How to calculate the limit $\lim\limits_{n\to\infty} \int\limits_0^1 \frac{n(2nx^{n-1}-(1+x))}{2(1+x)}\,dx$?

Question

How to calculate the limit $\lim\limits_{n\to\infty}\displaystyle\int_0^1\dfrac{n(2nx^{n-1}-(1+x))}{2(1+x)} dx$?

I have to calculate the limit when solving

Find $a,b$ for $\displaystyle\int_0^1 \dfrac{x^{n-1}}{x+1} dx=\dfrac{a}{n}+\dfrac{b}{n^2}+o(\dfrac{1}{n^2}) (n\to\infty)$

First I calculated that

$\lim\limits_{n\to\infty} \displaystyle\int_0^1 \dfrac{nx^{n-1}}{x+1} dx=\dfrac{1}{2}$, thus $a=\dfrac{1}{2}$, then $2b=\lim\limits_{n\to\infty}\displaystyle\int_0^1\dfrac{n(2nx^{n-1}-(1+x))}{2(1+x)} dx$.

However, I cannot find a good way to calculate it.

Answer 1

\begin{align*} 2b &= \lim_n n \int_0^1 \frac{nx^{n-1}} {x+1} \mathrm dx - \frac n 2 \\ &= \lim_n n \int_0^1 \frac {\mathrm d (x^n)} {x+1} - \frac n2\\ &= \lim_n n \left. \frac {x^n} {1+x}\right|_0^1 + n \int_0^1 \frac {x^n \mathrm dx} {(1+x)^2} - \frac n 2\\ &= \lim_n \frac n {n+1} \cdot \left.\frac {x^{n+1}}{(x+1)^2}\right|_0^1 + \frac {2n}{n+1}\int_0^1 \frac {x^{n+1}}{(x+1)^3} \mathrm d x\\ &= \frac 14 + 2\lim_n \int_0^1 \frac {x^{n+1} \mathrm dx} {(x+1)^3}\\ &= \frac 14, \end{align*} where $$ 0 \gets \frac 18 \int_0^1 x^{n+1} \mathrm dx \leqslant \int_0^1 \frac {x^{n+1} \mathrm dx} {(1+x)^3} \leqslant \int_0^1 x^{n+1} \mathrm d x \to 0. $$

UPDATE

The limit you gave at the very first is actually $b$, not $2b$.

Answer 2

Your evaluation of $a$ is correct. As regards $b$, note that, by integration by parts applied twice, we have that $$\begin{align} \int_0^1 \dfrac{x^{n-1}}{x+1} dx&=\frac{1}{2n}+\frac{1}{n}\int_0^1 \frac{x^{n}}{(x+1)^2} dx \\ &=\frac{1}{2n}+\frac{1}{4n(n+1)}+\frac{2}{n(n+1)}\int_0^1 \frac{x^{n+1}}{(x+1)^3} dx.\end{align}$$ Morever $$0\leq \int_0^1 \frac{x^{n+1}}{(x+1)^3} dx\leq \int_0^1 x^{n+1}dx=\frac{1}{n+2}.$$ Hence $$\int_0^1 \dfrac{x^{n-1}}{x+1} dx=\frac{1}{2n}+\frac{1}{4n^2}+O(1/n^3)$$ and it follows that $b=1/4$ (and $a=1/2$).

Answer 3

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

$\ds{\int_{0}^{1}{x^{n - 1} \over x + 1}\,\dd x = {a \over n} + {b \over n^{2}} + \mrm{o}\pars{1 \over n^{2}}}$ as $\ds{n \to \infty:\ {\Large ?}}$.

\begin{align} &\bbox[10px,#ffd]{\ds{\int_{0}^{1}{x^{n - 1} \over x + 1}\,\dd x}} = \int_{0}^{1}{x^{n - 1} - x^{n} \over 1 - x^{2}}\,\dd x = {1 \over 2}\int_{0}^{1}{x^{n/2 - 1} - x^{n/2 - 1/2} \over 1 - x}\,\dd x \\[5mm] = &\ {1 \over 2}\bracks{% \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}\bracks{\Psi\pars{{n \over 2} + {1 \over 2}} - \Psi\pars{n \over 2}}\quad\pars{~\Psi:\ Digamma\ Function~} \label{1}\tag{1} \end{align}

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As $\ds{n \to \infty}$: \begin{equation} \left\{\begin{array}{rcl} \ds{\Psi\pars{{n \over 2} + {1 \over 2}}} & \ds{\sim} & \ds{\ln\pars{n \over 2} + {1 \over 24\pars{n/2}^{2}} - {7 \over 960\pars{n/2}^{4}} + \cdots} \\[1mm] \ds{\Psi\pars{n \over 2}} & \ds{\sim} & \ds{\ln\pars{n \over 2} - {1 \over 2\pars{n/2}} - {1 \over 12\pars{n/2}^{2}} + {1 \over 120\pars{n/2}^{4}} + \cdots} \end{array}\right.\label{2}\tag{2} \end{equation}

\eqref{1} and \eqref{2} lead to

\begin{align} &\bbox[10px,#ffd]{\ds{\int_{0}^{1}{x^{n - 1} \over x + 1}\,\dd x}} \sim {\color{red}{1/2} \over n} + {\color{red}{1/4} \over n^{2}} + \mrm{o}\pars{1 \over n^{4}} \\[5mm] &\ \implies \bbx{a = {1 \over 2}\,,\quad b = {1 \over 4}} \end{align}

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ANOTHER METHOD: The Laplace one. Namely,

\begin{align} &\bbox[10px,#ffd]{\ds{\int_{0}^{1}{x^{n - 1} \over x + 1}\,\dd x}} = {1 \over 2}\int_{0}^{1}{\pars{1 - x}^{n - 1} \over 1 - x/2}\,\dd x = {1 \over 2}\int_{0}^{1}{\expo{\pars{n- 1}\ln\pars{1 - x}} \over 1 - x/2}\,\dd x \\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, {1 \over 2}\int_{0}^{\infty}{\expo{-\pars{n - 1}x}\expo{-\pars{n - 1}x^{2}/2} \over 1 - x/2}\,\dd x \\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, {1 \over 2}\int_{0}^{\infty}\expo{-\pars{n - 1}x} \bracks{1 + {x \over 2} + \pars{{3 \over 4} - {n \over 2}}\,x^{2}}\,\dd x = {1 \over 2}\,{4n^{2} - 10n + 8 \over 4\pars{n - 1}^{3}} \\[5mm] & \sim \bbx{{\color{red}{1/2} \over n} + {\color{red}{1/4} \over n^{2}}} + \cdots \end{align}