Answer
Diverges
Work Step by Step
$$\eqalign{
& \int_0^\infty {\frac{1}{{{x^2}}}} dx \cr
& {\text{Expressing the integral as a sum of improper integrals type 1}} \cr
& {\text{and type 2 as follows:}} \cr
& \int_0^\infty {\frac{1}{{{x^2}}}} dx = \int_0^1 {\frac{1}{{{x^2}}}} dx + \int_1^\infty {\frac{1}{{{x^2}}}} dx,{\text{ }}0 > 1 \cr
& {\text{Calculating }}\int_0^1 {\frac{1}{{{x^2}}}} dx \cr
& {\text{Using the definition of improper integrals }} \cr
& \int_0^1 {\frac{1}{{{x^2}}}} dx = \mathop {\lim }\limits_{a \to {0^ + }} \int_a^1 {\frac{1}{{{x^2}}}} dx \cr
& = \mathop {\lim }\limits_{a \to {0^ + }} \left[ { - \frac{1}{x}} \right]_a^1 \cr
& = \mathop {\lim }\limits_{a \to {0^ + }} \left[ { - \frac{1}{1} - \frac{1}{a}} \right] \cr
& = \mathop {\lim }\limits_{a \to {0^ + }} \left[ { - 1 + \frac{1}{a}} \right] \cr
& = - 1 + \underbrace {\mathop {\lim }\limits_{a \to {0^ + }} \left( {\frac{1}{a}} \right)}_{{\text{Tends to }}\infty } \cr
& {\text{Diverges}} \cr
& {\text{The improper integral }}\int_0^1 {\frac{1}{{{x^2}}}} dx{\text{ diverges}}{\text{, then}} \cr
& \int_0^\infty {\frac{1}{{{x^2}}}} dx = \underbrace {\int_0^1 {\frac{1}{{{x^2}}}} dx}_{{\text{diverges}}} + \int_1^\infty {\frac{1}{{{x^2}}}} dx \cr
& {\text{Diverges}} \cr} $$
