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If $$I_n=\int_0^1 \frac{x^n}{x^2+2019}\,\mathrm dx,$$ find $\lim_{n\to \infty} nI_n$. Can somebody help me, please?

I've only found that $$\frac{1}{2020} \le nI_n \le \frac{1}{2019}$$ knowing that $x^2 \in [0,1]$, but it doesn't help to evaluate the limit.

I want a proof without the Arzelà-Ascoli theorem or dominated convergence.

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The term $nx^n$ in the numerator of our integral is almost a derivative: $(x^n)' = nx^{n-1}$, so this suggests the following approach.

Notice that by the product rule, \begin{align} \frac{\mathrm d}{\mathrm dx}\bigg(x^n\cdot \frac{x}{x^2+2019}\bigg) &= nx^{n-1}\cdot\frac{x}{x^2+2019} + x^n\cdot\frac{\mathrm d}{\mathrm dx}\bigg(\frac{x}{x^2+2019}\bigg) \\ &= \frac{nx^n}{x^2+2019} + x^n\cdot \frac{-x^2+2019}{(x^2+2019)^2} \end{align} For $0 \leqslant x \leqslant 1$, the second term above is bounded above in absolute value by $|x|^n = x^n$. Integrating and applying the fundamental theorem of calculus gives \begin{align} \int_0^1\frac{\mathrm d}{\mathrm dx}\bigg(x^n\cdot \frac{x}{x^2+2019}\bigg)\,\mathrm dx &= n\int_0^1\frac{x^n}{x^2+2019}\,\mathrm dx + \int_0^1 x^n\cdot \frac{-x^2+2019}{(x^2+2019)^2}\,\mathrm dx \\ \leadsto\quad\frac{1}{2020} &= nI_n + \int_0^1 x^n\cdot \frac{-x^2+2019}{(x^2+2019)^2}\,\mathrm dx. \end{align} Hence, \begin{align} nI_n = \frac{1}{2020} - \int_0^1 x^n\cdot \frac{-x^2+2019}{(x^2+2019)^2}\,\mathrm dx. \end{align} By the triangle inequality and the fundamental theorem of calculus again, \begin{align*} \bigg|\int_0^1 x^n\cdot \frac{-x^2+2019}{(x^2+2019)^2}\,\mathrm dx\bigg| &\leqslant \int_0^1\bigg|x^n\cdot \frac{-x^2+2019}{(x^2+2019)^2}\bigg|\,\mathrm dx \\ &\leqslant \int_0^1 x^n\,\mathrm dx \\ &= \frac{1}{n+1} \to 0,\quad\text{as $n\to\infty$.} \end{align*} Thus, $$ \lim_{n\to\infty}nI_n = \frac{1}{2020}. $$

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For fixed $\delta\in(0,1)$, $$ \int_0^1 \frac{nx^n}{x^2+2019}dx=\int_0^\delta \frac{nx^n}{x^2+2019}dx+\int_\delta^1 \frac{nx^n}{x^2+2019}dx. $$ Note $$ 0\le\int_0^\delta \frac{nx^n}{x^2+2019}dx\le\int_0^\delta nx^ndx=\frac{n}{n+1}\delta^{n+1}. \tag{1}$$ and $$ \int_\delta^1 \frac{nx^n}{2020}dx\le\int_\delta^1 \frac{nx^n}{x^2+2019}dx\le\int_\delta^1 \frac{nx^n}{\delta^2+2019}dx. \tag{2}$$ But $$ \int_\delta^1 \frac{nx^n}{2020}dx=\frac{1}{2020}\frac{n}{n+1}(1-\delta^{n+1}), \int_\delta^1 \frac{nx^n}{\delta^2+2019}dx=\frac{1}{\delta^2+2019}\frac{n}{n+1}(1-\delta^{n+1}). \tag{3} $$ From (1),(2) and (3), one has $$ \frac{1}{2020}\frac{n}{n+1}(1-\delta^{n+1})\le nI_n\le \frac{n}{n+1}\delta^{n+1}+\frac{1}{\delta^2+2019}\frac{n}{n+1}(1-\delta^{n+1}).$$ Letting $n\to\infty$ gives $$ \frac{1}{2020}\le\liminf nI_n\le\limsup nI_n\le \frac{1}{\delta^2+2019}.$$ Letting $\delta\to1^-$ gives $$ \liminf nI_n=\limsup nI_n=\frac1{2020} $$ or $$ \lim_{n\to\infty}nI_n=\frac1{2020}.$$

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Let $J_n=n I_n$, notice that when you use integration by parts, you would get $$J_n=\left[ \frac{x^{n+1}}{x^2+2019}\right]_0^1 -\int_0^1\frac{x^{n}(2019-x^2)}{(x^2+2019)^2}\ dx= \frac{1}{2020}-\int_0^1\frac{x^{n}(2019-x^2)}{(x^2+2019)^2}\ dx$$

Now, you only have to show that the limit of the rightmost integral is zero, notice that the integrand can be Bounded by $Cx^n$.

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We can write \begin{align} {nx^n \over x^2+2019} &= {nx^n \over 2019} \cdot {1\over 1+{x^2\over 2019}} \\ &= {nx^n \over 2019} \cdot \left(1-{x^2\over 2019}+\left[{x^2\over 2019}\right]^2-\left[{x^2\over 2019}\right]^3+\cdots\right) \\ &={nx^n\over 2019}\cdot\sum_{k=0}^{\infty}{x^{2k}(-1)^k\over (2019)^k} \\ &={n\over2019}\cdot \sum_{k=0}^\infty {x^{n+2k}(-1)^k\over (2019)^k}. \end{align}

Therefore,

\begin{align} \int_0^1 {nx^n\over x^2+2019}\,dx &=\int_0^1 {n\over2019}\cdot \sum_{k=0}^{\infty}{x^{n+2k}(-1)^k\over (2019)^k}dx \\ &= {n\over2019}\cdot\sum_{k=0}^{\infty}{(-1)^k\over (2019)^k}\int_0^1 {x^{n+2k}}\,dx \\ &= \sum_{k=0}^\infty {n\over n+2k+1}\cdot{(-1)^k\over (2019)^{k+1}}. \end{align}

Finally,

\begin{align} \lim_{n\to \infty}\int_0^1 {nx^n\over x^2+2019}\,dx &= \lim_{n\to \infty}\sum_{k=0}^\infty {n\over n+2k+1}\cdot{(-1)^k\over (2019)^{k+1}}\\ &= \sum_{k=0}^\infty {(-1)^k\over (2019)^{k+1}}\\ &= {1\over 2020}. \end{align}

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\lim_{n \to \infty}\pars{n\int_{0}^{1}{x^{n} \over x^{2} + 2019}\,\dd x}}\,\,\,\stackrel{x\ \mapsto\ 1 - x}{=}\,\,\, \lim_{n \to \infty}\bracks{n\int_{0}^{1}{\pars{1 - x}^{n} \over x^{2} -2x + 2020}\,\dd x} \\[5mm] = &\ \lim_{n \to \infty}\bracks{n\int_{0}^{1}{\expo{n\ln\pars{1 - x}} \over x^{2} -2x + 2020}\,\dd x} \\[5mm] = &\ \lim_{n \to \infty}\pars{n\int_{0}^{\infty}{\expo{-nx} \over 0^{2} -2 \times 0 + 2020}\,\dd x}\qquad\pars{\text{Laplace's Method}} \\[5mm] = & \bbx{1 \over 2020} \approx 4.9505 \times 10^{-4} \end{align}

Laplace's Method.

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