Answer
$${\text{The improper integral diverges}}$$
Work Step by Step
$$\eqalign{
& \int_1^\infty {\frac{{\ln x}}{x}dx} \cr
& {\text{By the Definition of Improper Integrals with Infinite }} \cr
& {\text{Integration Limits}} \cr
& \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx,{\text{ so}} \cr
& \int_e^\infty {\frac{{\ln x}}{x}} dx = \mathop {\lim }\limits_{b \to \infty } \int_1^b {\ln x\left( {\frac{1}{x}} \right)} dx \cr
& {\text{Integrate, }}\int {{u^n}} du = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr
& \mathop {\lim }\limits_{b \to \infty } \int_1^b {\ln x\left( {\frac{1}{x}} \right)} dx = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{{\left( {\ln x} \right)}^2}}}{2}} \right]_1^b = \frac{1}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {{{\left( {\ln x} \right)}^2}} \right]_1^b \cr
& \frac{1}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {{{\left( {\ln x} \right)}^2}} \right]_1^b = \frac{1}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {{{\left( {\ln b} \right)}^2} - {{\left( {\ln 1} \right)}^2}} \right] \cr
& \frac{1}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {{{\left( {\ln b} \right)}^2} - 0} \right] \cr
& {\text{Evaluate the limit}} \cr
& \frac{1}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {{{\left( {\ln b} \right)}^2}} \right] = \frac{1}{2}\left[ {{{\left( {\ln \infty } \right)}^2}} \right] \cr
& {\text{Simplify}} \cr
& \frac{1}{2}\left[ \infty \right] \cr
& \infty \cr
& {\text{Therefore,}} \cr
& {\text{The improper integral diverges}} \cr} $$
