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$$ \lim_{n\to\infty} n^{1/2} \int_{0}^{1} \frac{1}{(1+x^2)^n}\mathrm{d}x=0 $$

Is my answer correct? But I am not sure of method by which I have done.

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  • $\begingroup$ What method did you use to get 0? $\endgroup$ – AHusain Nov 4 '18 at 16:42
  • $\begingroup$ ... and I'm getting $\sqrt{\pi}/2$. $\endgroup$ – metamorphy Nov 4 '18 at 17:45
  • $\begingroup$ try substitution $x=\tan\theta$ $\endgroup$ – hskimse Nov 3 '19 at 1:55
  • $\begingroup$ Actuallly this post asks the MCQ question that you are probably asked. $\endgroup$ – StubbornAtom Nov 3 '19 at 8:21
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No. With the change of variable $ t = \sqrt{n} x $ you get $$ \sqrt{n} \int_0^1 \frac{1}{(1+x^2)^n} \, dx = \int_0^\sqrt{n} \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \, dt = \int_0^{+\infty} \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \chi_{[0,\sqrt{n}]}(t) \, dt. $$Observe that $$ \lim_{n \to \infty} \left( 1 + \frac{t^2}{n} \right)^n \chi_{[0,\sqrt{n}]}(t) = e^{t^2}$$pointwise everywhere, say for $t>0$; also the sequence $ n \mapsto (1 + t^2 /n)^n$ is increasing for all $t \in \mathbb{R}$ which implies that $$\frac{1}{1+t^2} \ge \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \ge \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \chi_{[0,\sqrt{n}]}(t) $$ for all $n \in \mathbb{N}$ and for all $t \in \mathbb{R}$. By the Dominated Convergence Theorem, since $1/(1+t^2) \in L^1([0,+\infty))$, we have that $$\lim_{n \to \infty} \int_0^{+\infty} \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \chi_{[0,\sqrt{n}]}(t) \, dt = \int_0^{+\infty} e^{-t^2} \, dt = \frac{\sqrt{\pi}}{2}. $$

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  • $\begingroup$ If we take $x_0 \in (0,1)$ we have that $\frac{\sqrt{n}}{(1+x_0^2)^n} \to 0$ $\endgroup$ – Marios Gretsas Nov 4 '18 at 18:27
  • $\begingroup$ This question came in a test for high schoolers yesterday (I gave it too) and your approach is quite out of our reach. The question was actually simpler as it also stated that the limit existed, ie , the integral tends to $\frac{c}{\sqrt{n}}$ where c is positive. Maybe we could use this and dumber down the solution a bit? $\endgroup$ – N.S.JOHN Nov 4 '18 at 18:40
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    $\begingroup$ @N.S.JOHN: no matter how much you "dumb it down", you'll still need to get a $\sqrt{\pi}$ out of something. $\endgroup$ – Jack D'Aurizio Nov 5 '18 at 1:19
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    $\begingroup$ (+1) In regards of the Necky $\endgroup$ – TheOscillator Nov 5 '18 at 17:48
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Nope, the limit cannot be zero. In a right neighbourhood of the origin $\frac{1}{1+x^2}\approx e^{-x^2}$, and for large values of $n$ we have that $\int_{0}^{1}e^{-nx^2}\,dx$ is horribly close to $\int_{0}^{+\infty}e^{-nx^2}\,dx$, which scales like $\frac{K}{\sqrt{n}}$ for a positive constant $K$. This actually is the main idea of the Laplace/Hayman methods. In our case

$$ \int_{0}^{1}\frac{dx}{(1+x^2)^n}\stackrel{x\mapsto\tan\theta}{=}\int_{0}^{\pi/4}\cos^{2n-2}(\theta)\,d\theta $$ is at most $\frac{1}{2^{n-1}}\cdot\frac{\pi}{4}$ apart from $$ \int_{0}^{\pi/2}\cos^{2n-2}(\theta)\,d\theta = \frac{\pi}{2\cdot 4^{n-1}}\binom{2n-2}{n-1}=\frac{\pi n}{(2n-1)4^n}\binom{2n}{n}, $$ and since $\frac{1}{4^n}\binom{2n}{n}\sim\frac{1}{\sqrt{\pi n}}$ (by Wallis product or similar elementary manipulations) we have $$ \lim_{n\to +\infty}\sqrt{n}\int_{0}^{1}\frac{dx}{(1+x^2)^n}=\color{red}{\frac{\sqrt{\pi}}{2}}.$$

