Calculus of a Single Variable, Chapter 8, 8.8, Section 8.8, Problem 32

int_0^infty sin(x/2)dx=
Substitute u=x/2 => du=dx/2 => dx=2du, u_l=0/2=0, u_l=infty/2=infty (u_l and u_u are lower and upper bound respectively).
2int_0^infty sin u du=-2cos u|_0^infty=-2(lim_(u to infty)cos u-cos 0)
The integral does not converge (it diverges) because lim_(u to infty)cos u does not exist.
The image below shows the graph of the function (blue) and area between it and -axis representing the value of the integral (green positive and red negative). We can see that any such integral (with infinite bound(s)) of periodic function will diverge.