Problem source

Sketch the graph of 
\[ y= \dfrac1 { x \ln x} \text{ for $x>0$,  $x\ne1$}.\]
You may assume that $x\ln x \to 0$ as $x\to 0$.
The continuous random variable $X$ has probability density function 
\[
\f(x) = \begin{cases}
\dfrac \lambda {x\ln x}& \text{for $a\le x \le b$}\;, \\[3mm]
\ \ \  0 & \text{otherwise },
\end{cases}
\]
where $a$, $b$ and $\lambda$ are suitably chosen constants.
\begin{questionparts}
\item In the case $a=1/4$ and $b=1/2$, find $\lambda\,$.
\item In the case $\lambda=1$ and $a>1$, show that $b=a^\e$.
\item In the case $\lambda =1$ and $a=\e$, show that 
$\P(\e^{3/2}\le X \le \e^2)\approx \frac {31}{108}\,$.
\item In the case $\lambda =1$ and $a=\e^{1/2}$, find  
$\P(\e^{3/2}\le X \le \e^2)\;$.
\end{questionparts}

Solution source

\begin{questionparts}
\item 
\begin{align*}
1 &= \int_{1/4}^{1/2} \frac{\lambda}{x\ln x} \, dx \\
&= \lambda\left [ \ln |\ln x| \right ]_{1/4}^{1/2} \\
&= \lambda \l \ln |-\ln 2| - \ln |-2 \ln 2| \r \\
&= \lambda (-\ln 2)
\end{align*}
So $\lambda = -\frac{1}{\ln 2} = \frac{1}{\ln \frac12}$
\item \begin{align*}
1 &= \int_{a}^{b} \frac{1}{x\ln x} \, dx \\
&= \left [ \ln |\ln x| \right ]_{a}^{b} \\
&= \l \ln \ln b - \ln \ln a \r \\
&= \ln \l \frac{\ln b}{\ln a} \r \\
\end{align*}
So $b = e^{a}$

\item If $\lambda = 1, a = e, b = e^e$, then 
\begin{align*}
\P(\e^{3/2}\le X \le \e^2) &= \int_{e^{3/2}}^{e^2} \frac{1}{x \ln x} \, dx \\
&= \left [ \ln \ln x \right]_{e^{3/2}}^{e^2} \\
&= \ln 2 - \ln \frac{3}{2} \\
&= \ln \frac{4}{3} \\
&= \ln \l 1 + \frac{1}{3} \r \\
&\approx \frac{1}{3} - \frac{1}{2 \cdot 3^2} + \frac{1}{3 \cdot 3^3} - \frac{1}{4 \cdot 3^4} \\
&= \frac{31}{108}
\end{align*}

\item Note that $2 > e^{\frac12} > 1$ so $a = e^{\frac12}, b = e^{\frac{e}2}$. Since $3 > e \Rightarrow e^{3/2} > e^{\frac{e}{2}}$ this probability is out of range, therefore  $\P(\e^{3/2}\le X \le \e^2) = 0$
\end{questionparts}