The Dunford contour construction satisfies the semigroup law and strong continuity at the vertex

The double integral. Fix admissible contours ${\mathrm{\Gamma }}_1=\mathrm{\Gamma }(r_1,{\theta }_1)$ and ${\mathrm{\Gamma }}_2=\mathrm{\Gamma }(r_2,{\theta }_2)$ with $\pi \mathrm{/}2+{\delta }_0<{\theta }_1<{\theta }_2<\pi \mathrm{/}2+\delta$ and $r_2<r_1$; then every point of ${\mathrm{\Gamma }}_1$ lies outside the interior of ${\mathrm{\Gamma }}_2$ and every point of ${\mathrm{\Gamma }}_2$ lies inside the interior of ${\mathrm{\Gamma }}_1$. By [L1] and [L4] the product $T(z_1)T(z_2)$ is the norm limit of the truncated products, and for each truncation the finite double integral of $e^{{\lambda }_1{z}_1+{\lambda }_2{z}_2}R({\lambda }_1)R({\lambda }_2)$ equals its iterate, so $T(z_1)T(z_2)=\frac{1}{(2\pi i{)}^2}{\int }_{{\mathrm{\Gamma }}_2}{\int }_{{\mathrm{\Gamma }}_1}{e}^{{\lambda }_1{z}_1+{\lambda }_2{z}_2}R({\lambda }_1)R({\lambda }_2)d{\lambda }_{1}d{\lambda }_2$ with absolutely convergent iterated integrals.

The finite-keyhole winding calculation. For $R>r$, let $C_R(r,\theta )$ be the closed contour obtained by appending to ${\mathrm{\Gamma }}_R(r,\theta )$ the counterclockwise outer arc $R{e}^{i\alpha }$, $\theta \le \alpha \le 2\pi -\theta$, through the left half-plane. Its interior is $D_R=\{\lambda :\mathrm{\mid }\lambda \mathrm{\mid }<R,\mathrm{\mid }\lambda \mathrm{\mid }<r\text{ or }\mathrm{\mid }\arg \lambda \mathrm{\mid }>\theta \}$, with $0$ included by the first alternative; it is star-shaped about $0$. For $p$ off $C_R$, the winding number is one when $p\in D_R$ and zero when $p$ is outside $\overline{D_R}$; equivalently ${\oint }_{C_R(r,\theta )}(\lambda -p{)}^{-1}d\lambda$ is $2\pi i$ or $0$, respectively. At $p=0$, the two radial integrals of $d\lambda \mathrm{/}\lambda$ cancel, while the inner and outer arcs contribute $2i\theta$ and $i(2\pi -2\theta )$, so the index is $1$. Since $D_R$ is connected, [L5] gives the same index at every interior point. Every exterior point can move radially out beyond radius $R$ without meeting the trace, then along an outer circle, so it belongs to the unbounded component, where [L5] gives index $0$. If $z\in {\mathrm{\Sigma }}_{\delta }$ and $\theta$ is admissible for $z$, then on the outer arc $\mathrm{Re}(\lambda z)\le -cR\mathrm{\mid }z\mathrm{\mid }$ for some $c>0$, so ${\int }_{A_R}{e}^{\lambda z}(\lambda -p{)}^{-1}d\lambda \to 0$ for fixed $p$. Put $D_{\mathrm{\infty }}:={\bigcup }_{R>r}{D}_R$. If $p\in D_{\mathrm{\infty }}$, then $p\in D_R$ for every sufficiently large $R$; writing $e^{\lambda z}(\lambda -p{)}^{-1}=e^{pz}(\lambda -p{)}^{-1}+G(\lambda )$ with entire $G$ having a global primitive shows that the open contour integral tends to $2\pi i{e}^{pz}$. If instead $p\mathrm{\notin }\overline{D_{\mathrm{\infty }}}$, then $p$ lies outside every $\overline{D_R}$ and, for each finite $R$, choose $0<{\epsilon }_R<\mathrm{dist}(p,\overline{D_R})$ and set $U_{R,{\epsilon }_R}=\overline{D_R}+B(0,{\epsilon }_R)$. This is an open star-shaped neighborhood of $\overline{D_R}$ that avoids $p$; the integrand is holomorphic there, so its closed integral vanishes by [L3], and the outer arc tends to zero, giving an open contour integral equal to zero. The nested-contour cases evaluated below have $p$ either inside $D_{\mathrm{\infty }}$ or outside $\overline{D_{\mathrm{\infty }}}$, so no boundary-pole case is needed.

