Answer
$$\frac{1}{4}$$
Work Step by Step
$$\eqalign{
& \int_1^\infty {\frac{{\ln y}}{{{y^3}}}} dy \cr
& {\text{By the definition of the improper integrals, we have }} \cr
& \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx \cr
& {\text{then}}{\text{,}} \cr
& \int_1^\infty {\frac{{\ln y}}{{{y^3}}}} dy = \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{{\ln y}}{{{y^3}}}} \cr
& \cr
& {\text{Integrating }}\int {\frac{{\ln y}}{{{y^3}}}{\text{ by parts method}}} \cr
& {\text{Let }}\,\,\,\,\,u = \ln y,\,\,\,\,\,du = \frac{1}{y}dy \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = \frac{1}{{{y^3}}},\,\,\,\,\,v = - \frac{1}{{2{y^2}}} \cr
& \int {udv} = uv - \int {vdu} \,\,\,\, \cr
& \to \,\,\,\int {\frac{{\ln y}}{{{y^3}}}} dy = - \frac{{\ln y}}{{2{y^2}}} - \int {\left( { - \frac{1}{{2{y^2}}}} \right)\left( {\frac{1}{y}dy} \right)} \cr
& \,\,\,\,\,\,\, = - \frac{{\ln y}}{{2{y^2}}} + \frac{1}{2}\int {\frac{1}{{{y^3}}}dy} \cr
& \,\,\,\,\,\,\, = - \frac{{\ln y}}{{2{y^2}}} + \frac{1}{2}\left( { - \frac{1}{{2{y^2}}}} \right) + C \cr
& \,\,\,\,\,\,\, = - \frac{{\ln y}}{{2{y^2}}} - \frac{1}{{4{y^2}}} + C \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{{\ln y}}{{{y^3}}}} = \mathop {\lim }\limits_{b \to \infty } \left( { - \frac{{\ln y}}{{2{y^2}}} - \frac{1}{{4{y^2}}}} \right)_1^b \cr
& = \mathop {\lim }\limits_{b \to \infty } \left( { - \frac{{\ln b}}{{2{b^2}}} - \frac{1}{{4{b^2}}}} \right) - \mathop {\lim }\limits_{b \to \infty } \left( { - \frac{{\ln 1}}{{2{{\left( 1 \right)}^2}}} - \frac{1}{{4{{\left( 1 \right)}^2}}}} \right) \cr
& = \mathop {\lim }\limits_{b \to \infty } \left( { - \frac{{\ln b}}{{2{b^2}}} - \frac{1}{{4{b^2}}}} \right) + \mathop {\lim }\limits_{b \to \infty } \left( {\frac{1}{4}} \right) \cr
& {\text{Evaluate the limit when }}b \to \infty \cr
& = - \frac{{\ln \left( \infty \right)}}{{2{{\left( \infty \right)}^2}}} - \frac{1}{{4{{\left( \infty \right)}^2}}} + \frac{1}{4} \cr
& {\text{Use the L'Hopital's rule to }} - \mathop {\lim }\limits_{b \to \infty } \frac{{\ln b}}{{2{b^2}}} = - \mathop {\lim }\limits_{b \to \infty } \frac{{\frac{1}{b}}}{{4b}} = - \mathop {\lim }\limits_{b \to \infty } \frac{1}{{4{b^2}}} \cr
& = - \frac{1}{{4{{\left( \infty \right)}^2}}} - \frac{1}{{4{{\left( \infty \right)}^2}}} + \frac{1}{4} \cr
& = 0 + \frac{1}{4} \cr
& = \frac{1}{4} \cr} $$
