## Thomas' Calculus 13th Edition

$${\text{}}{\text{The improper integral diverges.}}$$
\eqalign{ & \int_1^\infty {\frac{{\ln z}}{z}} dz \cr & {\text{By the definition of the improper integrals, we have }} \cr & \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx \cr & {\text{then}}{\text{,}} \cr & \int_1^\infty {\frac{{\ln z}}{z}} dz = \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{{\ln z}}{z}dz} \cr & \cr & {\text{Integrating }}\int {\frac{{\ln z}}{z}} dz{\text{ using the substitution method}} \cr & \,\,u = \ln z,\,\,\,du = \frac{1}{z}dz \cr & \,\,\,\int {\frac{{\ln z}}{z}} dz = \int u du = \frac{{{u^2}}}{2} + C \cr & \,\,\,{\text{replace }}u = \ln z \cr & \,\,\,\, = \frac{{{{\left( {\ln z} \right)}^2}}}{2} + C \cr & \cr & \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{{\ln z}}{z}dz} = \mathop {\lim }\limits_{b \to \infty } \left( {\frac{{{{\left( {\ln z} \right)}^2}}}{2}} \right)_1^b \cr & \,\,\,\,\,\,\,{\text{ }}\mathop {\lim }\limits_{b \to \infty } \left( {\frac{{{{\left( {\ln b} \right)}^2}}}{2} - \frac{{{{\left( {\ln 1} \right)}^2}}}{2}} \right) \cr & \,\,\,\,\,\,\,{\text{ }}\mathop {\lim }\limits_{b \to \infty } \left( {\frac{{{{\left( {\ln b} \right)}^2}}}{2}} \right) \cr & {\text{Evaluate the limit when }}b \to \infty \cr & \,\,\,\,\,\,\,\,\, = \frac{{{{\left( {\ln \infty } \right)}^2}}}{2} \cr & \,\,\,\,\,\,\,\,\, = \infty \cr & {\text{Then}}{\text{, the improper integral diverges.}} \cr}