The resolvent identity and the inner integrals. By [L2], $R({\lambda }_1)R({\lambda }_2)=\frac{R({\lambda }_2)-R({\lambda }_1)}{{\lambda }_1-{\lambda }_2}$, so the double integral of [step 1.1] splits as $A_2-A_1$, where $A_1$ is the term with $R({\lambda }_1)$ and the inner ${\mathrm{\Gamma }}_2$ integral, and $A_2$ is the term with $R({\lambda }_2)$ and the inner ${\mathrm{\Gamma }}_1$ integral. The two infinite contours are disjoint and have positive separation $d:=\mathrm{dist}({\mathrm{\Gamma }}_1,{\mathrm{\Gamma }}_2)>0$: their finite arc pieces are disjoint compact sets, and their ray tails have distinct angles ${\theta }_1\mathrm{\ne }{\theta }_2$, so the distance between tails tends to infinity. Thus $\mathrm{\mid }{\lambda }_1-{\lambda }_2{\mathrm{\mid }}^{-1}\le d^{-1}$; together with exponential decay along the rays and the sectorial resolvent bound, this gives absolute integrability of each split double-integral term and justifies Fubini. For each fixed ${\lambda }_1\in {\mathrm{\Gamma }}_1$, the nesting in [step 1.1] puts ${\lambda }_1$ outside the closure of the full unbounded keyhole region $D_{\mathrm{\infty },2}:={\bigcup }_{R_2>r_2}{D}_{R_2}(r_2,{\theta }_2)$, so it is outside every truncated interior. For each $R_2$, the scalar integrand $e^{{\lambda }_2{z}_2}\mathrm{/}({\lambda }_1-{\lambda }_2)$ is holomorphic on an open star-shaped neighborhood of $\overline{D_{R_2}}$ avoiding its pole, so the integral on $C_{R_2}(r_2,{\theta }_2)$ is zero by [L3]. Its outer arc contribution tends to zero by exponential decay, hence the open inner ${\mathrm{\Gamma }}_2$ integral in $A_1$ vanishes. For each fixed ${\lambda }_2\in {\mathrm{\Gamma }}_2$, the nesting puts it inside the unbounded keyhole interior of ${\mathrm{\Gamma }}_1$, and therefore in $D_{R_1}$ for every sufficiently large $R_1$. On $C_{R_1}(r_1,{\theta }_1)$, the decomposition $e^{{\lambda }_1{z}_1}\mathrm{/}({\lambda }_1-{\lambda }_2)=e^{{\lambda }_2{z}_1}\mathrm{/}({\lambda }_1-{\lambda }_2)+[e^{{\lambda }_1{z}_1}-e^{{\lambda }_2{z}_1}]\mathrm{/}({\lambda }_1-{\lambda }_2)$ has an entire second term with a global primitive; by [step 1.2] the first term integrates to $2\pi i{e}^{{\lambda }_2{z}_1}$. The outer arc of the original exponential integrand tends to zero for this fixed ${\lambda }_2$, so the open inner ${\mathrm{\Gamma }}_1$ integral in $A_2$ is $2\pi i{e}^{{\lambda }_2{z}_1}$. These are pointwise evaluations of the inner improper integrals; no radius-limit is interchanged with the outer integration.

