\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\df}{\ds\frac}
\parindent=0pt
\begin{document}


{\bf Question}

By using the change of variables $\tau=\log t$ and integrating by
parts, show that as$x \to +\infty$

\begin{description}
\item[(a)]
$\ds\int_x^\infty \ds\frac{dt}{t^2\log t} \sim \ds\frac{1}{x\log
x}$

\item[(b)]
$\ds\int_2^x \ds\frac{dt}{t\log(\log(t))} \sim \ds\frac{\log
x}{\log(\log(x))}$
\end{description}

\vspace{.5in}

{\bf Answer}

\begin{description}
\item[(a)]
Let $\tau=\ln t\Rightarrow \,d\tau=\df{dt}{t}$

Therefore $\ds\int_x^\infty \df{dt}{t^2\ln t}=\ds\int_x^\infty
\df{d\tau}{t\ln t}=\ds\int_{\ln x}^\infty
\df{e^{-\tau}}{\tau}\,dt$

$x\to +infty \Rightarrow \ln x \to +\infty$

Use integration by parts

$\begin{array} {ll} dv=e^{-\tau} & u=\df{1}{\tau}\\ v=-e^{-\tau} &
du=-\df{1}{\tau^2} \end{array}$

Therefore $\begin{array}{rcl} \ds\int_x^\infty \df{dt}{\tau^2 \ln
t} & = & \left[\df{-e^{-\tau}}{\tau}\right]_{\ln
x}^\infty-\ds\int_{\ln x}^\infty \df{e^{-\tau}}{\tau^2}\,d\tau\\ &
= & \df{e^{-\ln x}}{\ln x}+R\\ & = & \df{1}{x\ln x}+R \end{array}$

\begin{eqnarray*} |R| & = & \left|\ds\int_{\ln x}^\infty \df{e^{-\tau}}{\tau^2}
\,d\tau\right|\\ & \leq & \ds\int_{\ln x}^\infty
\df{e^{-\tau}}{\tau^2}\,d\tau\\ & = & \df{1}{(\ln
x)^2}\ds\int_{\ln x}^\infty \left(\df{\ln
x}{\tau}\right)^2e^{-\tau}\, d\tau\\ & & \ln x > \tau\\ & \leq &
\df{1}{(\ln x)^2}\ds\int_{\ln x}^\infty e^{-\tau}\, d\tau\\ & = &
\df{1}{x(\ln x)^2}\\ & = & o\left(\df{1}{x\ln x}\right)\
\rm{since}\ \ln x\geq 1,\ x \to +\infty
\end{eqnarray*}

Therefore $\ds\int_x^\infty \sim \df{1}{x\ln x},\ x\to+\infty$

\newpage
\item[(b)]
This is more involves: $\tau=\ln t$ as above gives

$\ds\int_2^x \df{dt}{t\log(\log(t))}=\ds\int_{\ln 2}^{\ln
x}\df{d\tau}{\log\tau}$

set $\tau_1=\log\tau \Rightarrow d\tau_1=\df{d\tau}{\tau}$

Therefore $K=\ds\int_2^x
\df{dt}{t\log(\log(t))}=\ds\int_{\log(\log(2))}{\log(\log(x))}\df{d\tau_1
e^{\tau_1}}{\tau_1}$

Now integrate by parts:

$\begin{array}{ll} v=\df{1}{\tau_1} & du=e^{\tau_1}\\
dv=-\df{1}{tau_1^2} & u=e^{\tau_1}\end{array}$

\begin{eqnarray*} K & = &
\left[\df{e^{\tau_1}}{\tau_1}\right]_{\log(\log(2))}^{\log(\log(x))}+\ds\int
_{\log(\log(2))}^{\log(\log(x))}\df{e^{\tau_1}}{\tau_1^2}\,d\tau_1\\
& = &
\df{e^{\log(\log(x))}}{\log(\log(x))}-\df{e^{\log(\log(2))}}{\log(\log(2))}
+\ds\int_{\log(\log(2))}^{\log(\log(x))}
\df{e^{\tau_1}}{\tau_1^2}\, d\tau_1\\ & = & \df{\log
x}{\log(\log(x))}-\undb{{\log
2}{\log(\log(2))}}+\undb{O\left(\df{\log
x}{[\log(\log(x))]^2}\right)}\end{eqnarray*}

\hspace{2in} just a \ \ \ \ \ \ \ \ \ by similar arguments to

\hspace{1.8in}finite number\ \ \ \ \ \ \ \ \ \ \ \ \ \ part (a)

Now as $\begin{array} {rcl} x & \to & +\infty\\ \log x & \to &
+\infty\\ \log(\log(x)) & \to & +\infty\ \rm{(more\ slowly)}
\end{array}$

even so we still have

$K=\df{\log x}{\log(\log x)}+o\left(\df{\log x}{\log(\log
x)}\right)$

so $K \sim \df{\log x}{\log(\log x)},\ \ x\to +\infty$
\end{description}

\end{document}
