 Solution

Write ${\hat{u}}_0(k)=\hat{u}(k,0)$ and retain the left-to-right orientation of $L$. Substituting the previous relation into the deformed inverse Fourier transform gives the data-only representationThe formally remaining integral is ${\int }_L{e}^{ikx}\hat{u}(-k+i\beta ,t)dk$. Above $L$, its reflected argument lies strictly in the lower half-plane, where the transform is analytic. At large radius, $L$ approaches the rays of angles $\pi /4$ and $3\pi /4$. Close it by the upper arc between these rays. Along that arc $\mathrm{Im}k$ is bounded below by a positive multiple of its radius, whereas the reflected transform is bounded by ${\int }_0^{\infty }|u(s,t)|ds$. Thus the arc integral is bounded by a constant times $R{e}^{-cRx}$ and tends to zero for $x>0$. There are no enclosed singularities, so Cauchy integral theorem provesThis is the needed closure argument; it does not assume the unknown transform vanishes. The representation holds for smooth decaying compatible data, and extends to weaker data by the usual limiting heat-kernel interpretation.