= Solution
For real $s$, the multiplier $e^{ias}$ has modulus one and
$$
|ia+\log x|=\sqrt{a^2+\log^2x}.
$$
The <triangle inequality> applied to the <integral> therefore gives
$$
\boxed{|g(s;t,a,\sigma)|\le\frac1a\int_0^\infty
\frac{x^{s+\sigma-1}e^{-tx}}{\sqrt{a^2+\log^2x}}\,dx}.
$$
The 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:
$$
\begin{aligned}
g_s(s;t,a,\sigma)
&=-\frac1a e^{ias}\int_0^\infty
\frac{(ia+\log x)x^{s+\sigma-1}e^{-tx}}{ia+\log x}\,dx\\
&=-\frac1a e^{ias}\frac{\Gamma(s+\sigma)}{t^{s+\sigma}}.
\end{aligned}
$$
Here the scaled <Gamma function> <integral> is obtained by $u=tx$. Integrating this <derivative> from zero to $s$ yields
$$
\boxed{g(0;t,a,\sigma)=\frac1a\int_0^s
e^{iar}\Gamma(r+\sigma)t^{-r-\sigma}\,dr+g(s;t,a,\sigma)}.
$$
The same modulus bound also gives the useful <parameter-shift remainder bound for a logarithmic Laplace integral>
$$
\boxed{|g(s;t,a,\sigma)|\le\frac{\Gamma(s+\sigma)}{a^2t^{s+\sigma}}},
$$
because the denominator magnitude is at least $a$.
Back to article page
