 Solution

For $k=a+ib$, $\mathrm{Re}\omega =a^2-b^2+\beta b$. The upper contour is therefore the graphoriented from left to right. Between the real axis and $L$, $b\ge 0$ and $\mathrm{Re}\omega \ge 0$. Hence $e^{ikx-\omega (t-s)}$ is bounded by $e^{-bx}$ for $0\le s\le t$. The boundary transforms are entire in $k$; with the standard smooth, compatible boundary data, integration by parts in the time integral gives the required polynomial bounds. Closing the intervening sectors and using exponential decay for $x>0$ deforms their real-line integrals onto $L$. Small endpoint truncations, followed by dominated limits, handle the sector boundaries.

For $k^{\prime }=-k+i\beta$, $\omega (k^{\prime })=\omega (k)$ and $\mathrm{Im}{k}^{\prime }\le 0$ on $L$. Thus the already valid transformed equation at $k^{\prime }$ saysIt eliminates the unknown boundary derivative through