# 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?

• The main contribution should be from around $\pi/2$. where then we can approximate $\cot^2 \theta$ by $(\theta-\pi/2)^2$ – nir Dec 17 '16 at 13:40
• laplace method should do it. i think, which will give the $O(1/\sqrt{n})$ but now I want an even finite aproximation valid for all $n$ – nir Dec 17 '16 at 13:42

• A behaviour as $n \to \infty$. Laplace's method ($3.$, p. $2$) may be applied here, \begin{align} \int_a^bf(x)e^{-\lambda g(x)}dx\sim f(a)e^{-\lambda g(a)}\sqrt{\frac{\pi}{2\lambda g''(a)}},\qquad \lambda \to \infty, \tag1 \end{align} with $g'(a)=0$, $g''(a)>0$, $f(a)\neq 0$.

One may write \begin{align} \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 \\\\&=\int_{0}^{\infty} \frac{1}{1+x^2}e^{-\large \frac{n-1}{2}\cdot \ln(1 + t^2 x^2)} dx \end{align} here $\lambda =\frac{n-1}{2}$, $g(x)=\ln(1 + t^2 x^2)$, $g(0)=0$, $g'(0)=0$, $g''(0)=2t^2>0$, $f(0)=1\neq 0$, we then get

$$\int_{0}^{\large \frac \pi 2} \left( 1 + t^2 \cot^2\theta \right )^{-\large\frac{n-1}{2}} d \theta \sim \frac1t\sqrt{\frac{\pi}{2}}\cdot \frac1{\sqrt{n}}=\mathcal{O}\left(\frac1{\sqrt{n}}\right),\qquad n \to \infty. \tag2$$

• A closed form. One may evaluate $$\int_{0}^{\large \frac \pi 2} \left( a +1+ \cot^2\theta \right )^{-\frac{1}{2}} d \theta=\frac{\arctan \sqrt{a}}{\sqrt{a}},\qquad a>0$$ then differentiating $n$ times with respect to $a$ setting $t=\dfrac1{\sqrt{a+1}}$, gives

$$\int_{0}^{\large \frac \pi 2} \left(1+t^2 \cot^2\theta \right )^{-\frac{2n+1}{2}} d \theta=\frac{(-1)^n}{t^{2n+1}} \frac{(2n)!}{2^{2n-1}n!}\left[\frac{d^n}{da^n}\left(\frac{\arctan \sqrt{a}}{\sqrt{a}}\right)\right]_{t=\frac1{\sqrt{a+1}}}.\tag3$$

Similarly, from the evaluation $$\int_{0}^{\large \frac \pi 2} \left(a+1+\cot^2\theta\right)^{-1} d \theta=\dfrac1{a+1+\sqrt{a+1}},\qquad a>0$$ then differentiating $n$ times with respect to $a$ setting $t=\dfrac1{\sqrt{a+1}}$ we get

$$\int_{0}^{\large \frac \pi 2} \left(1+t^2 \cot^2\theta \right )^{\large-\frac{2n+2}2} d \theta=\frac{(-1)^n}{t^{2n+2}} \frac{1}{n!}\left[\frac{d^n}{da^n}\left(\dfrac1{a+1+\sqrt{a+1}}\right)\right]_{t=\frac1{\sqrt{a+1}}}. \tag4$$

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.