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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.

In the right panel of part a, $p$ enters the fermion line, $p^{\prime }=p+q$ leaves it, and a virtual photon of momentum $k$ bridges the external photon vertex. Write ${\Gamma }^{\mu }={\gamma }^{\mu }+{\Lambda }^{\mu }$ and $F_1=1+F_1^{(1)}+\cdots$. The one-loop amplitude isThis Feynman integral has a logarithmic superficial degree of divergence. Its ultraviolet pole is independent of the external momenta and $m$, because every such numerator insertion or denominator correction lowers the large-$k$ degree. Preserve an auxiliary positive mass $M$ while extracting the ultraviolet pole; setting everything to zero would create a scaleless integral and conceal the cancellation of ultraviolet and infrared divergences.

The leading numerator is ${\gamma }^{\alpha }/k{\gamma }^{\mu }/k{\gamma }_{\alpha }$. Symmetric integration replaces $k_{\rho }{k}_{\sigma }$ by $g_{\rho \sigma }{k}^2/d$. Two contractions of gamma matrices then giveFor the auxiliary-mass extraction of ultraviolet poles,The second term is ultraviolet convergent. ThusThe ultraviolet pole of the QED vertex is therefore ${\Lambda }^{\mu }{|}_{\mathrm{UV}}=e^2{\gamma }^{\mu }/(16{\pi }^2ϵ)$. The Pauli structure has zero trace against ${\gamma }_{\mu }$ and no ultraviolet pole at this order.

Form-factor qualification. The two electron electromagnetic vertex form factors in the source describe matrix elements between external on shell Dirac spinors. A general off shell vertex has more structures, so the unprojected trace is not a projector for the complete finite on-shell $F_1$. It does correctly isolate the divergent local ${\gamma }^{\mu }$ coefficient, which is all that is required here.

A Feynman integral with no nonzero mass or external momentum scale is assigned zero by dimensional regularization. This does not imply separate absence of ultraviolet divergences and infrared divergences. For a logarithmically scaleless example, zero can encode cancellation of the two pole contributions. Keeping a scale or using auxiliary-mass extraction of ultraviolet poles is necessary when only the ultraviolet part is sought.