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How do I use Lebesgue Dominated Convergence Theorem to evaluate

$$\lim_{n \to \infty}\int_{[0,1]}\frac{n\sin(x)}{1+n^2\sqrt x}dx$$

What dominating function to use here?

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How did you come upon this particular integral? What background results and methods are you used to? –  Carl Mummert Oct 24 '10 at 23:16
    
Carl - I am going through a couple of books on probability and asking questions from exercises and things I do not understand. I am still in the very basics. If you need reference to the book/article let me know. –  user957 Oct 25 '10 at 0:17

3 Answers 3

up vote 7 down vote accepted

Note that for $x\in(0,1]$ and $n\geq 1$, we have: $$\frac{n\sin(x)}{1+n^2\sqrt{x}} \leq \frac{n}{1+n^2\sqrt{x}} \leq \frac{n}{n^2\sqrt{x}} \leq \frac{1}{n\sqrt{x}} \leq \frac{1}{\sqrt{x}}.$$

Thus, if $g(x) = \frac{1}{\sqrt{x}}$ for $0\lt x\leq 1$ and $g(x)=0$ if $x=0$, then $0\leq f_n(x)\leq g(x)$ almost everywhere in $[0,1]$, where $f_n(x) = \frac{n\sin(x)}{1+n^2\sqrt{x}}$. Since $$\int g(x)\,d\mu = \int_{[0,1]}\frac{1}{\sqrt{x}}\,d\mu= 2\sqrt{x}\Biggm|_0^1 = 2\lt \infty$$ by Lebesgue's Dominated Convergence Theorem you know that if $f(x)$ equals the pointwise limit of the $f_n(x)$ almost everywhere on $[0,1]$, then $$\lim_{n\to\infty}\int_{[0,1]}\frac{n\sin(x)}{1+n^2\sqrt{x}}\,d\mu = \int_{[0,1]}f(x)\,d\mu.$$ So to use Dominated Convergence you now need to figure out an $f(x)$.

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Thanks Arturo. The function $f_n(x)$ converges to 0 as $n\to\infty$. So it converges point wise to 0. So the integral as $n\to\infty$ should evaluate to 0 by DCT. –  user957 Oct 25 '10 at 0:50

Consider this for a solution (it does not use dominated convergence). In $[0,1]$, we have the following inequality

\begin{equation} 0 \leq \frac{n \sin(x)}{1+ n^2 \sqrt{x}} < \frac{n}{n^2 \sqrt{x}} = \frac{1}{n \sqrt{x}} \end{equation}

Taking integral over $[0,1]$ on both sides, we get

\begin{equation} \int_{[0,1]} \frac{n \sin(x)}{1+ n^2 \sqrt{x}} < \int_{[0,1]} \frac{1}{n \sqrt{x}} = \left[ \frac{2 \sqrt{x}}{n} \right]_0^1 \end{equation}

taking limit $n \to \infty$ on both sides, you can show that

\begin{equation} \lim_{n \to \infty} \int_{[0,1]} \frac{n \sin(x)}{1+ n^2 \sqrt{x}} = 0 \end{equation}

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In fact, with some work you can show that the integrand is at most 1 across the interval (0,1), for all n. –  Yuval Filmus Oct 24 '10 at 23:31

$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{{\rm e}^{#1}}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\pp}{{\cal P}}% \newcommand{\sgn}{{\rm sgn}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}$

$$ 0 < \int_{0}^{1}{n\sin\pars{x} \over 1 + n^{2}\,\sqrt{x\,}}\,\dd x < \int_{0}^{1}{n \over 1 + n^{2}\,\sqrt{x\,}\,}\,\dd x = 2n\int_{0}^{1}{x \over 1 + n^{2}x}\,\dd x = {2 \over n}\,\ln\pars{1 + n^{2}} $$
$$ \lim_{n \to \infty}{2 \over n}\,\ln\pars{1 + n^{2}} = 2\lim_{n \to \infty}{2n/\pars{1 + n^{2}} \over 1} = 0 $$

$$ \lim_{n \to \infty}\int_{0}^{1}{n\sin\pars{x} \over 1 + n^{2}\,\sqrt{x\,}}\,\dd x = 0 $$

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