Answer
Divergent
Work Step by Step
Let \[I=\int_{1}^{\infty}\frac{\ln x}{x}dx\]
\[I=\lim_{t\rightarrow \infty}\int_{1}^{t}\frac{\ln x}{x}dx\;\;\;\ldots (1)\]
\[I=\lim_{t\rightarrow \infty}\left[\frac{(\ln x)^2}{2}\right]_{1}^{t}\]
\[I=\lim_{t\rightarrow \infty}\left[\frac{(\ln t)^2}{2}-0\right]\]
$\;\;\;\;\;\;\;\;\;\;$does not exist
Since limit on R.H.S. of (1) does not exist
So $I$ is divergent.
