Chapter 7 - Principles Of Integral Evaluation - 7.8 Improper Integrals - Exercises Set 7.8 - Page 554: 7

$$\frac{1}{2}$$

$$\eqalign{ & \int_e^{ + \infty } {\frac{1}{{x{{\ln }^3}x}}} dx \cr & \cr & {\text{using the definition 7}}{\text{.8}}{\text{.1 of improper integrals}} \cr & \,\,\,\int_a^{ + \infty } {f\left( x \right)} dx = \mathop {\lim }\limits_{b \to + \infty } \int_a^b {f\left( x \right)} dx \cr & \cr & {\text{then}} \cr & \,\,\,\int_e^{ + \infty } {\frac{1}{{x{{\ln }^3}x}}} dx = \mathop {\lim }\limits_{b \to + \infty } \int_e^b {\frac{1}{{x{{\ln }^3}x}}} dx \cr & = \mathop {\lim }\limits_{b \to + \infty } \int_e^b {\left( {\frac{1}{{{{\ln }^3}x}}} \right)\left( {\frac{1}{x}} \right)} dx \cr & = \mathop {\lim }\limits_{b \to + \infty } \int_e^b {{{\left( {\ln x} \right)}^{ - 3}}\left( {\frac{1}{x}} \right)} dx \cr & \cr & {\text{integrate by tables using the formula }}\int {{u^n}du = \frac{{{u^{n + 1}}}}{{n + 1}}} + C \cr & = \mathop {\lim }\limits_{b \to + \infty } \left[ {\frac{{{{\left( {\ln x} \right)}^{ - 2}}}}{{ - 2}}} \right]_e^b \cr & = - \frac{1}{2}\mathop {\lim }\limits_{b \to + \infty } \left[ {\frac{1}{{{{\ln }^2}x}}} \right]_e^b \cr & = - \frac{1}{2}\mathop {\lim }\limits_{b \to + \infty } \left[ {\frac{1}{{{{\ln }^2}b}} - \frac{1}{{{{\ln }^2}e}}} \right] \cr & = - \frac{1}{2}\mathop {\lim }\limits_{b \to + \infty } \left[ {\frac{1}{{{{\ln }^2}b}} - 1} \right] \cr & \cr & {\text{calculate the limit when }}b \to + \infty \cr & = - \frac{1}{2}\left[ {\frac{1}{{{{\ln }^2}\left( \infty \right)}} - 1} \right] \cr & = - \frac{1}{2}\left( {\frac{1}{\infty } - 1} \right) \cr & = - \frac{1}{2}\left( { - 1} \right) \cr & = \frac{1}{2} \cr & \cr & {\text{then}}{\text{,}} \cr & {\text{The integral converges to }}\frac{1}{2} \cr} $$