Answer
$1$
Work Step by Step
$$\eqalign{
& \int_1^\infty {2{x^{ - 3}}} dx \cr
& {\text{Using the definition of improper integrals }} \cr
& \underbrace {\int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{t \to \infty } } \int_a^t {f\left( x \right)} dx}_ \Downarrow \cr
& \int_1^\infty {2{x^{ - 3}}} dx = \mathop {\lim }\limits_{t \to \infty } \int_1^t {2{x^{ - 3}}} dx \cr
& {\text{Integrating}} \cr
& = \mathop {\lim }\limits_{t \to \infty } \left[ {\frac{{2{x^{ - 2}}}}{{ - 2}}} \right]_1^t \cr
& = - \mathop {\lim }\limits_{t \to \infty } \left[ {\frac{1}{{{x^2}}}} \right]_1^t \cr
& = - \mathop {\lim }\limits_{t \to \infty } \left[ {\frac{1}{{{t^2}}} - \frac{1}{{{{\left( 1 \right)}^2}}}} \right] \cr
& = - \mathop {\lim }\limits_{t \to \infty } \left[ {\frac{1}{{{t^2}}} - 1} \right] \cr
& {\text{Evaluate the limit when }}t \to \infty \cr
& = - \left( {0 - 1} \right) \cr
& = 1 \cr
& {\text{Therefore}}{\text{, the integral converges.}} \cr} $$
