How do you show whether the improper integral $\int{\lim \dfrac{\ln x}{x}dx}$ converges or diverges from 1 to infinity?

Hint: To check whether $\int{\lim \dfrac{\ln x}{x}dx}$ converges or diverges from 1 to infinity, we need to apply the intervals so that we will get $\int\limits_{1}^{\infty }{\dfrac{\ln x}{x}dx}$ . Now, we will take the limits to get $\displaystyle \lim_{K \to \infty }\int\limits_{1}^{K}{\dfrac{\ln x}{x}dx}$ . Then we will do the integration to get \[\displaystyle \lim_{K \to \infty }\left( \dfrac{{{\ln }^{2}}K}{2} \right)\] . We will then apply the limit. If the limit exists and is a finite number, that is it's not plus or minus infinity, then we can say that the given integral is convergent. If the limit either doesn't exist or is plus or minus infinity, then we can say that the integral is divergent.

Complete step-by-step solution:
We need to check whether $\int{\lim \dfrac{\ln x}{x}dx}$ converges or diverges from 1 to infinity. We can say that an integral of a function is convergent if the associated limit exists and is a finite number, that is it's not plus or minus infinity. We can say that an integral of a function is divergent if the associated limit either doesn't exist or is plus or minus infinity.
We are given that $\int{\lim \dfrac{\ln x}{x}dx}$ . Let us apply the given intervals here. We will get
$\int\limits_{1}^{\infty }{\dfrac{\ln x}{x}dx}$
 Let us write this in terms of limits.
$\int\limits_{1}^{\infty }{\dfrac{\ln x}{x}dx}=\displaystyle \lim_{K \to \infty }\int\limits_{1}^{K}{\dfrac{\ln x}{x}dx}...\left( i \right)$
Now, we have to integrate this. Let us consider $u=\ln x$ . When differentiating this with respect to x, we will get
$\begin{align}
  & \dfrac{du}{dx}=\dfrac{1}{x} \\
 & \Rightarrow du=\dfrac{1}{x}dx \\
\end{align}$
When we substitute $x=1$ in $u=\ln x$ we will get $u=\ln 1=0$ .
Also when $x=K$ , $u=\ln K$ .
Let us substitute these values in (i).
$\displaystyle \int\limits_{1}^{K}{\lim \dfrac{\ln x}{x}dx}=\displaystyle \lim_{K \to \infty }\int\limits_{0}^{\ln K}{udu}$
We know that integral of ${{x}^{n}}$, that is, $\int\limits_{a}^{b}{{{x}^{n}}dx}=\left[ \dfrac{{{x}^{n+1}}}{n+1} \right]_{a}^{b}$ . Therefore, the above integral becomes
$\displaystyle \lim_{K \to \infty }\int\limits_{0}^{\ln K}{udu}=\displaystyle \lim_{K \to \infty }\left[ \dfrac{{{u}^{2}}}{2} \right]_{0}^{\ln K}$
Let us now give the intervals. We know that after integration, $\left[ x \right]_{a}^{b}=\left( b-a \right)$
\[\displaystyle \lim_{K \to \infty }\left[ \dfrac{{{u}^{2}}}{2} \right]_{0}^{\ln K}=\displaystyle \lim_{K \to \infty }\left( \dfrac{{{\left( \ln K \right)}^{2}}}{2}-\dfrac{{{0}^{2}}}{2} \right)\]
We can solve this to
\[\displaystyle \lim_{K \to \infty }\left( \dfrac{{{\left( \ln K \right)}^{2}}}{2}-\dfrac{{{0}^{2}}}{2} \right)=\displaystyle \lim_{K \to \infty }\left( \dfrac{{{\ln }^{2}}K}{2} \right)\]
Now, let us apply the limit. We will get
\[\displaystyle \lim_{K \to \infty }\left( \dfrac{{{\ln }^{2}}K}{2} \right)=\left( \dfrac{{{\ln }^{2}}\infty }{2} \right)=\infty \]
Hence the improper integral $\int{\lim \dfrac{\ln x}{x}dx}$ diverges from 1 to infinity.

Note: Students must be thorough with the concept of convergence and divergence. They must know all the rules and identities of integration. They have a chance of making mistakes when writing the function $\int{\lim \dfrac{\ln x}{x}dx}$ with the intervals. They may write it as $\int\limits_{1}^{\infty }{\lim \dfrac{\ln x}{x}dx}$ instead of $\int\limits_{1}^{\infty }{\dfrac{\ln x}{x}dx}$ .