Answer
$$\frac{2}{\pi }$$
Work Step by Step
$$\eqalign{
& \int_1^\infty {\frac{1}{{{z^2}}}\sin \frac{\pi }{z}} dz \cr
& {\text{definition of improper integral}} \cr
& \int_1^\infty {\frac{1}{{{z^2}}}\sin \frac{\pi }{z}} dz = \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{1}{{{z^2}}}\sin \frac{\pi }{z}} dz \cr
& {\text{evaluate the integral}} \cr
& = \frac{1}{\pi }\mathop {\lim }\limits_{b \to \infty } \left. {\left( {\cos \frac{\pi }{z}} \right)} \right|_1^b \cr
& = \frac{1}{\pi }\mathop {\lim }\limits_{b \to \infty } \left( {\cos \frac{\pi }{b} - \cos \pi } \right) \cr
& = \frac{1}{\pi }\mathop {\lim }\limits_{b \to \infty } \left( {\cos \frac{\pi }{b} - \left( { - 1} \right)} \right) \cr
& = \frac{1}{\pi }\mathop {\lim }\limits_{b \to \infty } \left( {\cos \frac{\pi }{b} + 1} \right) \cr
& {\text{evaluate the limit}} \cr
& = \frac{1}{\pi }\left( {\cos \frac{\pi }{\infty } + 1} \right) \cr
& = \frac{1}{\pi }\left( {\cos 0 + 1} \right) \cr
& = \frac{1}{\pi }\left( {1 + 1} \right) \cr
& = \frac{2}{\pi } \cr} $$
