= Solution
{c}
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 $\eta>\beta$. Thus
$$
\boxed{\overline f\text{ is analytic for }\operatorname{Im}k<\alpha,\qquad
\overline g\text{ is analytic for }\operatorname{Im}k>\beta.}
$$
On 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.
Back to article page
