= Solution
Let $\eta=\operatorname{Im}k$. 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 $k$, proving
$$
\boxed{Y_+\text{ holomorphic for }\operatorname{Im}k>-\gamma,\qquad Y_-\text{ holomorphic for }\operatorname{Im}k<\gamma.}
$$
Thus the half-planes guaranteed directly by the given common decay bound have $\alpha=-\gamma$ and $\beta=\gamma$, and the equation holds on their overlap
$$
\boxed{-\gamma<\operatorname{Im}k<\gamma.}
$$
This 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 $\operatorname{Im}k>-\operatorname{Im}a$ and $\operatorname{Im}k<\operatorname{Im}b$, with maximal common strip $-\operatorname{Im}a<\operatorname{Im}k<\operatorname{Im}b$.
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