How fast does the sequence $a_n := \int_{1/8}^{1/2} \Bigl(\frac{1}{2}+x^2\Bigr)^{1/2+n} dx$ decays as $n \to \infty$? The question is as in the title.
The sequence $\{a_n \}$ define by
\begin{equation}
a_n := \int_{1/8}^{1/2} \Bigl(\frac{1}{2}+x^2\Bigr)^{1/2+n} dx
\end{equation}
decays to $0$ as $n \to \infty$, which is a direct consequence of the Dominated Convergence Theorem.
But, when I run the Wolfram  Alpha to evluate each $a_n$, I obtain the hypergeometric functions:
\begin{equation}
a_n=2^{-n-7/2}\Bigl(4 {_2}{F}_1\bigl(\frac{1}{2}, -n -\frac{1}{2}; \frac{3}{2} ; -\frac{1}{2}\bigr)-{_2}{F}_1\bigl(\frac{1}{2}, -n, -\frac{1}{2}; \frac{3}{2} ; -\frac{1}{32}\bigr) \Bigr)
\end{equation}
But I cannot find information on the asymptotics of the above hypergeometric functions as $n \to \infty$.
Could anyone please help me?
 A: If you perform the substitution $u = -\ln(1/2+x^2)$ you have that
\begin{equation}
a_n = \int_{1/8}^{1/2} \left( \frac{1}{2} + x^2\right)^{1/2+n} = \frac{1}{2} \int_{\ln(4/3)}^{\ln(64/33)} \frac{e^{-3u/2}}{\sqrt{e^{-u}-1/2}} e^{-nu} du
\end{equation}
This can be solved by Laplace's method. The minimum of $p(u)=u$ is at $\ln(4/3)$. Hence giving that at this point,
\begin{equation}
q(u) = \frac{e^{-3u/2}}{\sqrt{e^{-u}-1/2}} = \frac{3\sqrt{3}}{4} + \frac{9\sqrt{3}}{8}\left( u-\ln(4/3) \right)^2 +...
\end{equation}
and $p(u) = u \Rightarrow p(u) = \ln(4/3) + (u - \ln(4/3))$,
the two first terms are given by
\begin{equation}
a_n = \frac{1}{2} \int_{\ln(4/3)}^{\ln(64/33)} \frac{e^{-3u/2}}{\sqrt{e^{-u}-1/2}} e^{-nu} du = \left(\frac{3}{4}\right)^{n+1}\frac{\sqrt{3}}{2} \left(\frac{1}{n} + \frac{3}{n^3} + O(n^{-4}) \right)
\end{equation}
when $n \rightarrow \infty$.
A: With the substitution $t = \log \left( {\frac{3}{4}} \right) - \log \left( {x^2  + \frac{1}{2}} \right)$, your integral becomes
$$
a_n  = \left( {\frac{3}{4}} \right)^{n + 3/2} \int_0^{\log (16/11)} {{\rm e}^{ - nt} \frac{{{\rm e}^{ - t} }}{{\sqrt {3 - 2{\rm e}^t } }}{\rm d}t} .
$$
Now
$$
\frac{{{\rm e}^{ - t} }}{{\sqrt {3 - 2{\rm e}^t } }}= 
1 + \frac{3}{2}t^2  + \frac{5}{2}t^3  + \frac{{23}}{4}t^4  +  \ldots  
$$
when $|t|<\log(3/2)$. Thus, by Watson's lemma,
$$
a_n  \sim \left( {\frac{3}{4}} \right)^{n + 3/2} \frac{1}{n}\left(
1 + \frac{3}{{n^2 }} + \frac{{15}}{{n^3 }} + \frac{{138}}{{n^4 }} + \frac{{1545}}{{n^5 }} +  \ldots \right)
$$
as $n\to +\infty$.
Alternatively,
$$
\frac{1}{{\sqrt {3 - 2{\rm e}^t } }} = \sum\limits_{k = 0}^\infty  {c_k \frac{{t^k }}{{k!}}}
$$
for $|t|<\log (3/2)$, where
$$
c_k  = \sum\limits_{j = 0}^k {(2j - 1)!!S(k,j)}.
$$
Here $j!!$ is the double factorial and $S(k,j)$ denotes the Stirling numbers of the second kind (cf. $\mathrm{A}305404$). Thus, by Watson's lemma,
\begin{align*}
a_n & \sim \left( {\frac{3}{4}} \right)^{n + 3/2} \frac{1}{n+1}\sum\limits_{k = 0}^\infty  {\frac{{c_k }}{{(n + 1)^k }}} \\ & = \left( {\frac{3}{4}} \right)^{n + 3/2} \frac{1}{n+1}\left( {1 + \frac{1}{{n + 1}} + \frac{4}{{(n + 1)^2 }} + \frac{{25}}{{(n + 1)^3 }} +  \ldots } \right)
\end{align*}
as $n\to +\infty$.
