Limit exercise solved by using monotone convergence theorem (Lebesgue , Measure Theory )

Exercise:

Show that , $\underset{n\to \infty }{\mathop{\lim }}\,\int_{0}^{1}{\frac{\arctan \left( x \right)}{1+{{x}^{n}}}dx}=\frac{1}{4}\left( \pi -\ln 4 \right)$

Solution: Let’s show that ${{f}_{n}}\left( x \right)=\frac{1}{1+{{x}^{n}}}$ is monotone

So we need to compare ${{f}_{n+1}}$ and${{f}_{n}}$

Observe that $x\in \left[ 0,1 \right]\Leftrightarrow 0\le x\le 1$ multiply both sides by ${{x}^{n}}$ to get

$0\le {{x}^{n+1}}\le {{x}^{n}}$ Adding 1 to get $1\le 1+{{x}^{n+1}}\le 1+{{x}^{n}}$

Thus $1\ge \frac{1}{1+{{x}^{n+1}}}\ge \frac{1}{1+{{x}^{n}}}\Leftrightarrow {{f}_{n+1}}\left( x \right)\ge {{f}_{n}}\left( x \right)$

Hence ${{f}_{n}}$ is decreasing  thus ${{f}_{n}}$ is monotone

Now lets prove $\underset{n\to \infty }{\mathop{\lim }}\,\frac{\arctan x}{1+{{x}^{n}}}=\arctan x$ a.e

Let $x\in \left[ 0,1 \right]$ so we have to consider two cases are when $x=1\,\,\And \,\,0\le x<1$

Case 1: if $x=1\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\frac{\arctan x}{1+{{x}^{n}}}=\frac{\arctan \left( 1 \right)}{1+1}=\frac{\pi /4}{2}=\frac{\pi }{8}$

Case 2  if $0\le x<1\Leftrightarrow {{x}^{n}}\to 0,n\to \infty $ so $\underset{n\to \infty }{\mathop{\lim }}\,\frac{\arctan x}{1+{{x}^{n}}}=\arctan x$

Note that $M=\left\{ 1 \right\}\,\,\And \,\,\mu \left( M \right)=0$ thus ${{f}_{n}}\left( x \right)\xrightarrow[n\to \infty ]{}\arctan x$ a.e

So by Monotone convergence theorem we get

$\underset{n\to \infty }{\mathop{\lim }}\,\int_{0}^{1}{\frac{\arctan x}{1+{{x}^{n}}}dx}=\int_{0}^{1}{\underset{n\to \infty }{\mathop{\lim }}\,\frac{\arctan x}{1+{{x}^{n}}}dx}=\int_{0}^{1}{\arctan x\,dx}$

Let $u=\arctan x\,\,\,\,\And \,\,\,dv=\int{dx}\Leftrightarrow du=\frac{1}{1+{{x}^{2}}}dx\,\,\,\And \,\,\,v=x$

So $\int_{0}^{1}{\arctan x\,dx}=\left[ x\arctan x \right]_{0}^{1}-\int_{0}^{1}{\frac{x}{1+{{x}^{2}}}dx}=\frac{\pi }{4}-\int_{0}^{1}{\frac{x}{1+{{x}^{2}}}dx}$

Let $w=1+{{x}^{2}}\Leftrightarrow dw=2xdx\Leftrightarrow \frac{dw}{2}=xdx$

So \(\int_{0}^{1}{\frac{x}{1+{{x}^{2}}}dx}=\frac{1}{2}\int_{w\left( 0 \right)}^{w\left( 1 \right)}{\frac{dw}{w}=\frac{1}{2}\ln \left| w \right|_{1}^{2}}=\frac{1}{2}\ln 2\)

Hence $\int_{0}^{1}{\arctan x\,dx}=\frac{\pi }{4}-\frac{1}{2}\ln 2=\frac{\pi -2\ln 2}{4}=\frac{\pi -\ln 4}{4}=\frac{1}{4}\left( \pi -\ln 4 \right)$