<s>[INST] 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?
 [/INST] 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 $n \rightarrow \infty$.
</s>