## Calculus: Early Transcendentals (2nd Edition)

$= \frac{1}{{\,\left( {p - 1} \right)\,{{\left( {\ln 2} \right)}^{p - 1}}}}$
$\begin{gathered} \int_2^\infty {\frac{{dx}}{{x{{\ln }^p}x}}} \hfill \\ \hfill \\ use\,\,the\,\,definition\,\,of\,\,improper\,\,{\text{integrals}} \hfill \\ \hfill \\ \int_2^\infty {\frac{{dx}}{{x{{\ln }^p}x}}} = \mathop {\lim }\limits_{b \to \infty } \int_2^b {\frac{{dx}}{{x{{\ln }^p}x}}} \hfill \\ \hfill \\ set \hfill \\ \ln x = u\,\,\,then\,\,\,\frac{{dx}}{x} = du \hfill \\ \hfill \\ {\text{substituting}}\,\,{\text{and}}\,\,{\text{integrating}} \hfill \\ \hfill \\ \mathop {\lim }\limits_{b \to \infty } \int_2^b {\frac{{dx}}{{x{{\ln }^p}x}}} = \mathop {\lim }\limits_{b \to \infty } d\,\,\left[ {\frac{1}{{\,\left( {1 - p} \right){{\ln }^{p - 1}}x}}} \right]_2^b \hfill \\ \hfill \\ use\,\,the\,\,ftc \hfill \\ \hfill \\ = \mathop {\lim }\limits_{b \to \infty } \,\,\left[ {\frac{1}{{\,\left( {1 - p} \right){{\ln }^{p - 1}}b}} - \frac{1}{{\,\left( {1 - p} \right){{\ln }^{p - 1}}2}}} \right] \hfill \\ \hfill \\ evaluate\,\,the\,\,\lim it \hfill \\ \hfill \\ = \frac{1}{{\,\left( {p - 1} \right)\,{{\left( {\ln 2} \right)}^{p - 1}}}} \hfill \\ \end{gathered}$