Answer
$${\text{The integral is divergent}}$$
Work Step by Step
$$\eqalign{
& \int_{10}^\infty {{x^{ - 1}}} dx \cr
& = \int_{10}^\infty {\frac{1}{x}} dx \cr
& {\text{by the definition of an improper integral}}{\text{}} \cr
& \int_{10}^\infty {{x^{ - 1}}} dx = \mathop {\lim }\limits_{b \to \infty } \int_{10}^b {{x^{ - 1}}} dx \cr
& {\text{integrate by the logarithmic rule}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left( {\ln \left| x \right|} \right)_{10}^b \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\left( {{\text{see page 388}}} \right) \cr
& = \mathop {\lim }\limits_{b \to \infty } \left( {\ln \left| b \right| - \ln \left| {10} \right|} \right) \cr
& {\text{evaluate the limit when }}b \to \infty \cr
& = \ln \left( \infty \right) - \ln \left( {10} \right) \cr
& = \infty \cr
& {\text{the integral is divergent}} \cr} $$
