Answer
$$\frac{1}{\pi }$$
Work Step by Step
$$\eqalign{
& \int_2^\infty {\frac{{\cos \left( {\pi /x} \right)}}{{{x^2}}}} dx \cr
& {\text{definition of improper integral}} \cr
& \int_2^\infty {\frac{{\cos \left( {\pi /x} \right)}}{{{x^2}}}} dx = \mathop {\lim }\limits_{b \to \infty } \int_2^b {\frac{{\cos \left( {\pi /x} \right)}}{{{x^2}}}} dx \cr
& {\text{evaluate the integral}} \cr
& = - \frac{1}{\pi }\mathop {\lim }\limits_{b \to \infty } \left. {\left( {sin\left( {\pi /x} \right)} \right)} \right|_2^b \cr
& = - \frac{1}{\pi }\mathop {\lim }\limits_{b \to \infty } \left( {sin\left( {\pi /b} \right) - \sin \left( {\pi /2} \right)} \right) \cr
& = - \frac{1}{\pi }\mathop {\lim }\limits_{b \to \infty } \left( {sin\left( {\frac{\pi }{b}} \right) - 1} \right) \cr
& {\text{evaluate the limit}} \cr
& = - \frac{1}{\pi }\left( {sin\left( {\frac{\pi }{\infty }} \right) - 1} \right) \cr
& = - \frac{1}{\pi }\left( {sin\left( 0 \right) - 1} \right) \cr
& = \frac{1}{\pi } \cr} $$
