Finding limit of I(n)..

Question

Given: $$\ I(n) = \int_{0}^1 \frac {nx^{n-1}}{1+x}dx$$ Find $\lim_{n \to\infty} I(n)$.

Please help me with this. I cannot find any way to do this.

Answer 1

Hint Use integration by parts. Indeed, $$ \int_{0}^1\frac{nx^{n-1}}{1+x}\,dx=\left[\frac{x^n}{1+x}\right]_{0}^1+\int_{0}^1\frac{x^n}{(1+x)^2}\,dx $$ Let $n\to\infty$ and apply the dominated convergence theorem to the second integral on the RHS.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\Large\left. a\right)}$

With Laplace's Method:

\begin{align} \lim_{n \to \infty}\int_{0}^{1}{nx^{n - 1} \over 1 + x}\,\dd x & = \lim_{n \to \infty}\bracks{n\int_{0}^{1}{\pars{1 - x}^{n - 1} \over 2 - x} \,\dd x} = \lim_{n \to \infty}\bracks{n\int_{0}^{1} {\exp\pars{\bracks{n - 1}\ln\pars{1 - x}} \over 2 - x}\,\dd x} \\[5mm] & = \lim_{n \to \infty}\bracks{n\int_{0}^{\infty} {\exp\pars{-\bracks{n - 1}x} \over 2 - 0}\,\dd x} = {1 \over 2}\lim_{n \to \infty}\pars{n\,{1 \over n - 1}} = \bbx{1 \over 2} \end{align}

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$\ds{\Large\left. b\right)}$ \begin{align} \lim_{n \to \infty}\int_{0}^{1}{nx^{n - 1} \over 1 + x}\,\dd x & = \lim_{n \to \infty}\bracks{n\int_{0}^{1}{x^{n - 1} - x^{n} \over 1 - x^{2}} \,\dd x} = \lim_{n \to \infty}\bracks{n\int_{0}^{1}{x^{n/2 - 1} - x^{n/2 - 1/2} \over 1 - x}\,{1 \over 2}\dd x} \\[5mm] & = {1 \over 2}\lim_{n \to \infty}\bracks{n\pars{% \int_{0}^{1}{1 - x^{n/2 - 1/2} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{n/2 - 1} \over 1 - x}\,\dd x}} \\[5mm] & = {1 \over 2}\lim_{n \to \infty}\bracks{n\pars{H_{n/2 - 1/2} - H_{n/2 - 1}}} \qquad\pars{~H_{z}:\ Harmonic\ Number~} \\[5mm] & = \bbx{1 \over 2} \end{align}

Note that $\ds{H_{z} \sim \ln\pars{z} + \gamma + {1 \over 2z} - {1 \over 12z^{2}}}$ as $\ds{\verts{z} \to \infty.\quad}$ $\ds{\gamma}$ is the Euler-Mascheroni Constant.