Integral with limit exercise asked by Dan Sitrua in many math groups

Exercise:

Determine$\underset{n\to \infty }{\mathop{\lim }}\,\int_{1}^{e}{\left( \sum\nolimits_{k=1}^{n}{\frac{1}{x\left( k+\ln x \right)\left( k+1+\ln x \right)}} \right)}\,dx$

Solution: We have $\sum\nolimits_{k=1}^{n}{\frac{1}{x\left( k+\ln x \right)\left( k+1+\ln x \right)}}=\frac{1}{x}\sum\nolimits_{k=1}^{n}{\frac{1}{\left( k+\ln x \right)\left( k+1+\ln x \right)}}$

But $\frac{1}{\left( k+\ln x \right)\left( k+1+\ln x \right)}=\frac{A}{k+\ln x}+\frac{B}{k+1+\ln x}$

                                    $=\frac{A\left( k+1+\ln x \right)+B\left( k+\ln x \right)}{\left( k+\ln x \right)\left( k+1+\ln x \right)}=\frac{\ln x\left( A+B \right)+k\left( A+B \right)+A}{\left( k+\ln x \right)\left( k+1+\ln x \right)}$

So $A+B=0\,\,\And \,A=1\Leftrightarrow B=-1\,\,\,\And \,A=1\,$

Thus $\sum\nolimits_{k=1}^{n}{\frac{1}{\left( k+\ln x \right)\left( k+1+\ln x \right)}}=\sum\nolimits_{k=1}^{n}{\frac{1}{k+\ln x}-\frac{1}{k+1+\ln x}}$

Observe that ,

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

So $\sum\nolimits_{k=1}^{n}{\frac{1}{k+\ln x}-\frac{1}{k+1+\ln x}}=\frac{1}{1+\ln x}-\frac{1}{n+1+\ln x}$

Thus $\underset{n\to \infty }{\mathop{\lim }}\,\int_{1}^{e}{\left( \sum\nolimits_{k=1}^{n}{\frac{1}{x\left( k+\ln x \right)\left( k+1+\ln x \right)}} \right)dx=\underset{n\to \infty }{\mathop{\lim }}\,\int_{1}^{e}{\left( \frac{1}{1+\ln x}-\frac{1}{n+1+\ln x} \right)\frac{dx}{x}}}$

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

Let $u=\ln x\Leftrightarrow du=\frac{dx}{x}$ , $u\left( 1 \right)=0\,\,\And \,\,u\left( e \right)=1$

So $\int_{1}^{e}{\frac{dx}{x\left( 1+\ln x \right)}=\int_{0}^{1}{\frac{du}{1+u}=\ln \left| 1+u \right|_{0}^{1}=\ln \left( 2 \right)}}$ hence $\underset{n\to \infty }{\mathop{\lim }}\,\int_{1}^{e}{\frac{dx}{x\left( 1+\ln x \right)}=\ln 2}$

$\int_{1}^{e}{\frac{dx}{x\left( n+1+\ln x \right)}=\int_{0}^{1}{\frac{du}{n+1+u}}}=\ln \left| \left( n+1 \right)+u \right|_{0}^{1}=\ln \left( n+2 \right)-\ln \left( n+1 \right)=\ln \frac{n+2}{n+1}$

Thus $\underset{n\to \infty }{\mathop{\lim }}\,\int_{1}^{e}{\frac{dx}{x\left( n+1+\ln x \right)}=\underset{n\to \infty }{\mathop{\lim }}\,\ln \left( \frac{n+2}{n+1} \right)=}\ln \left( \underset{n\to \infty }{\mathop{\lim }}\,\frac{n+2}{n+1} \right)=\ln 1=0$ as ln is continuous function

So $\underset{n\to \infty }{\mathop{\lim }}\,\int_{1}^{e}{\left( \sum\nolimits_{k=1}^{n}{\frac{1}{x\left( k+\ln x \right)\left( k+1+\ln x \right)}} \right)dx=\ln 2}$