Calculus of a Single Variable, Chapter 8, 8.8, Section 8.8, Problem 31

int_0^infty cos (pi x)dx=
Substitute u=pi x => du=pi dx => dx=(du)/pi, u_l=pi cdot 0=0, u_u=pi cdot infty=infty. (u_l and u_u are lower and upper bound respectively).
1/pi int_0^infty cos u du=1/pi sin u|_0^infty=1/pi(lim_(u to infty)sin u-sin0)
The integral does not converge (it diverges) because lim_(u to infty) sin u does not exist.
The image below shows the graph of the function (blue) and area between it and x-axis representing the value of integral (green positive and red negative). We can see that any such integral (with infinite bound(s)) of periodic function will diverge.