Strong continuity with the quantitative estimate. Let $x\in D(A)$ and fix $0<{\delta }^{\mathrm{\prime }}<\delta$. Choose once and for all an angle $\theta$ with $\pi \mathrm{/}2+{\delta }^{\mathrm{\prime }}<\theta <\pi \mathrm{/}2+\delta$, and choose $\epsilon >0$ so that $\theta <\pi \mathrm{/}2+\delta -\epsilon$; this same contour angle is admissible for every $z\in {\mathrm{\Sigma }}_{{\delta }^{\mathrm{\prime }}}$. Every $\mu$ on the contour is nonzero, so [L2] gives $R(\mu ,A)x-{\mu }^{-1}x={\mu }^{-1}R(\mu ,A)Ax$, and the scalar identity $\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}{e}^{\mu z}\frac{d\mu }{\mu }=1$ for $z\in {\mathrm{\Sigma }}_{{\delta }^{\mathrm{\prime }}}$ follows by closing the truncated keyhole: the closed integral is $2\pi i$ by [step 1.2] at $p=0$ plus the entire quotient $(e^{\mu z}-1)\mathrm{/}\mu$, whose integral is zero, and the outer arc of $e^{\mu z}\mathrm{/}\mu$ tends to zero. Hence $T(z)x-x=\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}{e}^{\mu z}{\mu }^{-1}R(\mu ,A)Axd\mu$. Use this fixed $\theta$ and the contour inner radius $\varrho =1\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$. Since $\mathrm{\mid }\arg z\mathrm{\mid }<{\delta }^{\mathrm{\prime }}$, on both rays $\mathrm{Re}(\mu z)\le -cs\mathrm{\mid }z\mathrm{\mid }$ with a single $c=c(\theta ,{\delta }^{\mathrm{\prime }})>0$; the two rays contribute at most $2M_{\epsilon }\mathrm{\parallel }Ax\mathrm{\parallel }\mathrm{\mid }z\mathrm{\mid }{\int }_1^{\mathrm{\infty }}{e}^{-c\sigma }{\sigma }^{-2}d\sigma$ after $\sigma =s\mathrm{\mid }z\mathrm{\mid }$, and the arc contributes at most $2\theta e{M}_{\epsilon }\mathrm{\mid }z\mathrm{\mid }\mathrm{\parallel }Ax\mathrm{\parallel }$. Thus $\mathrm{\parallel }T(z)x-x\mathrm{\parallel }\le C_{{\delta }^{\mathrm{\prime }},r}\mathrm{\mid }z\mathrm{\mid }\mathrm{\parallel }Ax\mathrm{\parallel }$ with one constant for all $z\in {\mathrm{\Sigma }}_{{\delta }^{\mathrm{\prime }}}$ and $\mathrm{\mid }z\mathrm{\mid }\le r$, and this tends to $0$ as $z\to 0$ in that smaller sector.

The semigroup law. Substituting [step 2.1] into [step 1.1] gives $T(z_1)T(z_2)=\frac{1}{(2\pi i{)}^2}{\int }_{{\mathrm{\Gamma }}_2}{e}^{{\lambda }_2{z}_2}2\pi i{e}^{{\lambda }_2{z}_1}R({\lambda }_2,A)d{\lambda }_2=\frac{1}{2\pi i}{\int }_{{\mathrm{\Gamma }}_2}{e}^{{\lambda }_2(z_1+z_2)}R({\lambda }_2,A)d{\lambda }_2=T(z_1+z_2)$, the last equality by the independence of the contour [L1], since ${\mathrm{\Gamma }}_2$ is admissible for $z_1+z_2\in {\mathrm{\Sigma }}_{{\delta }_0}$ when ${\theta }_2>\pi \mathrm{/}2+{\delta }_0$.

Extension to all $x$ and assembly. $D(A)$ is dense in $X$, and $\mathrm{\parallel }T(z)\mathrm{\parallel }\le M_{{\delta }^{\mathrm{\prime }}}$ on ${\mathrm{\Sigma }}_{{\delta }^{\mathrm{\prime }}}$ by [L1], so the uniform estimate of [step 2.2] on the dense set extends the strong limit $T(z)x\to x$ to every $x\in X$. Together with the semigroup law [step 3.1], the holomorphy and boundedness of [L1], and $T(0)=I$, this exhibits $T$ as a bounded analytic semigroup of angle $\delta$; the argument used only the resolvent identity, the contour computations and norm estimates, so no choice principle beyond Dependent Choice was used.