Answer
Convergent $\;,\;$ 1
Work Step by Step
Let \[I=\int_{1}^{\infty}\frac{\ln x}{x^2}dx\]
\[I=\lim_{t\rightarrow \infty}\int_{1}^{t}\frac{\ln x}{x^2}dx\;\;\;\ldots (1)\]
Let \[I_1=\int\frac{\ln x}{x^2}dx\]
Substitute $\ln x=r \;\;\;\ldots (2)$
\[\Rightarrow \frac{1}{x}dx=dr\]
\[I_1=\int\frac{r}{e^r}dr\]
\[I_1=\int re^{-r}dr\]
Using integration by parts
\[I_1=r\int e^{-r}dr-\int \left((r)'\int e^{-r}dr\right)dr\]
\[I_1=-re^{-r}+\int e^{-r}dr\]
\[I_1=-re^{-r}-e^{-r}\]
From (2)
\[I_1=\frac{-(\ln x)}{x}-\frac{1}{x}\;\;\;\ldots (3)\]
Using (3) in (1)
\[I=\lim_{t\rightarrow \infty}\left[\frac{-(\ln x)}{x}-\frac{1}{x}\right]_{1}^{t}\]
\[I=\lim_{t\rightarrow \infty}\left[\frac{-(\ln t)}{t}-\frac{1}{t}-(0-1)\right]\]
\[I=\lim_{t\rightarrow \infty}\left[\frac{-(\ln t)}{t}+1\right]\]
Using L' Hopital's rule
\[I=\lim_{t\rightarrow \infty}\left[\frac{\frac{-1}{t}}{1}+1\right]\]
\[I=1\]
Since limit on R.H.S. of (1) exists so $I$ is convergent and $ I=1$.
