= Solution
Let $D_1,D_2,D_3,D_4$ denote the open quadrants, counterclockwise from the first. Write $\theta=kx+2k^2t$. The <Lax pair> has the <Volterra integral equation> kernel $e^{-i[k(x-x')+2k^2(t-t')]\widehat\sigma_3}$, obtained by integrating its closed matrix differential form from a normalization point.
In column one, the off-diagonal factor is $e^{2i[k(x-x')+2k^2(t-t')]}$, whose modulus is $e^{-2\operatorname{Im}[k(x-x')+2k^2(t-t')]}$. Column two has the reciprocal factor. For $\mu_2$, choose the path $(0,0)\to(0,t)\to(x,t)$, so both differences are nonnegative. Column one requires $\operatorname{Im}k\geq0$ and $\operatorname{Im}k^2\geq0$, giving $D_1$; column two requires the opposite inequalities, giving $D_4$.
For $\mu_1$, the time path starts at $T$ and runs backward, while the spatial path runs rightward. Its column-one conditions are $\operatorname{Im}k\geq0$, $\operatorname{Im}k^2\leq0$, giving $D_2$; column two is bounded in $D_3$.
Finally, $\mu_3$ integrates spatially from infinity to $x$, where $x-x'\leq0$. Column one is bounded in the lower half-plane and column two in the upper half-plane. Normalization at $(\infty,T)$ is equivalent to normalization at $(\infty,t)$ because the decaying potential vanishes on the vertical path at infinity. Thus
$$
\boxed{\mu_1:(D_2,D_3),\qquad\mu_2:(D_1,D_4),\qquad\mu_3:(\mathbb C_-,\mathbb C_+).}
$$
The <Volterra integral equation> Neumann series converges locally uniformly under the usual integrable smooth scattering assumptions, proving holomorphic dependence in the open domains and continuous boundary values. These are the <Volterra analyticity sectors for a half-line NLS Lax pair>; at $x=0$, absence of a spatial segment allows the larger time-only domains used for $A,B$.
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