Originally Posted by
HallsofIvy You say you "tried u substitution" but don't say what substitution. Was it "u= ln(x)"? That would seem the obvious substitution. Of course then $\displaystyle du= (1/x)dx$ so the integral becomes $\displaystyle \int_1^\infty \frac{ln(x)}{x}(1/x)dx= \int_0^\infty \frac{u}{x}du$.
And are you saying "it didn't seem to get me anywhere" because of that "x" still in the integrand? Well, if u= ln(x), then $\displaystyle x= e^u$ so the integral becomes
$\displaystyle \int_0^\infty \frac{u}{e^u}du= \int_0^\infty ue^{-u}du$ which can be integrated "by parts".