and one photon propagator have overall factor $(-ie{)}^3{i}^2(-i)=-e^3$. Thus the supplied integrand gives $V_{\mathrm{supplied}}^{\mu }=-e^3\xi I^{\mu }$.

First isolate the numerator term $l^2$. Its contribution isUnder $l\mapsto -l-k$ the two denominators exchange and the numerator changes sign. A shift-invariant dimensional regularization therefore gives $\boxed{I_0^{\mu }=0}$, including the finite part.

For the remaining numerator $2(l\cdot k)(2l+k{)}^{\mu }$, only its term $4l^{\mu }(l\cdot k)$ contributes to the ultraviolet pole. Corrections from expanding the denominators, and the numerator $2k^{\mu }(l\cdot k)$, are ultraviolet finite by power counting. Use auxiliary-mass extraction of ultraviolet poles with any $M>0$ rather than literally integrating a massless scaleless integral. Lorentz covariance then givesTODO (source gap): The algebra of the supplied expression is complete, but its graphical identification with Figure 1(b), including the routing at each internal vertex, cannot be visually verified from the unreadable panel.

The complete renormalization constant can nevertheless be obtained independently. Here is a short scalar QED wave-function pole calculation, to avoid assuming either missing panel's value. For an external four-momentum $p$, the exchange contribution to the scalar self-energy has numeratorIts denominator is $l^2[(p-l{)}^2-m^2+i0]$. The necessary ultraviolet poles, with the measure $d^{d}l/(2\pi {)}^d$, are respectivelyThe first two follow by a Feynman parameter and a shift of $l$; the third is a tadpole integral; the last follows by the same large-$l$ extraction used above. The seagull vertex contributes only a photon tadpole integral, which is a scaleless integral, and in any case cannot affect the coefficient of $p^2$. Multiplication by the exchange graph's factor $-e^2$ gives the scalar self-energy insertionThe kinetic counterterm $i{\delta }_2{p}^2$ therefore requires ${\delta }_2=e^2(3-\xi )C_{ϵ}$. The scalar QED Ward identity sets ${\delta }_1={\delta }_2$, soIndeed the full unrenormalized scalar electrodynamics three-point vertex has ultraviolet pole $i{e}^3(3-\xi )C_{ϵ}{k}^{\mu }$, which is cancelled by $-ie{\delta }_1{k}^{\mu }$. Subtracting the supplied contribution gives the gauge-independent sum stated in part (ii), without assigning either unreadable panel an integrand.