 Parameter-shift remainder bound for a logarithmic Laplace integral

For $g(s)=-(1/a)e^{ias}{\int }_0^{\infty }{x}^{s+\sigma -1}{e}^{-tx}/(ia+\log x)dx$ with positive $a,t,\sigma$ and $s\ge 0$, differentiation cancels the logarithmic denominator and gives $g_s=-(1/a)e^{ias}\Gamma (s+\sigma )t^{-s-\sigma }$. Integrating this derivative expresses $g(0)$ as a finite parameter-Laplace integral plus $g(s)$. The elementary lower bound $|ia+\log x|\ge a$ gives the displayed remainder, smaller than every logarithmic term at fixed $s>0$. Watson's lemma then determines all coefficients from the finite integral's endpoint Taylor series.