Answer
$${\text{The integral converges to }}\frac{1}{{p - 1}}{\text{ if }}p > 1$$
Work Step by Step
$$\eqalign{
& \int_1^\infty {\frac{1}{{{x^p}}}} dx \cr
& {\text{By the Definition of Improper Integrals with Infinite }} \cr
& {\text{Integration Limits}} \cr
& \int_1^\infty {\frac{1}{{{x^p}}}} dx = \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{1}{{{x^p}}}} dx \cr
& = \mathop {\lim }\limits_{b \to \infty } \int_1^b {{x^{ - p}}} dx \cr
& {\text{Integrate}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left( {\frac{{{x^{ - p + 1}}}}{{ - p + 1}}} \right)_1^b \cr
& = \frac{1}{{1 - p}}\mathop {\lim }\limits_{b \to \infty } \left( {{x^{ - p + 1}}} \right)_1^b \cr
& = \frac{1}{{1 - p}}\mathop {\lim }\limits_{b \to \infty } \left[ {{x^{1 - p}} - {1^{ - p + 1}}} \right] \cr
& {\text{Evaluate for }}p = 1,{\text{ }}p < 1,{\text{ }}p > 1 \cr
& {\text{For }}p = 1 \cr
& = \frac{1}{{1 - 1}}\mathop {\lim }\limits_{b \to \infty } \left( {{x^{1 - 1}} - {1^0}} \right) = \infty \cr
& {\text{For }}p < 1 \cr
& \frac{1}{{1 - p}}\left( {{{\left( \infty \right)}^{1 - p}} - 1} \right) = \infty \cr
& {\text{For }}p > 1 \cr
& \frac{1}{{1 - p}}\left( {\frac{1}{\infty } - 1} \right) = 0 \cr
& \frac{1}{{1 - p}}\left( {0 - 1} \right) = \frac{1}{{p - 1}} \cr
& {\text{Therefore, the integral converges to }}\frac{1}{{p - 1}}{\text{ if }}p > 1 \cr} $$
