 Solution

For real $s$, the multiplier $e^{ias}$ has modulus one andThe triangle inequality applied to the integral therefore givesThe hypotheses $s+\sigma >0$ and $t>0$ ensure endpoint convergence. Differentiation in $s$ is justified on compact parameter intervals by integrable bounds with an additional factor $1+|\log x|$. Differentiating both the exponential multiplier and the power of $x$ makes the logarithmic denominator cancel:Here the scaled Gamma function integral is obtained by $u=tx$. Integrating this derivative from zero to $s$ yieldsThe same modulus bound also gives the useful parameter-shift remainder bound for a logarithmic Laplace integralbecause the denominator magnitude is at least $a$.