 Solution

Let $D_1,D_2,D_3,D_4$ denote the open quadrants, counterclockwise from the first. Write $\theta =kx+2k^{2}t$. The Lax pair has the Volterra integral equation kernel $e^{-i[k(x-x^{\prime })+2k^2(t-t^{\prime })]{\hat{\sigma }}_3}$, obtained by integrating its closed matrix differential form from a normalization point.

In column one, the off-diagonal factor is $e^{2i[k(x-x^{\prime })+2k^2(t-t^{\prime })]}$, whose modulus is $e^{-2\mathrm{Im}[k(x-x^{\prime })+2k^2(t-t^{\prime })]}$. 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 $\mathrm{Im}k\ge 0$ and $\mathrm{Im}{k}^2\ge 0$, 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 $\mathrm{Im}k\ge 0$, $\mathrm{Im}{k}^2\le 0$, giving $D_2$; column two is bounded in $D_3$.

Finally, ${\mu }_3$ integrates spatially from infinity to $x$, where $x-x^{\prime }\le 0$. 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. ThusThe 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 $A,B$.