Answer
$${\text{Diverges}}$$
Work Step by Step
$$\eqalign{
& \int_0^{ + \infty } {\ln x} dx \cr
& {\text{The integrand }}\ln x{\text{ is not continuous to }}x = 0,{\text{ then}} \cr
& {\text{Improper Integral with an Infinite Limit of Integration}} \cr
& \int_0^{ + \infty } {\ln x} dx = \mathop {\lim }\limits_{a \to {0^ + }} \int_a^1 {\ln x} dx + \mathop {\lim }\limits_{b \to + \infty } \int_1^b {\ln x} dx \cr
& {\text{*Computing }}\mathop {\lim }\limits_{a \to {0^ + }} \int_a^1 {\ln x} dx \cr
& \mathop {\lim }\limits_{a \to {0^ + }} \int_a^1 {\ln x} dx = \mathop {\lim }\limits_{a \to {0^ + }} \left[ {x\ln x - x} \right]_a^1 \cr
& = \mathop {\lim }\limits_{a \to {0^ + }} \left[ {\left( {1\ln 1 - 1} \right) - \left( {a\ln a - a} \right)} \right] \cr
& = \mathop {\lim }\limits_{a \to {0^ + }} \left[ { - 1 - \left( {a\ln a - a} \right)} \right] \cr
& {\text{Find when }}a \to {{\text{0}}^ + } \cr
& = - 1 - 0 \cr
& = - 1 \cr
& {\text{*Computing }}\mathop {\lim }\limits_{b \to + \infty } \int_1^b {\ln x} dx \cr
& \mathop {\lim }\limits_{b \to + \infty } \int_1^b {\ln x} dx = \mathop {\lim }\limits_{b \to + \infty } \left[ {x\ln x - x} \right]_1^b \cr
& = \mathop {\lim }\limits_{b \to + \infty } \left[ {\left( {b\ln b - b} \right) - \left( {1\ln 1 - 1} \right)} \right] \cr
& = \mathop {\lim }\limits_{b \to + \infty } \left[ {\left( {b\ln b - b} \right) + 1} \right] \cr
& = \left( {b\ln b - b} \right) + 1 \cr
& {\text{Find when }}b \to + \infty \cr
& = + \infty \cr
& {\text{Then,}} \cr
& \int_0^{ + \infty } {\ln x} dx{\text{ diverges}} \cr} $$