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  • $\begingroup$ Can the solution be simpler if we were provided that the limit exists and is positive? It was provided so in the actual question. $\endgroup$ – N.S.JOHN Nov 4 '18 at 18:59
  • $\begingroup$ @N.S.JOHN: what do you mean by simpler? Both answers now prove that the limit is precisely $\int_{0}^{+\infty}e^{-x^2}\,dx$. $\endgroup$ – Jack D'Aurizio Nov 4 '18 at 19:01
  • $\begingroup$ Simpler in the sense that you can assume the limit exists. This question came in a test today for high schoolers (which I gave) with this additional info that the OP probably forgot to add. $\endgroup$ – N.S.JOHN Nov 4 '18 at 19:03
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    $\begingroup$ The fact that the limit is positive is fairly obvious, but in order to compute it you need to know the value of the Gaussian integral or, equivalently, the value of $$\lim_{n\to +\infty}n\left[\frac{1}{4^n}\binom{2n}{n}\right]^2,$$ i.e. Wallis' product or some form of the central limit theorem. $\endgroup$ – Jack D'Aurizio Nov 4 '18 at 19:08
  • $\begingroup$ Sorry . Your solution is actually within my grasp. I never actually read the complete solution because the first few lines did not make sense to me. $\endgroup$ – N.S.JOHN Nov 8 '18 at 15:55
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Let \begin{align*} I_n&=\int_0^\infty\frac{dx}{(1+x^2)^n}=\left.\frac{x}{(1+x^2)^n}\right|_{x=0}^\infty+\int_0^\infty\frac{2nx^2\,dx}{(1+x^2)^{n+1}}\\ &=2n\int_0^\infty\frac{dx}{(1+x^2)^n}-2n\int_0^\infty\frac{dx}{(1+x^2)^{n+1}}=2nI_n-2nI_{n+1} \end{align*} So we get \begin{align*}I_{n+1}&=\frac{2n-1}{2n}I_n=\frac{2n-1}{2n}\frac{2n-3}{2n-2}I_{n-1}=\cdots\\ &=\frac{(2n-1)!!}{(2n)!!}I_1=4^{-n}\binom{2n}{n}\frac \pi 2 \end{align*} By Stirling's formula, $$\int_0^1\frac{dx}{(1+x^2)^{n+1}}=I_{n+1}+O(\tfrac 1 n)\sim \frac{1}{\sqrt{\pi n}}\cdot\frac\pi 2=\frac{\sqrt\pi}{2\sqrt n}$$ Thus the original limit equals $$\sqrt n\int_0^1\frac{dx}{(1+x^2)^n}\sim\frac{\sqrt{\pi n}}{2\sqrt{n-1}}\to\frac{\sqrt \pi}{2}=0.8862\ldots$$

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  • $\begingroup$ Did you mean for the first few integrals to be from $0$ to $1$ instead of $0$ to $\infty$? $\endgroup$ – JimmyK4542 Nov 3 '19 at 2:45
  • $\begingroup$ @JimmyK4542 They were meant to be $0$ to $\infty$ $\endgroup$ – Edward H Nov 3 '19 at 3:16
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    $\begingroup$ Oh nvm. I see how your solution works now. $\int_{1}^{\infty}(1+x^2)^{-(n+1)}\,dx \le \int_{1}^{\infty}x^{-(2n+2)}\,dx = O(1/n)$, which is how you got $\int_{0}^{1}(1+x^2)^{-(n+1)}\,dx = I_{n+1}+O(1/n)$. $\endgroup$ – JimmyK4542 Nov 3 '19 at 3:36
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ This can be evaluated by means of Laplace Method: \begin{align} &\bbox[10px,#ffd]{\lim_{n \to \infty}\bracks{n^{1/2}\int_{0}^{1}{\dd x \over \pars{1 + x^{2}}^{n}}}} \\[5mm] = &\ \lim_{n \to \infty}\bracks{n^{1/2}\int_{0}^{1} \exp\pars{-n\ln\pars{1 + x^{2}}}\,\dd x} \\[5mm] = &\ \lim_{n \to \infty}\bracks{n^{1/2}\int_{0}^{\infty} \exp\pars{-nx^{2}}\,\dd x} \\[5mm] = &\ \int_{0}^{\infty}\exp\pars{-x^{2}}\,\dd x = \bbx{\root{\pi} \over 2} \approx 0.8862 \end{align}

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If you are fine with using the Gamma function (specifically $\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$) and the dominated convergence theorem (DCT), then there is another way of finding your limit:

  • Having in mind to get an $e$-function in the denominator, using the substitution $x^2=\frac{u}{n}$ you get $$I_n = \sqrt{n}\int_0^1 \frac{1}{(1+x^2)^n} dx = \frac 12 \int_0^{\color{blue}{n}}\frac{du}{\sqrt u\left(1+\frac un\right)^n}$$
  • Now, in order to apply the DCT, we need to bound the integrand properly:

$$I_n = \frac{1}{2}\int_0^{\color{blue}{1}}\frac{du}{\sqrt u\left(1+\frac un\right)^n} + \frac{1}{2}\int_{\color{blue}{1}}^{\color{blue}{n}}\frac{du}{\sqrt u\left(1+\frac un\right)^n}$$ $$ < \frac 12 \int_0^{\color{blue}{1}}\frac{du}{\sqrt u} + \frac{1}{2}\int_{\color{blue}{1}}^{\color{blue}{n}}\frac{du}{\sqrt u(1+u)}$$ $$< 1 + \int_{\color{blue}{1}}^{\color{blue}{\infty}}\frac{du}{u^{\frac{3}{2}}} = 2$$

So, we are allowed to apply the DCT and get

$$\lim_{n\to\infty}I_n = \frac{1}{2}\int_0^{\infty}\frac{du}{\sqrt{u}e^u}=\frac{1}{2}\int_0^{\infty}u^{-\frac{1}{2}}e^{-u}\;du$$ $$= \frac{1}{2}\Gamma\left(\frac{1}{2}\right) = \boxed{\frac{\sqrt{\pi}}{2}}$$

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