The limit of $\int_{0}^{1}\frac{\sqrt{n}}{1+n\ln(1+x^2)}dx$ as $n\to\infty$

Question

The task is to calculate $$\lim_{n\to\infty}\int_{0}^{1}\frac{\sqrt{n}}{1+n\ln(1+x^2)}dx$$ I tried various estimates I know to find the dominating integrable function and nothing worked. Does anyone have any idea? Is this even an application of DCT or something else?

Answer 1

DCT as Discrete Cosine Transform? I cannot understand what you are meaning.

Anyway, your integral converges towards $\frac{\pi}{2}$, since: $$\int_{0}^{1}\frac{\sqrt{n}}{1+n\log(1+x^2)}\,dx \geq \int_{0}^{1}\frac{\sqrt{n}}{1+nx^2}\,dx = \arctan(\sqrt{n})=\frac{\pi}{2}+O\left(\frac{1}{\sqrt{n}}\right),$$ while the difference between the first and the second integral is bounded by: $$\begin{eqnarray*}&&n^{3/2}(1-\log 2)\int_{0}^{1}\frac{x^4\,dx}{(1+nx^2)(1+n\log(1+x^2))}\\&\leq& n^{3/2}(1-\log 2)\int_{0}^{1}\frac{x^4 \,dx}{(1+n\log(1+x^2))^2}\\&\leq&n^{3/2}(1-\log 2)\int_{0}^{1}\frac{x^4 \,dx}{(1+nx^2-\frac{n}{2}x^4)^2}\end{eqnarray*}$$ where: $$\frac{x^4}{(1+nx^2-\frac{n}{2}x^4)^2}\leq \frac{1}{n^2}\text{ for }x\in[0,1/\sqrt{n}],$$ $$\frac{x^4}{(1+nx^2-\frac{n}{2}x^4)^2}\leq \frac{1}{n^2(1-\frac{1}{2}x^2)^2}\leq\frac{4}{n^2}\text{ for }x\in[1/\sqrt{n},1],$$ hence:

$$\int_{0}^{1}\frac{\sqrt{n}}{1+n\log(1+x^2)}\,dx =\frac{\pi}{2}+O\left(\frac{1}{\sqrt{n}}\right).$$

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

Heuristically, when $\ds{n \ggg 1}$, the main contribution to the integral comes from $\ds{x \gtrsim 0}$ such that

\begin{align} &\int_{0}^{1}{\root{n} \over 1 + n\ln\pars{1 + x^{2}}}\,\dd x \sim\int_{0}^{1}{\root{n} \over 1 + nx^{2}}\,\dd x =\arctan\pars{\root{n}} = {\pi \over 2} - \arctan\pars{1 \over \root{n}} \\[3mm]&\stackrel{n \to \infty}{\to} \color{#66f}{\large{\pi \over 2}} \approx {\tt 1.5708} \end{align}