nice exercise to understand some analysis like MVT and limit application

Exercise:

Consider a function $f$ defined by $f\left( x \right)=\frac{x\ln x}{{{x}^{2}}-1}\,\,,\,x\in \left( 0,1 \right)$

1) Show that, $1\le \frac{\left( x+1 \right)f\left( x \right)}{x}\le \frac{1}{x}\,\,,\,\,\forall x\in \left( 0,1 \right)$

2) Deduce the bound for ${{R}_{n}}=\int_{0}^{1}{\frac{{{x}^{n}}\ln x}{{{x}^{2}}-1}dx}$

3) Find $\underset{n\to \infty }{\mathop{\lim }}\,{{R}_{n}}$

4) Calculate $\int_{\alpha }^{1}{{{x}^{n}}\ln x\,dx\,\,\,,\alpha \in \left( 0,1 \right)}$ then deduce the value of ${{I}_{n}}=\int_{0}^{1}{{{x}^{n}}\ln x\,dx}\,\,\,,n\ge 1$

5) Express $\int_{0}^{1}{f\left( x \right)\,dx}$ in terms of ${{I}_{1}},{{I}_{3}},...,{{I}_{2n+1}}\,\,\And \,\,{{R}_{2n+1}}\,\,\,\,,\,\,\,\,n\ge 2$

Solution:

1) Let $f\left( x \right)=\ln x$ which is continuous on $\left[ x,1 \right]$ and differentiable on $\left( x,1 \right)$

So by MVT $\exists c\in \left[ x,1 \right]$ such that $f'\left( c \right)=\frac{\ln x-\ln 1}{x-1}$

Notice that $f'\left( x \right)=\frac{1}{x}\Leftrightarrow f'\left( c \right)=\frac{1}{c}$  and Observe that $x\le c\le 1\Leftrightarrow 1\le \frac{1}{c}\le \frac{1}{x}$

Thus $1\le \frac{\ln x}{x-1}\le \frac{1}{x}\Leftrightarrow 1\le \frac{\left( x+1 \right)f\left( x \right)}{x}\le \frac{1}{x}$

2) From part (1) we have

$1\le \frac{\left( x+1 \right)f\left( x \right)}{x}\le \frac{1}{x}\Leftrightarrow 1\le \frac{\ln x}{x-1}\le \frac{1}{x}\Leftrightarrow {{x}^{n}}\le \frac{{{x}^{n}}\ln x}{x-1}\le {{x}^{n-1}}\Leftrightarrow \frac{{{x}^{n}}}{x+1}\le \frac{{{x}^{n}}\ln x}{{{x}^{2}}-1}\le \frac{{{x}^{n-1}}}{x+1}$

But $\frac{{{x}^{n-1}}}{x+1}\ge {{x}^{n-1}}\,\,\,\,\And \,\,\,\,\,\,\,\,\frac{{{x}^{n}}}{x+1}\le {{x}^{n}}\,\,\,for\,\,n\ge 1$

Hence $\frac{{{x}^{n}}}{x+1}\le {{x}^{n}}\le \frac{{{x}^{n}}\ln x}{{{x}^{2}}-1}\le {{x}^{n-1}}\le \frac{{{x}^{n-1}}}{x+1}\Leftrightarrow {{x}^{n}}\le \frac{{{x}^{n}}\ln x}{{{x}^{2}}-1}\le {{x}^{n-1}}$

Thus $\int_{0}^{1}{{{x}^{n}}dx}\le \int_{0}^{1}{\frac{{{x}^{n}}\ln x}{{{x}^{2}}-1}dx\le \int_{0}^{1}{{{x}^{n-1}}dx}}$

$\Leftrightarrow \left[ \frac{{{x}^{n+1}}}{n+1} \right]_{0}^{1}\le {{R}_{n}}\le \left[ \frac{{{x}^{n-1+1}}}{n-1+1} \right]_{0}^{1}\Leftrightarrow \frac{1}{n+1}\le {{R}_{n}}\le \frac{1}{n}$

3) From part (2) we have $\frac{1}{n+1}\le {{R}_{n}}\le \frac{1}{n}\Leftrightarrow \underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n+1}\le \underset{n\to \infty }{\mathop{\lim }}\,{{R}_{n}}\le \underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\Leftrightarrow 0\le \underset{n\to \infty }{\mathop{\lim }}\,{{R}_{n}}\le 0$

Hence by squeeze theorem $\underset{n\to \infty }{\mathop{\lim }}\,{{R}_{n}}=0$

4) Let $u=\ln x\,\,\And \,\,\,\,dv={{x}^{n}}dx\Leftrightarrow du=\frac{1}{x}dx\,\,\And \,v=\frac{{{x}^{n+1}}}{n+1}$

So $\int_{\alpha }^{1}{{{x}^{n}}\ln x\,dx}=\left[ \frac{{{x}^{n+1}}}{n+1}\ln x \right]_{\alpha }^{1}-\int_{\alpha }^{1}{\frac{{{x}^{n+1}}}{n+1}\times \frac{1}{x}dx}$

