 Solution

Let $\eta =\mathrm{Im}k$. On $x>0$, the modulus of the Fourier factor is $e^{-\eta x}$, so the given decay bound makes $Y_{+}$ converge absolutely when $\gamma +\eta >0$. On $x<0$, write $x=-r$; the bound becomes $e^{-(\gamma -\eta )r}$, so $Y_{-}$ converges absolutely when $\eta <\gamma$. Uniform domination on compact subsets permits differentiation with respect to $k$, provingThus the half-planes guaranteed directly by the given common decay bound have $\alpha =-\gamma$ and $\beta =\gamma$, and the equation holds on their overlapThis is a statement about the domains guaranteed by that bound, not necessarily their largest possible domains. The exact solution below decays at different rates on the two half-lines and gives the sharper convergence half-planes $\mathrm{Im}k>-\mathrm{Im}a$ and $\mathrm{Im}k<\mathrm{Im}b$, with maximal common strip $-\mathrm{Im}a<\mathrm{Im}k<\mathrm{Im}b$.