Answer
$${\text{diverges}}$$
Work Step by Step
$$\eqalign{
& \int_0^\infty {\sin \left( {\frac{x}{2}} \right)} dx \cr
& {\text{By the Definition of Improper Integrals with Infinite }} \cr
& {\text{Integration Limits}} \cr
& \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx,{\text{ so}} \cr
& \int_0^\infty {\sin \left( {\frac{x}{2}} \right)} dx = \mathop {\lim }\limits_{b \to \infty } \int_0^b {\sin \left( {\frac{x}{2}} \right)dx} \cr
& {\text{Integrating}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ { - 2\cos \left( {\frac{x}{2}} \right)} \right]_0^b \cr
& = - 2\mathop {\lim }\limits_{b \to \infty } \left[ {\cos \left( {\frac{b}{2}} \right) - \cos \left( {\frac{0}{2}} \right)} \right] \cr
& = - 2\mathop {\lim }\limits_{b \to \infty } \left[ {\cos \left( {\frac{b}{2}} \right) - 1} \right] \cr
& {\text{Evaluate the limit}} \cr
& = - 2\mathop {\lim }\limits_{b \to \infty } \left[ {\cos \left( {\frac{b}{2}} \right)} \right] + 2\mathop {\lim }\limits_{b \to \infty } 1 \cr
& {\text{The }}\mathop {\lim }\limits_{x \to \infty } \cos x{\text{ does not exist, so the integral diverges}} \cr} $$