But $\int_{\alpha }^{1}{\frac{{{x}^{n+1}}{{x}^{-1}}}{n+1}dx}=\int_{\alpha }^{1}{\frac{{{x}^{n}}}{n+1}dx}=\frac{1}{n+1}\int_{\alpha }^{1}{{{x}^{n}}dx}=\left[ \frac{{{x}^{n+1}}}{{{\left( n+1 \right)}^{2}}} \right]_{\alpha }^{1}$

So $\int_{\alpha }^{1}{{{x}^{n}}\ln x\,dx}=\left[ \frac{{{x}^{n+1}}}{n+1}\ln x \right]_{\alpha }^{1}-\left[ \frac{{{x}^{n+1}}}{{{\left( n+1 \right)}^{2}}} \right]_{\alpha }^{1}=-\frac{{{\alpha }^{n+1}}}{n+1}\ln \alpha -\left( \frac{1}{{{\left( n+1 \right)}^{2}}}-\frac{{{\alpha }^{n+1}}}{{{\left( n+1 \right)}^{2}}} \right)$

                    $=\frac{{{\alpha }^{n+1}}}{{{\left( n+1 \right)}^{2}}}-\frac{1}{{{\left( n+1 \right)}^{2}}}-\frac{{{\alpha }^{n+1}}\ln \alpha }{n+1}$

So $\int_{0}^{1}{{{x}^{n}}\ln x\,dx}=\underset{\alpha \to 0}{\mathop{\lim }}\,\int_{\alpha }^{1}{{{x}^{n}}\ln x\,dx}=\underset{\alpha \to 0}{\mathop{\lim }}\,\left( \frac{{{\alpha }^{n+1}}}{{{\left( n+1 \right)}^{2}}}-\frac{1}{{{\left( n+1 \right)}^{2}}}-\frac{{{\alpha }^{n+1}}\ln \alpha }{n+1} \right)$

                     $=\underset{\alpha \to 0}{\mathop{\lim }}\,\frac{{{\alpha }^{n+1}}}{{{\left( n+1 \right)}^{2}}}-\underset{\alpha \to 0}{\mathop{\lim }}\,\frac{{{\alpha }^{n+1}}\ln \alpha }{n+1}-\underset{\alpha \to 0}{\mathop{\lim }}\,\frac{1}{{{\left( n+1 \right)}^{2}}}$

But $0<\alpha <1\Leftrightarrow 0<{{\alpha }^{n+1}}<{{\alpha }^{n}}$ as $\alpha \to 0\,,\,{{\alpha }^{n}}\to 0$ thus by squeeze theorem ${{\alpha }^{n+1}}\to 0$

So $\underset{\alpha \to 0}{\mathop{\lim }}\,\frac{{{\alpha }^{n+1}}\ln \alpha }{n+1}=\frac{1}{n+1}\underset{\alpha \to 0}{\mathop{\lim }}\,{{\alpha }^{n}}\left( \alpha \ln \alpha  \right)=\frac{1}{n+1}\left( 0 \right)=0$

Thus $\int_{0}^{1}{{{x}^{n}}\ln x\,dx}=0-0-\frac{1}{{{\left( n+1 \right)}^{2}}}=\frac{-1}{{{\left( n+1 \right)}^{2}}}$

5) $\int_{0}^{1}{f\left( x \right)dx}=\int_{0}^{1}{\frac{x\ln x}{{{x}^{2}}-1}dx}$ the trick here is to add zeros for $n\ge 2$

So we have $\frac{x\ln x}{{{x}^{2}}-1}=\frac{x\ln x-0}{{{x}^{2}}-1}$

$=\frac{x\ln x+{{x}^{3}}\ln x-{{x}^{3}}\ln x+{{x}^{5}}\ln x-{{x}^{5}}\ln x+....+{{x}^{2n+1}}\ln x-{{x}^{2n+1}}\ln x+{{x}^{2n+3}}\ln x-{{x}^{2n+3}}\ln x}{{{x}^{2}}-1}$

$=\frac{x\ln x\left( 1-{{x}^{2}} \right)+{{x}^{3}}\ln x\left( 1-{{x}^{2}} \right)+.....+{{x}^{2n+1}}\ln \left( 1-{{x}^{2}} \right)+{{x}^{2n+3}}\ln x}{{{x}^{2}}-1}$

$=\frac{\left( 1-{{x}^{2}} \right)\left[ x\ln x+{{x}^{3}}\ln x+....+{{x}^{2n+1}}\ln x \right]+{{x}^{2n+3}}\ln x}{{{x}^{2}}-1}$

$=-\left( x\ln x+{{x}^{3}}\ln x+.....+{{x}^{2n+1}}\ln x \right)+\frac{{{x}^{2n+3}}\ln x}{{{x}^{2}}-1}$

$=-\left( {{I}_{1}}+{{I}_{3}}+.....+{{I}_{2n+1}} \right)+{{R}_{2n+1}}$

$=-\left( \frac{1}{4}+\frac{1}{16}+....+\frac{1}{4{{\left( n+1 \right)}^{2}}} \right)+{{R}_{2n+1}}$

$=-\frac{1}{4}\sum\limits_{i=0}^{n}{\frac{1}{{{\left( i+1 \right)}^{2}}}+{{R}_{2n+1}}}$