The limit and Asymptotic Behavior Wallis-like Integral It is known that $$I_n=\int^{\frac{\pi}{2}}_{0} x \cos^n xdx \quad (n=0,1,2,3,\dots),$$ then
$$\lim_{n \to \infty} nI_{n}=1.$$
My question is: does there exist $k$ such that
$$\lim_{n \to \infty} n^k(nI_{n}-1)$$
converges to a non-zero real number?
 A: You can use Laplace's method to establish the asymptotics
$$
\int_0^{\pi /2} {x\cos ^n xdx}  = \int_0^{\pi /2} {xe^{n\log \cos x} dx}  = \frac{1}{n} - \frac{2}{3} \frac{1}{n^2} + \mathcal{O}\!\left( {\frac{1}{{n^{5/2} }}} \right).
$$
Thus
$$
n\cdot (nI_n  - 1) \to  - \frac{2}{3} ,
$$
i.e., $k=1$.
Addendum 1. A more straightforward way is to use Watson's lemma. By the change of integration variable $x = \arccos e^{ - t}$,
$$
I_n  = \int_0^{ + \infty } {e^{ - nt} \frac{{\arccos (e^{ - t} )}}{{\sqrt {e^{2t}  - 1} }} dt} .
$$
Near the origin,
$$
\frac{{\arccos (e^{ - t} )}}{{\sqrt {e^{2t}  - 1} }} = 1 - \frac{2}{3}t + \frac{2}{{15}}t^2  + \frac{4}{{315}}t^3  +  \cdots .
$$
Therefore, by Watson's lemma,
$$
I_n  \sim \frac{1}{n} - \frac{2}{3}\frac{1}{{n^2 }} + \frac{4}{{15}}\frac{1}{{n^3 }} + \frac{8}{{105}}\frac{1}{{n^4 }} +  \cdots 
$$
as $n\to +\infty$.
Addendum 2. Just for fun. You can rewrite $I_n$ in the form
$$
I_n  = \int_0^{ + \infty } {e^{ - (n + 1)t} \frac{{\arcsin (\sqrt {1 - e^{ - 2t} } )}}{{\sqrt {1 - e^{ - 2t} } }}dt} .
$$
Substituting the Taylor series of the inverse sine around the origin yields the series
\begin{align*}
I_n  & = \sum\limits_{k = 0}^\infty  {\frac{{(2k)!}}{{(2^k k!)^2 }}\frac{1}{{2k + 1}}\int_0^{ + \infty } {e^{ - (n + 1)t} (1 - e^{ - 2t} )^k dt} } \\ & = \sum\limits_{k = 0}^\infty  {\frac{{(2k)!}}{{2^k k!}}\frac{1}{{2k + 1}}\frac{1}{{(n + 1)(n + 3) \cdots (n + 2k + 1)}}} ,
\end{align*}
which coverges for all positive integer $n$.
Addendum 3. An elementary argument. You make the substitution as in Addendum 1. Because the function tends to $0$ exponentially fast at infinity, taking into account its Taylor series, it is readily seen that there is a $C>0$ such that $$
\left| \frac{\arccos (e^{ - t} )}{\sqrt {e^{2t}  - 1} } - 1 + \frac{2}{3}t\right| \le Ct^2 
$$ for all $t>0$. Then
\begin{align*}
\left| {I_n  - \frac{1}{n} + \frac{2}{3}\frac{1}{{n^2 }}} \right| & = \left| {\int_0^{ + \infty } {e^{ - nt} \left( {\frac{{\arccos (e^{ - t} )}}{{\sqrt {e^{2t}  - 1} }} - 1 + \frac{2}{3}t} \right)dt} } \right| \\ & \le \int_0^{ + \infty } {e^{ - nt} \left| {\frac{{\arccos (e^{ - t} )}}{{\sqrt {e^{2t}  - 1} }} - 1 + \frac{2}{3}t} \right|dt}  \le C\int_0^{ + \infty } {e^{ - nt} t^2 dt}  = \frac{{2C}}{{n^3 }}
\end{align*}
for all $n\geq 1$.
