Lemma

The family $u_{\zeta }$ admits a distributional meromorphic continuation near $\zeta =0$ against compact smooth tests, with a simple pole. On tests supported in the open set $U$, the pairing at $\zeta =0$ equals $⟨(Q-i0{)}^{-2},f⟩$. The pole part is local,

and the residue, obtained only after the holomorphy premise below, is

The extension ${u_{\mu }}^R$ is Lorentz covariant and even under $z\mapsto -z$. Its scaling degree satisfies $\mathrm{sd}({u_{\mu }}^R)\le 4$. Residual contact freedom is frozen by excluding any added $c\delta$. XVI-J, prop:extension, is freedom only and is not this contact term.

Proof. Do not argue from the zero of $4\zeta (\zeta -1)$ alone, and do not argue from the Wick Jacobian alone. Let $F(\zeta )=(Q-i0{)}^{\zeta }$. This is a holomorphic family of distributions near $\zeta =0$ by instantiating Dütsch–Fredenhagen–Keller–Rejzner, arXiv:1311.5424, Definition 4.2 and Sections 4.1–4.2, with hypotheses $Q-i0$, the principal logarithm, tests in $\mathcal{D}({\mathbb{R}}^4)$, and the Gate 0 normalization. Classical support is the Gelfand–Shilov and Hörmander fact that complex powers $(x\pm i0{)}^{\lambda }$ are entire in $\lambda$. Constant-coefficient differentiation under the pairing gives $\square {\varphi }_{\lambda }=4\lambda (\lambda +1){\varphi }_{\lambda -1}$, hence $\square F(\zeta )=4\zeta (\zeta +1){\varphi }_{-1+\zeta }$. Since $F(0)=1$, one has $\square F(0)=0$, so holomorphy of $F$ yields $\square F(\zeta )=\zeta G(\zeta )$ with $G$ holomorphic in ${\mathcal{D}}^{\prime }$. Therefore

is holomorphic near $\zeta =0$. Only after that premise, Gate 0 gives $D_F=-A(Q-i0{)}^{-1}$ with $A=1/(4{\pi }^2)$ and $(-\square )D_F=i\delta$, hence $\square (Q-i0{)}^{-1}=4{\pi }^{2}i\delta$. The second recurrence $\square {\varphi }_{-1+\zeta }=4\zeta (-1+\zeta ){\varphi }_{-2+\zeta }$ then produces the residue $r=-i{\pi }^2$. The scale ${\mu }^{2\zeta }$ affects only the finite part. The operational Laurent identity is

Near a nonzero null point, ordinary absolute integrability of $(Q-i0{)}^{-2+\zeta }$ requires $\mathrm{Re}\zeta >1$. The pairing at $\zeta =0$ on $U$ is analytic continuation, not an ordinary integral. A Euclidean radial integral with $|S^3|=2{\pi }^2$ is an independent sign check, not the justification.