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Question

<span class="math-container">n de $$ \lim{n\to\infty} ^ {1/2} \int{0}^{1} \frac{1} {(1 + x ^ 2) ^ n} \mathrm {d} x = 0 $</span>

¿Es correcta mi respuesta? Pero no estoy seguro del método por el que he hecho.

Answer 1

Jajaja Con el cambio de la variable <span class="math-container">$ t = \sqrt{n} x $</span> tienes <span class="math-container">$$ \sqrt{n} \int_0^1 \frac{1}{(1+x^2)^n} \, dx = \int_0^\sqrt{n} \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \, dt = \int0^{+\infty} \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \chi{[0,\sqrt{n}]}(t) \, dt. $$Observe that <span class="math-container">$$ \lim{n \to \infty} \left( 1 + \frac{t^2}{n} \right)^n \chi{[0,\sqrt{n}]}(t) = e^{t^2}$$</span>pointwise everywhere, say for <span class="math-container">$t # > 0 $</span>; also the sequence <span class="math-container">$ \mapsto n (1 + t ^ 2 /n)^n$</span> is increasing for all <span class="math-container">$t \in \mathbb{R}$</span> which implies that <span class="math-container">$$\frac{1}{1+t^2} \ge \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \ge \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \chi{[0,\sqrt{n}]}(t) $$</span> for all <span class="math-container">$n \in \mathbb{N}$</span> and for all <span class="math-container">$t \in \mathbb{R}$</span>. By the Dominated Convergence Theorem, since <span class="math-container">$1/(1+t^2) \in L^1([0,+\infty)) $</span>, we have that <span class="math-container">$% $ $\lim{n \to \infty} \int0^{+\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}. $</span></span>

Answer 2

Nope, el límite no puede ser cero. En un buen barrio de el origen $\frac{1}{1+x^2}\approx e^{-x^2}$, y para valores grandes de a$n$ tenemos que $\int_{0}^{1}e^{-nx^2}\,dx$ es terriblemente cerca de $\int_{0}^{+\infty}e^{-nx^2}\,dx$, que las escalas como $\frac{K}{\sqrt{n}}$ para una constante positiva $K$. Esta realidad es la idea principal de la Laplace/Hayman métodos. En nuestro caso

$$ \int_{0}^{1}\frac{dx}{(1+x^2)^n}\stackrel{x\mapsto\tan\theta}{=}\int_{0}^{\pi/4}\cos^{2n-2}(\theta)\,d\theta $$ es en la mayoría de las $\frac{1}{2^{n-1}}\cdot\frac{\pi}{4}$ aparte de $$ \int_{0}^{\pi/2}\cos^{2n-2}(\theta)\,d\theta = \frac{\pi}{2\cdot 4^{n-1}}\binom{2n-2}{n-1}=\frac{\pi n}{(2n-1)4^n}\binom{2n}{n}, $$ y desde $\frac{1}{4^n}\binom{2n}{n}\sim\frac{1}{\sqrt{\pi n}}$ (por Wallis producto o similares de primaria manipulaciones) tenemos $$ \lim_{n\to +\infty}\sqrt{n}\int_{0}^{1}\frac{dx}{(1+x^2)^n}=\color{red}{\frac{\sqrt{\pi}}{2}}.$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove armada]{\displaystyle{#1}}\,} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Esto puede ser evaluado por medio de Laplace Método: \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}