Answer
$$\frac{1}{{\left( {p - 1} \right){2^{p - 1}}}}$$
Work Step by Step
$$\eqalign{
& \int_{{e^2}}^\infty {\frac{{dx}}{{x{{\ln }^p}x}}} \cr
& {\text{definition of improper integral}} \cr
& \int_{{e^2}}^\infty {\frac{{dx}}{{x{{\ln }^p}x}}} = \mathop {\lim }\limits_{b \to \infty } \int_{{e^2}}^b {\frac{{dx}}{{x{{\ln }^p}x}}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \int_{{e^2}}^b {\frac{{dx}}{{x{{\ln }^p}x}}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \int_{{e^2}}^b {{{\left( {\ln x} \right)}^{ - p}}\left( {\frac{1}{x}} \right)dx} \cr
& {\text{evaluate the integral}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left( {\frac{{{{\left( {\ln x} \right)}^{ - p + 1}}}}{{ - p + 1}} - \frac{{{{\left( {\ln {e^2}} \right)}^{ - p + 1}}}}{{ - p + 1}}} \right) \cr
& {\text{simplify}} \cr
& p > 1 \cr
& = \mathop {\lim }\limits_{b \to \infty } \left( {\frac{1}{{\left( { - p + 1} \right){{\left( {\ln x} \right)}^{p - 1}}}} - \frac{{{{\left( {\ln {e^2}} \right)}^{ - p + 1}}}}{{ - p + 1}}} \right) \cr
& {\text{evaluate the limit}} \cr
& = \frac{1}{{\left( { - p + 1} \right){{\left( {\ln \infty } \right)}^{p - 1}}}} - \frac{{{{\left( {\ln {e^2}} \right)}^{ - p + 1}}}}{{ - p + 1}} \cr
& = - \frac{{{{\left( 2 \right)}^{ - p + 1}}}}{{ - p + 1}} \cr
& = \frac{{{{\left( 2 \right)}^{ - p + 1}}}}{{p - 1}} \cr
& = \frac{1}{{\left( {p - 1} \right){2^{p - 1}}}} \cr} $$
