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Substitute part (b) into the spectral representation. The terms containing $\psi (ik)$ areThe transform $\psi (ik)$ is analytic for $\mathrm{Im}k>0$. Write $k=a+ib$. In the first quadrant,which decays since $x,y>0$. The Cauchy integral theorem in that quadrant equates the two integrals in the first parentheses. In the second quadrant,which decays since $a<0$, $b>0$, and $y<\ell$. The same contour integration equates the two integrals in the second parentheses. Therefore $\psi (ik)$ makes no contribution. This is upper-quadrant cancellation of reflected boundary transforms, and it uses the whole paired expression rather than attempting to discard an individual integral.

The terms involving $c$ cancel by the same two quadrant arguments with the analytic factors $k\pm \gamma$ and $\psi$ omitted. There is a sign defect in the printed reconstruction prefactor. The sides specified in part (a) form a clockwise boundary, so the Cauchy integral theorem gives a negative reconstruction sign. Indeed, integrating each spectral ray first yields $i/(z-\zeta )$, and hence the sum of the three ray integrals is $i{\oint }_{\mathrm{clockwise}}{q}_z(\zeta )/(z-\zeta )d\zeta =-2\pi q_z(z)$. Thus the printed positive prefactor reconstructs $-q_z$. The requested cancellation above holds with either overall sign, but the actual data-dependent derivative isDecay at infinity fixes the integration constant when recovering $q$ from its Wirtinger derivative.

For completeness, the missing condition at infinity and the unrestricted printed $\gamma$ deserve an explicit solvability of a decaying Robin strip check. Let $Y_n$ be orthonormal eigenfunctions from a Sturm-Liouville problem of $-d^2/d{y}^2$ with $Y_n^{\prime }(0)=\gamma Y_n(0)$ and $Y_n^{\prime }(\ell )=-\gamma Y_n(\ell )$, and eigenvalues ${\lambda }_n$. With $g_n={\int }_0^{\ell }g{Y}_{n}dy$, separation of variables givesIndeed, each coefficient obeys $q_n^{\prime \prime }-{\lambda }_n{q}_n=0$ and $q_n^{\prime }(0)=g_n$. A positive eigenvalue has one decaying exponential; a zero eigenvalue has only affine solutions and a negative eigenvalue only oscillatory solutions, neither of which decays unless its coefficient vanishes. This also proves uniqueness in the decay class and justifies reflection symmetry there. For $\gamma >0$, all eigenvalues are positive. For $\gamma =0$, the constant eigenfunction requires ${\int }_0^{\ell }gdy=0$; the apparent zero of $D$ at the origin is then removable. For $\gamma <0$, compatibility with every nonpositive eigenmode is necessary. These restrictions cannot be inferred from smoothness and reflection symmetry alone. The printed boundary value problem without any far-field condition permits additional growing or nondecaying homogeneous solutions, whereas the spectral transforms select the decaying branch.