Problem source

Find the area of the region between the curve 
$\displaystyle y = {\ln x \over x}\,$ and the $x$-axis, for $1 \le x \le a$.
What happens to this area as $a$ tends to infinity?
Find the volume of the solid obtained when the region  between the curve 
$\displaystyle y = {\ln x \over x}\,$ and the $x$-axis, for $1 \le x\le a$,
 is rotated through $2 \pi$ radians about the $x$-axis.
What happens to this volume as $a$ tends to infinity?

Solution source

\begin{align*}
&& \int_1^a \frac{\ln x}{x} \d x &= \left [ \ln x \cdot \ln x\right ]_1^a - \int_1^a \frac{\ln x}{x} \d x \\
\Rightarrow && \int_1^a \frac{\ln x}{x} \d x &= \frac12 \left ( \ln a \right) ^2 \\
&& \int_1^\infty \frac{\ln x}{x} \d x &= \lim_{a \to \infty} \frac12 (\ln a)^2 \\
&&&= \infty
\end{align*}

\begin{align*}
&& \pi \int_1^a \left ( \frac{\ln x}{x} \right)^2 \d x &= \pi \int_{u=0}^{u=\ln a} \left ( \frac{u}{e^u} \right)^2 e^u \d u \\
&&&= \pi \int_0^{\ln a} u^2 e^{-u} \d u \\
&&&= \pi \left [-u^2e^{-u} \right]_0^{\ln a} +\pi \int_0^{\ln a} 2u e^{-u} \d u \\
&&&= -\frac{\pi}{a} (\ln a)^2 + \pi \left [-2u e^{-u} \right]_0^{\ln a} + \pi \int_0^{\ln a} e^{-u} \d u \\
&&&= -\frac{\pi}{a} (\ln a)^2- \frac{2 \pi}{a}\ln a+\pi \left (1 - \frac{1}{a} \right) \\
\\
&& \pi \int_1^{\infty} \left ( \frac{\ln x}{x} \right)^2 \d x &= \lim_{a \to \infty} \left ( -\frac{\pi}{a} (\ln a)^2- \frac{2 \pi}{a}\ln a+\pi \left (1 - \frac{1}{a} \right)  \right) \\
&&&= \pi
\end{align*}