= Solution
Fix a small $s>0$, for example $s<\sigma/2$, independently of $t$. Taking imaginary parts of the identity in part (a) gives
$$
F(t,a,\sigma)=t^{-\sigma}\int_0^s e^{-Lr}h(r)\,dr
+\operatorname{Im}g(s;t,a,\sigma),\qquad
L=\log t,\quad h(r)=\frac{\sin(ar)}a\Gamma(\sigma+r).
$$
The <parameter-shift remainder bound for a logarithmic Laplace integral> makes the last term $O(t^{-\sigma-s})$, smaller than $t^{-\sigma}L^{-k}$ for every fixed $k$.
The endpoint function $h$ is analytic near zero. For any fixed $N$, its <Taylor expansion> is
$$
h(r)=\sum_{n=0}^N\frac{h^{(n)}(0)}{n!}r^n+O(r^{N+1}).
$$
Integrating this expansion against $e^{-Lr}$ uses
$$
\int_0^\infty e^{-Lr}r^n\,dr=\frac{n!}{L^{n+1}},
$$
while the error is $O(L^{-N-2})$ and the extension from $s$ to infinity contributes an exponentially small tail. This is <Watson's lemma>, and gives
$$
\boxed{F(t,a,\sigma)=t^{-\sigma}\left[
\sum_{n=0}^N\frac{\beta_n(a,\sigma)}{(\log t)^{n+1}}
+O\bigl((\log t)^{-N-2}\bigr)\right],\qquad
\beta_n=h^{(n)}(0)}.
$$
Equivalently the coefficients have the stated exponential generating function $h(r)=\sum_{n\ge0}\beta_nr^n/n!$. In particular,
$$
\beta_0=0,\quad\beta_1=\Gamma(\sigma),\quad
\beta_2=2\Gamma'(\sigma),\quad
\beta_3=3\Gamma''(\sigma)-a^2\Gamma(\sigma).
$$
Thus the apparent $t^{-\sigma}(\log t)^{-1}$ term vanishes: the first nonzero terms are $t^{-\sigma}[\Gamma(\sigma)(\log t)^{-2}+2\Gamma'(\sigma)(\log t)^{-3}+\cdots]$. The <derivatives of the gamma function> supply the successive logarithmic coefficients.
Back to article page
