Answer
Convergent $\;,\;$ 1
Work Step by Step
Let \[I=\int_{e}^{\infty}\frac{1}{x(\ln x)^2}dx\]
\[I=\lim_{t\rightarrow \infty}\int_{e}^{t}\frac{1}{x(\ln x)^2}dx\;\;\;\ldots (1)\]
Let \[I_1=\int\frac{1}{x(\ln x)^2}dx\]
Substitute $\;r=\ln x\;\;\;\ldots (2)$
\[\Rightarrow dr=\frac{1}{x}dx\]
\[I_1=\int\frac{dr}{r^2}=\frac{r^{-1}}{-1}\]
Using (2)
\[I_1=\frac{-1}{\ln x}\;\;\;\ldots (3)\]
Using (3) in (1)
\[I=\lim_{t\rightarrow \infty}\left[\frac{-1}{\ln x}\right]_{e}^{t}\]
\[I=\lim_{t\rightarrow \infty}\left[\frac{-1}{\ln t}+\frac{1}{1}\right]\]
\[I=1\]
Since limit on R.H.S. of (1) exists so $I$ is convergent and $I=1$.
