Calculus of a Single Variable, Chapter 8, 8.8, Section 8.8, Problem 26

int_1^infty (ln x)/x dx=
Substitute u=lnx => du=1/x dx, u_l=ln 1=0, u_u=ln infty=infty (u_l and u_u denote lower and upper bound respectively).
int_0^infty u du=u^2/2|_0^infty=lim_(u to infty) u^2/2-0^2/2=infty-0=infty
As we can see the integral does not converge.
The image below shows graph of the function and area under it representing the value of the integral. Looking at the image we can see that the graph approaches the x-axis (function converges to zero). However, this convergence is "relatively slow" and gives us infinite area under the graph.