How fast does $a_n:=\int_{1/\sqrt{2}}^1 \frac{dx}{\left(\frac{1}{2}+x^2\right)^{n+1/2}}$ decay as $n \to \infty$? Let
\begin{equation}
a_n:=\int_{1/\sqrt{2}}^1 \frac{dx}{\left(\frac{1}{2}+x^2\right)^{n+1/2}}
\end{equation}
for $n \in \mathbb{N}$.
Then, $a_n$ is clearly a monotone decreasing sequence of positive numbers and by the Dominated Convergence Theorem, $a_n \to 0^+$ as $n \to \infty$.
However, I have some difficulty estimating how fast $a_n$ decays. For example, $\frac{a_n}{f(n)}=O(1)$ as $n \to \infty$ for some polynomial $f$?
Could anyone please provide insight into the decay rate of $a_n$?
 A: We can also try to find the asymptotics of the integral directly.
Making the substitution $x=\frac{t}{\sqrt2}$
$$a_n=\int_{1/\sqrt{2}}^1 \frac{dx}{\left(\frac{1}{2}+x^2\right)^{n+1/2}}=2^n\int_1^\sqrt2\frac{dt}{(1+t^2)^{n+1/2}}=2^n\int_1^\sqrt2\frac{e^{-n\ln(1+t^2)}}{\sqrt{1+t^2}}dt$$
Making one more substitution $t=1+x$
$$a_n=\frac{1}{\sqrt2}\int_0^{\sqrt{2}-1} \frac{e^{-n\ln(1+x+x^2/2)}}{\sqrt{1+x+x^2/2}}dx$$
It is straightforward to see that the function $f(x)=\ln(1+x+x^2/2)$ does not have an extremum on the interval $[0;\sqrt2-1]$, but has the minimal value at $x=0$. Decomposing the integrand near this point
$$\ln(1+x+x^2/2)=f(0)+f'(0)\,x+f''(0)\frac{x^2}{2}+...=x+O(x^3)$$
$$\frac{1}{\sqrt{1+x+x^2/2}}=1-\frac{1}{2}\,x+O(x^2)$$
$$a_n=\frac{1}{\sqrt2}\int_0^{\sqrt2-1}e^{-nx}\Big(1-\frac{x}{2}+O(x^2)\Big)dx$$
With the accuracy up to exponentially small corrections we can extend the interval of integration till $\infty$. Performing integration,
$$\boxed{\,\,a_n=\frac{1}{\sqrt2}\frac{1}{n}\Big(1-\frac{1}{2n}+O\big(1/n^2\big)\Big)\,\,}$$
Numerical check confirms the asymptotics:
$$n=10\quad a_n=0.0667968\quad \text{approximation}=\frac{1}{10\sqrt2}(1-1/20)=0.067175$$
$$n=30\quad a_n=0.0232066\quad \text{approximation}=\frac{1}{30\sqrt2}(1-1/60)=0.023177$$
$$n=100\quad a_n=0.0070366\quad \text{approximation}=\frac{1}{100\sqrt2}(1-1/200)= 0.00703571$$
$$n=1000\quad a_n=0.000706754\quad \text{approximation}=\frac{1}{1000\sqrt2}(1-1/2000)= 0.000706753$$
A: Here is a more elementary answer to close the debate. By the change of variable $t = n\,(x^2-1/2)$ (which gives $x^2 + 1/2 = 1+t/n$) one gets
$$
\int_{\frac{1}{\sqrt 2}}^1 \frac{\mathrm d x}{\left(\tfrac{1}{2}+x^2\right)^{n+1/2}} = \frac{1}{2n}\int_0^\frac{n}{2} \frac{\mathrm d t}{\left(1+\frac{t}{n}\right)^{n+1/2}\left(\frac{1}{2}+\frac{t}{n}\right)^{1/2}} = \frac{I_n}{2n}
$$
where $I_n = \int_0^\infty f_n$ with
$$
f_n(t) = \frac{\mathbb{1}_{[0,n]}(t)}{\left(1+\frac{t}{n}\right)^n\left(1+\frac{t}{n}\right)^{1/2}\left(\frac{1}{2}+\frac{t}{n}\right)^{1/2}} \underset{n\to\infty}\longrightarrow \frac{1}{e^t \left(\tfrac{1}{2}\right)^{1/2}}
$$
since $(1+t/n)^n \to e^t$. Hence, by dominated convergence,
$$
I_n = \int_0^\infty f_n \underset{n\to\infty}\longrightarrow\int_0^\infty \sqrt{2} \,e^{-t}\,\mathrm d t = \sqrt 2
$$
that is $I_n = \sqrt{2} + o(1)$. Thus,
$$
\int_{1/\sqrt 2}^1 \frac{\mathrm d x}{(\tfrac{1}{2}+x^2)^{n+1/2}} = \frac{1}{\sqrt 2}\frac{1}{n} + o\left(\frac{1}{n}\right).
