Find differents methods

Question

Let $I_n=\int_0^1 \frac 1{(1+x^2)^n} dx$

With Laplace's method, i find easlly that $I_n\sim \frac 12\sqrt{\frac \pi n}$

I would like to find differents methods to prove it

Answer 1

We will prove that \begin{equation*} \lim_{n\to \infty}\dfrac{I_n}{\frac{1}{2}\sqrt{\frac{\pi}{n}}} =1. \end{equation*} After the substitution $x=\frac{y}{\sqrt{n}}$ we have that \begin{equation*} I_n = \dfrac{1}{\sqrt{n}}\int_{0}^{\sqrt{n}}\dfrac{1}{\left(1+\frac{y^2}{n}\right)^{n}}\, \mathrm{d}y . \end{equation*}

Put \begin{equation*} f_n(y) = \begin{cases} \dfrac{1}{\left(1+\frac{y^2}{n}\right)^{n}}&\mbox{ if } 0<y<\sqrt{n}\\ 0&\mbox{ if } y\ge \sqrt{n} \end{cases} \end{equation*} It is well known that $\left(1+\frac{y^2}{n}\right)^{n}$ is increasing to $e^{y^2}$. Consequently $0\le f_n(y) \le \dfrac{1}{1+y^2} \in L_1(\mathbb{R}_{+})$ and $\displaystyle \lim_{n\to \infty}f_n(y) = e^{-y^2}$ pointwise. According to Lebesgue's dominated convergence theorem we get that \begin{gather*} \dfrac{I_n}{\frac{1}{2}\sqrt{\frac{\pi}{n}}} =\dfrac{2}{\sqrt{\pi}}\int_{0}^{\sqrt{n}}\dfrac{1}{\left(1+\frac{y^2}{n}\right)^{n}}\, \mathrm{d}y = \dfrac{2}{\sqrt{\pi}}\int_{0}^{\infty}f_n(y)\, \mathrm{d}y \to \\[2ex] \dfrac{2}{\sqrt{\pi}}\int_{0}^{\infty}e^{-y^2}\, \mathrm{d}y =1 , \mbox{ as } n\to \infty . \end{gather*}

Answer 2

First set $x=\tan u$. Then your integral becomes $$ I_n=\int_0^{\pi/4}\cos^{2n-2}(u)\,du. $$ It would have been nicer to have the integral from $0$ to $\pi/2$. Since $$ \Bigl|\int_{\pi/4}^{\pi/2}\cos^{2n-2}(u)\,du\Bigr|\leq\frac{\pi}{4}\Bigl(\frac{1}{\sqrt{2}}\Bigr)^{2n-2}=\frac{\pi}{2^{n+1}} $$ this integral is exponentially small, and we can add it without changing the asymptotics (as long as we find that the result is not exponential small). The integral from $0$ to $\pi/2$ can be calculated explicitly, $$ \int_0^{\pi/2}\cos^{2n-2}(u)\,du=\frac{\sqrt{\pi}\,\Gamma(n-1/2)}{2\Gamma(n)}. $$ The suggested result follows by using the Stirling approximation of $\Gamma$.