Recursive integral converges to Gaussian integral

Question

We define $I_{n,a} = \int\limits_{-\infty}^\infty\frac{dx}{(1+\frac{x^2}{a})^n}.$
A. Show that for all $n\ge1, a>0, \space I_{n,a}$ converges and $I_{n+1,a}=\frac{2n-1}{2n}\cdot I_{n,a}.$
B. Prove that $I_{n,n}\xrightarrow[n \to \infty]{}\int\limits_{-\infty}^{\infty}e^{-x^2}.$
I managed to show that for all $n\ge1, a>0, \space I_{n,a}$ converges. It can be done in several ways (e.g. change of variables). However I failed to show that $I_{n+1,a}=\frac{2n-1}{2n}\cdot I_{n,a}.$ I tried integration by parts but got stuck.
Moreover, I did not manage to prove B. One can notice that the integrand $\frac{1}{(1+\frac{x^2}{a})^n}$ converges to $e^{-x^2}$ when $n\to\infty$, but I am not sure how to proceed.
Help would be much appreciated.

Answer 1

A. Use the change of variable $x=\sqrt{a}\tan{\theta}$. \begin{align} I_{n,a}=\int_{-\infty}^{+\infty}{dx \over (1+{x^2 \over a})^n} & =\sqrt{a}\int_{\theta=-{\pi \over 2}}^{\theta=+{\pi \over 2}}{d\tan{\theta} \over \sec^{2n}{\theta}} \\ & = \sqrt{a}\bigg({\tan{\theta} \over \sec^{2n}{\theta}}\bigg|_{\theta=-{\pi \over 2}}^{\theta=+{\pi \over 2}}-\int_{\theta=-{\pi \over 2}}^{\theta=+{\pi \over 2}}{\tan{\theta}d\bigg({1 \over \sec^{2n}{\theta}}\bigg)}\bigg) \\ & = \sqrt{a}\cdot 2n \int_{-{\pi \over 2}}^{+{\pi \over 2}}{\tan^2{\theta}\bigg({1 \over \sec^{2n}{\theta}}\bigg)d\theta} \\ & = \sqrt{a}\cdot 2n \int_{\theta=-{\pi \over 2}}^{\theta=+{\pi \over 2}}{(\sec^2{\theta}-1)\bigg({1 \over \sec^{2(n+1)}{\theta}}\bigg)d{\tan{\theta}}} \\ & = 2nI_{n,a}-2nI_{n+1,a} \end{align} Therefore, $I_{n+1,a}={2n-1 \over 2n}I_{n,a}$.

B. In case you are not familiar with the uniform convergence mentioned in the comment, I will show how it works in an explicit way. Consider the interval $[-m,+m]$ where $m>0$, we show the integrand $f_{n}(x)={1 \over (1+{x^2 \over n})^n}$ converges uniformly to $e^{-x^2}$ on $[-m,+m]$. This would be enough to argue $$\lim_{n \rightarrow +\infty}{\int_{-m}^{+m}{f_n(x)dx}}=\int_{-m}^{+m}{\lim_{n \rightarrow +\infty}{f_n(x)}dx}=\int_{-m}^{+m}{e^{-x^2}dx}$$ The swapping of limit and integral is legitimate under the condition of uniform convergence. We will then immediately show $B$ by extending $m \rightarrow +\infty$.

To show uniform convergence, we have to prove that given some $\varepsilon>0$, there is some large enough $k \in \mathbb{N}$ such that for any $n>k$ and $x \in [-m,+m]$, $|f_n(x)-e^{-x^2}|<\varepsilon$. The convergence is "uniform" in the sense that $f_n(x)$ is close to $e^{-x^2}$ for all $x \in [-m,+m]$. Assume $m>1$ and $\varepsilon<1$, and let $l=\lceil 8m^2\varepsilon^{-1} \rceil$. Take $k=\lceil 2l^3m^{2j}\varepsilon^{-1} \rceil$, and for any $n>k$, we have \begin{align} \bigg|(1+{x^2 \over n})^n-e^{x^2}\bigg| & =\bigg|\sum_{j=1}^{l}{({1 \over n^j}{n! \over j!(n-j)!}-{1 \over j!})x^{2j}}+\sum_{j=l+1}^{n}{({1 \over n^j}{n! \over j!(n-j)!})x^{2j}}-\sum_{j=l+1}^{+\infty}{{1 \over j!}x^{2j}}\bigg| \\ & \leq \bigg|\sum_{j=1}^{l}{{1 \over j!}({n! \over n^j(n-j)!}-1)x^{2j}}\bigg|+2\bigg|\sum_{j=l+1}^{+\infty}{{1 \over j!}x^{2j}}\bigg| \\ & \leq \sum_{j=1}^{l}{{1 \over j!}m^{2j}\bigg|\bigg({n-l+1 \over n}\bigg)^l-1\bigg|}+2\bigg|{1 \over (l+1)!}\sum_{j=0}^{+\infty}{\bigg({m^2 \over l+2}\bigg)^{j}}\bigg| \\ & \leq \sum_{j=1}^{l}{m^{2j}\bigg|\bigg(1-l\cdot{l \over n}\bigg)-1\bigg|}+4{1 \over (l+1)!} \\ & \leq {m^{2j}l^3 \over n} + 4{1 \over (l+1)!} < \varepsilon \end{align} This implies $$\bigg|f_n(x)-e^{-x^2}\bigg| = \bigg|{e^{x^2}-(1+{x^2 \over n})^n \over e^{x^2}(1+{x^2 \over n})^n}\bigg| \leq \bigg|e^{x^2}-(1+{x^2 \over n})^n\bigg| < \varepsilon$$ And it completes the proof.

Edit:

In A., I just found it unnecessary to change the variable: \begin{align} I_{n,a} & = \int_{-\infty}^{+\infty}{dx \over (1+{x^2 \over a})^n} = {x \over (1+{x^2 \over a})^n}\bigg|_{-\infty}^{+\infty}+\int_{-\infty}^{+\infty}{2nx^2dx \over a(1+{x^2 \over a})^{n+1}} \\ & = 2n\int_{-\infty}^{+\infty}{(1+{x^2 \over a}-1)dx \over (1+{x^2 \over a})^{n+1}}=2nI_{n,a}-2nI_{n+1,a} \end{align} Apparently an easier way to solve the problem.

In addition, combining A and B, we have $$I_{n,n}=\Pi_{k=1}^{n-1}{\big({2k-1 \over 2k}\big)}I_{1,n}=\Pi_{k=1}^{n-1}{\big({2k-1 \over 2k}\big)}\int_{-\infty}^{+\infty}{dx \over 1+{x^2 \over n}}=\sqrt{n}\pi\Pi_{k=1}^{n-1}{\big({2k-1 \over 2k}\big)}$$ Since $\int_{-\infty}^{+\infty}{e^{-x^2}dx}=\sqrt{\pi}$, there is this identity $${1 \over \pi}\lim_{n \rightarrow +\infty}{I_{n,n}}=\lim_{n \rightarrow +\infty}{\Pi_{k=1}^{n-1}{\big({2k-1 \over 2k}\big)}\sqrt{n}}={1 \over \sqrt{\pi}}$$