$$
In particular, this agrees with the coefficient given by Svyatoslav.
A: If we perform the substitution $t = \sqrt {{\rm 2e}^s  - 1}$ in Svyatoslav's formula
$$
a_n  = 2^n \int_1^{\sqrt 2 } {\frac{{{\rm e}^{ - n\log (1 + t^2 )} }}{{\sqrt {1 + t^2 } }}{\rm d}t} ,
$$
we obtain
$$
a_n  = \frac{1}{{\sqrt 2 }}\int_0^{\log (3/2)} {{\rm e}^{ - ns} \frac{{{\rm d}s}}{{\sqrt {2 - {\rm e}^{ - s} } }}} .
$$
Now
$$
\frac{1}{{\sqrt {2 - {\rm e}^{ - s} } }} = \sum\limits_{k = 0}^\infty  {c_k \frac{{s^k }}{{k!}}} 
$$
for $|s|<\log 2$, where
$$\boxed{
c_k  = ( - 1)^k \sum\limits_{j = 0}^k {\frac{{(2j - 1)!!}}{{2^j }}S(k,j)} .}
$$
Here $j!!$ is the double factorial and $S(k,j)$ denotes the Stirling numbers of the second kind (cf. $\mathrm{A}352117$). Thus, by Watson's lemma,
$$\boxed{
a_n  \sim \frac{1}{{\sqrt 2 n}}\sum\limits_{k = 0}^\infty  {\frac{{c_k }}{{n^k }}}  = \frac{1}{{\sqrt 2 n}}\left( {1 - \frac{1}{{2n}} + \frac{5}{{4n^2 }} - \frac{{37}}{{8n^3 }} + \frac{{377}}{{16n^4 }} -  \ldots } \right)}
$$
as $n\to +\infty$. Note that, in fact,
$$
a_n  = \frac{1}{{\sqrt 2 n}}\sum\limits_{k = 0}^\infty  { P (k + 1,n\log (3/2))\frac{c_k}{{n^k }}} 
$$
where $P(a,z)$ is the normalised incomplete gamma function.
A: Alternative solution:
With the substitution $x = \sqrt y$, we have
$$a_n = \int_{1/2}^1 \frac{1}{(1/2 + y)^{n+1/2}}\frac{1}{2\sqrt y}\,\mathrm{d}y.$$
It is easy to prove that, for all $y \in [1/2, 1]$,
$$\frac{\sqrt 2}{2} - \frac{\sqrt 2}{2}(y - 1/2) \le \frac{1}{2\sqrt y} \le \frac{\sqrt 2}{2}.$$
We have
$$\int_{1/2}^1 \frac{\frac{\sqrt 2}{2}}{(1/2 + y)^{n+1/2}}\,\mathrm{d}y - \int_{1/2}^1 \frac{\frac{\sqrt 2}{2}(y - 1/2)}{(1/2 + y)^{n+1/2}}\,\mathrm{d}y \le a_n \le \int_{1/2}^1 \frac{\frac{\sqrt 2}{2}}{(1/2 + y)^{n+1/2}}\,\mathrm{d}y$$
which results in
\begin{align*}
 &\frac{\sqrt 2}{2n-1} - \frac{\sqrt 2}{2n-1}(2/3)^{n-1/2} - \frac{2\sqrt 2}{4n^2 - 8n + 3} + \frac{(n + 3/2)\sqrt 2}{4n^2 - 8n + 3}(2/3)^{n-1/2}\\ 
 \le{}& a_n \le \frac{\sqrt 2}{2n-1} - \frac{\sqrt 2}{2n-1}(2/3)^{n-1/2}.
\end{align*}
Thus,
$$a_n = \frac{\sqrt 2}{2n} + O\left(\frac{1}{n^2}\right).$$
