 Solution

Let $k=s+i\eta$. On the negative half-line, the integrand magnitude is bounded at infinity by a constant times $e^{(\alpha -\eta )x}$, which is integrable as $x\to -\infty$ when $\eta <\alpha$. On the positive half-line the corresponding bound is $e^{(\beta -\eta )x}$, integrable when $\eta >\beta$. ThusOn compact subsets strictly inside these half-planes, the same exponential domination absorbs every power of $|x|$. Differentiation under the integral is therefore valid and proves holomorphy, rather than only absolute convergence. The bars here denote the printed one-sided Fourier transforms, not complex conjugation.