Consider the sequence
$$a_n=\int_0^\infty \frac{s^{n-1}}{(n-1)!}e^{-g(s)}ds, \quad n \in\Bbb N,$$
where $g\colon [0,\infty)\to[0,\infty)$ is an increasing function. I'm trying to understand how its behavior influences the asymptotics of $a_n$ for $n\to\infty$.
If $g(s)=cs$ for a non-negative constant $c$, then transforming the integral and using the gamma function shows that $a_n=c^{-n}$, so the sequence either goes to zero very fast ($c>1$), or it is constant ($c=1$), or it goes to infinity very fast ($c<1$).
$c=1$ is the critical case here, and the behavior changes drastically there: If $g$ is linear, then there is no scenario in which $a_n$ decays slowly. So what if $g(s)=s+as^b$ with $a>0$ and $b\in(0,1)$? Clearly, $a_n$ is still bounded by 1 then, but I have a hard time figuring out whether it goes to zero and how fast, or whether it is summable.
If anyone could give me some advice or refer me to a helpful result in this direction, I'd be very thankful.
Substituting $s = ne^t$ and splitting the integral at $t=0$ into two parts, we have \begin{align*} \int_0^{ + \infty } {s^{n - 1} e^{ - s - as^b } ds} = &\; n^n e^{ - n - an^b } \int_{ - \infty }^{ + \infty } {\exp \left( { - n(e^t - t - 1) + n^b (a - ae^{bt} )} \right)dt} \\ = &\; n^n e^{ - n - an^b } \int_0^{ + \infty } {\exp \left( { - n(e^t - t - 1) + n^b (a - ae^{bt} )} \right)dt} \\ & + n^n e^{ - n - an^b } \int_0^{ + \infty } {\exp \left( { - n(e^{ - t} + t - 1) + n^b (a - ae^{ - bt} )} \right)dt} . \end{align*} Employing Theorem 2.1 on page 326 of F. W. J. Olver's book Asymptotics and Special Functions, we find $$ \int_0^{ + \infty } {s^{n - 1} e^{ - s - as^b } ds} \sim n^{n - 1/2} e^{ - n - an^b } \sqrt {2\pi } $$ as $n\to +\infty$ provided that $0 \leq b<\frac{1}{2}$. Thus, by Stirling's formula $$ a_n \sim e^{ - an^b } $$ as $n\to +\infty$ provided that $0 \leq b<\frac{1}{2}$.
If $b=\frac{1}{2}$, we apply Theorem 4.1 on page 332 of Olver's book instead and find $$ \int_0^{ + \infty } {s^{n - 1} e^{ - s - as^{1/2} } ds} \sim n^{n - 1/2} e^{ - n - an^{1/2} } \sqrt {2\pi } e^{a^2/8} $$ as $n\to +\infty$, i.e, $$ a_n \sim e^{a^2/8} e^{- an^{1/2} } $$ as $n\to +\infty$. Let me add that in Olver's theorems the intervals of integration are finite, but since in our cases the tails are asymptotically negligible, this is not an issue.
For the remaining case $\frac{1}{2}<b<1$, I give a rough estimate showing the dacay to $0$. With $t = \frac{x}{{\sqrt n }}$, \begin{align*} & n^n e^{ - n - an^b } \int_0^{ + \infty } {\exp \left( { - n(e^t - t - 1) + n^b (a - ae^{bt} )} \right)dt} \\ & < n^n e^{ - n - an^b } \int_0^{ + \infty } {\exp \left( { - n\frac{{t^2 }}{2} - abn^b t} \right)dt} \\ & = n^{n - 1/2} e^{ - n - an^b } \int_0^{ + \infty } {\exp \left( { - \frac{{x^2 }}{2} - abn^{b - 1/2} x} \right)dx} \\ & < n^{n - 1/2} e^{ - n - an^b } \int_0^{ + \infty } {\exp \left( { - abn^{b - 1/2} x} \right)dx} < n^{n - 1/2} e^{ - n - an^b } \frac{1}{{abn^{b - 1/2} }}. \end{align*} Also, \begin{align*} & n^n e^{ - n - an^b } \int_0^{ + \infty } {\exp \left( { - n(e^{ - t} + t - 1) + n^b (a - ae^{ - bt} )} \right)dt} \\ & < n^n e^{ - n - an^b } \int_0^{ + \infty } {\exp \left( { - nt + an^b } \right)dt} = n^{n - 1} e^{ - n} . \end{align*} Applying the inequality $$\frac{1}{{(n - 1)!}} < \frac{1}{{n^{n - 1/2} e^{ - n} \sqrt {2\pi } }} $$ gives $$ a_n < \frac{{e^{ - an^b } }}{{\sqrt {2\pi } }}\frac{1}{{abn^{b - 1/2} }} + \frac{1}{{\sqrt {2\pi n} }}. $$