Nice exercise posted by me collected from different math bloggers

Exercise:

Evaluate the following,

 $\int_{0}^{1}{\left( \sqrt{x\sqrt[3]{x\sqrt[4]{x......}}}+\sqrt{x+\sqrt{x+\sqrt{x+.....}}} \right)dx}$

Solution:

We have, $\int_{0}^{1}{\left( \sqrt{x\sqrt[3]{x\sqrt[4]{x......}}}+\sqrt{x+\sqrt{x+\sqrt{x+.....}}} \right)dx}$

           $=\int_{0}^{1}{\sqrt{x\sqrt[3]{x\sqrt[4]{x.....}}}dx}+\int_{0}^{1}{\sqrt{x+\sqrt{x+\sqrt{x+.....}}}dx}$

           $={{A}_{1}}+{{A}_{2}}$

Notice that, $\sqrt{x\sqrt[3]{x}}={{\left( x{{\left( x \right)}^{1/3}} \right)}^{1/2}}={{x}^{1/2}}.{{x}^{1/6}}={{x}^{\frac{1}{2!}+\frac{1}{3!}}}$

Thus, we may generate this result to be

                 $\sqrt{x\sqrt[3]{x\sqrt[4]{x.....}}}={{\left( x{{\left( x{{\left( x \right)}^{1/4}}.... \right)}^{1/3}} \right)}^{1/2}}={{x}^{\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+....}}$

Remember that, ${{e}^{x}}=\sum\nolimits_{n=0}^{\infty }{\frac{{{x}^{n}}}{n!}}=1+x+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}+....$

Hence $e=2+\frac{1}{2!}+\frac{1}{3!}+....\Leftrightarrow \sum\nolimits_{n=2}^{\infty }{\frac{1}{n!}=e-2}$

Thus, ${{A}_{1}}=\int_{0}^{1}{\sqrt{x\sqrt[3]{x\sqrt[4]{x......}}}dx}=\int_{0}^{1}{{{x}^{e-2}}dx}=\left[ \frac{{{x}^{e-1}}}{e-1} \right]_{0}^{1}=\frac{1}{e-1}$

To compute ${{A}_{2}}$ we need to start by defining a recursive sequence ${{\left( {{f}_{n}}\left( x \right) \right)}_{n\ge 1}}$

with ${{f}_{1}}\left( x \right)=\sqrt{x}$ and ${{f}_{n+1}}\left( x \right)=\sqrt{x+{{f}_{n}}\left( x \right)}$ for all $0\le x\le 1$

We have ${{f}_{n+1}}\left( x \right)=\sqrt{x+{{f}_{n}}\left( x \right)}<{{f}_{n}}\left( x \right)$ hence ${{f}_{n}}\left( x \right)>{{f}_{n+1}}\left( x \right)$.

So that, ${{\left( {{f}_{n}} \right)}_{n}}$ is a monotone increasing sequence bounded above by $x>0$.

By the monotone convergent theorem${{f}_{n}}\to f$ as $n\to \infty $ hence

It follows that, $f\left( x \right)=\sqrt{x+f\left( x \right)}$ then ${{f}^{2}}\left( x \right)-f\left( x \right)+x=0$

$\Rightarrow {{f}^{2}}\left( x \right)-2\frac{1}{2}f\left( x \right)+\frac{1}{4}-\frac{1}{4}-x=0\Leftrightarrow {{\left( f\left( x \right)-\frac{1}{2} \right)}^{2}}=\frac{1+4x}{4}\Leftrightarrow f\left( x \right)=\frac{1}{2}\pm \sqrt{x+\frac{1}{4}}$

Now we are not sure which sign is corrected one

For any $0<x\le 1$ ,
we have $\frac{1}{4}<x+\frac{1}{4}\le 1+\frac{1}{4}\Leftrightarrow \frac{1}{2}<\sqrt{x+\frac{1}{4}}\le \frac{\sqrt{5}}{2}\Leftrightarrow 0<-\frac{1}{2}+\sqrt{x+\frac{1}{4}}\le \frac{\sqrt{5}}{2}\Rightarrow 0<-{{f}_{1}}\left( x \right)$

but $f\left( x \right)>0$ hence the accepted root is $f\left( x \right)=\frac{1}{2}+\sqrt{x+\frac{1}{4}}$

it follows that, ${{A}_{2}}=\int_{0}^{1}{\underset{n\to \infty }{\mathop{\lim }}\,{{f}_{n}}\left( x \right)dx}=\int_{0}^{1}{f\left( x \right)dx}=\int_{0}^{1}{\frac{1}{2}+\sqrt{x+\frac{1}{4}}\,dx}$

Let $u=x+\frac{1}{4}\Leftrightarrow du=dx$ and $u\left( 0 \right)=\frac{1}{4}\And u\left( 1 \right)=\frac{5}{4}$

Hence, \[{{A}_{2}}=\int_{0}^{1}{f\left( x \right)dx}=\left[ \frac{1}{2}x \right]_{0}^{1}+\int_{0}^{1}{\sqrt{x+\frac{1}{4}}\,dx}=\frac{1}{2}+\int_{1/4}^{5/4}{\sqrt{u}\,du}=\frac{1}{2}+\frac{2}{3}\left( {{u}^{3/2}} \right)_{1/4}^{5/4}=\frac{5}{12}\left( 1+\sqrt{5} \right)\]