<s>[INST] Asymptotic of integral I have this definite integral:
$$ \int_{0}^{\frac \pi 2} \left( 1 + t^2 \cot^2\theta \right )^{-\frac{n-1}{2}} d \theta , t\in (0,1)$$
where $n$ is an integer. I need to find the asymptotic as a function on $n$. 
I suspect it should be $O(\frac{1}{\sqrt{n}})$ but wasn't able to complete the calculation.
Any ideas?
 [/INST] One may write:
$$ 
\begin{align} 
I(n)=\int_{0}^{\large \frac \pi 2} \left( 1
 + t^2 \cot^2\theta \right )^{-\frac{n-1}{2}} d \theta&=\int_{0}^{\large \frac \pi 2} \left( 1 + t^2 \tan^2\theta
   \right )^{-\frac{n-1}{2}} d \theta \\\\
&=\int_{0}^{\infty}
   \frac{1}{1+x^2}\left( 1 + t^2 x^2 \right )^{-\frac{n-1}{2}} dx
\end{align}
$$
Upper bound ($t\leq1$):
$$
\begin{align}
I(n) &=\int_{0}^{\infty}
   \frac{1}{1+x^2}\left( 1 + t^2 x^2 \right )^{-\frac{n-1}{2}} dx \\\\
&\leq\int_{0}^{\infty}
   \frac{1}{1+t^2x^2}\left( 1 + t^2 x^2 \right )^{-\frac{n-1}{2}} dx \\\\
&= \int_{0}^{\infty}
   \left( 1 + t^2 x^2 \right )^{-\frac{n+1}{2}} dx \\\\
&= \frac1t \int_{0}^{\infty}
   \left( 1 + x^2 \right )^{-\frac{n+1}{2}} dx = \frac1t f(n)
\end{align}
$$
Lower bound(for $t\leq 1$):
$$
\begin{align}
I(n) &\geq \int_{0}^{\infty}
   \frac{1}{1+x^2}\left( 1 + x^2 \right )^{-\frac{n-1}{2}} dx \\\\
&= \int_{0}^{\infty}
   \left( 1 + x^2 \right )^{-\frac{n+1}{2}} dx =f(n)\\\\
\end{align}
$$
Thus:
$$ f(n) \leq I(n) \leq \frac1t f(n)$$
which is valid as a finite approximation when $t$ is strictly larger than 0.
A recursion can be obtain for $f(n)$. Laplace approximation for $f(n)$ would give that $f(n) \approx \sqrt{\pi \over 2n}$ which show that the upper bound can be tight. 
</s>