\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}

This question requires some ingenuity, rather than any standard

method.

Remembering that

$$\left(1+\ds\frac{1}{t}\right)^t=\exp\left(t\log\left[1+\ds\frac{1}{t}\right]\right)$$

and expanding the integral as a power series in $\ds\frac{1}{t}$
which is convergent for $t>1$, show that

$$\ds\int_1^x \left(1+\ds\frac{1}{t}\right)^t\, dt \sim
xe-\ds\frac{1}{2}e\log x+O(1)$$

\vspace{.5in}

{\bf Answer}

$\left(1+\df{1}{t}\right)^t=e^{t\log(1+\frac{1}{t})}$

$\ds\int_1^x \left(1+\df{1}{t}\right)^t\,dt=\ds\int_1^x
e^{t\log(1+\frac{1}{t})}\,dt$

Expand $\begin{array}{rcl} t\log\left(1+\df{1}{t}\right) & = &
t\left[\df{1}{t}-\df{1}{2t^2}+\df{1}{3t^3}-\cdots\right]\\ & &
\mbox{\un{convergent} for} |t|>1 (|frac{1}{t}|<1)\\ & = &
1-\df{1}{2t}+\df{1}{3t^2}-\cdots \end{array}$

Therefore

$\begin{array}{rcl} e^{t\log(1+\frac{1}{t})} & = &
e^{1-\frac{1}{2t}+\frac{1}{3t^2}-\cdots}\\ & = & e \cdot
e^{-\frac{1}{2t}+\frac{1}{3t^2}}-\cdots\\ & = &
e\left[1-\df{1}{2t}+\df{1}{3t^2}-\cdots+\df{1}{2!}\left(-\df{1}{2t}+\df{1}{3t^2}+\cdots\right)^2+\cdots\right]\\
& = & e\left[1-\df{1}{2t}+O\left(\df{1}{t^2}\right)\right]\
\rm{for}\ |t|>1 \end{array}$

Thus

$\begin{array}{rcl} \ds\int_1^x \left(1+\df{1}{t}\right)^t\, dt &
= & e\ds\int_1^x
\left[1-\df{1}{2t}+O\left(\df{1}{t^2}\right)\right]\,dt\\ & = &
e\left[t-\df{1}{2}\log t+O\left(\df{1}{t}\right)\right]_1^x\\ & =
& e\left[x-\df{1}{2}\log x+O91)\right]\ \rm{as}\ x \to +\infty\\ &
\sim & ex-\df{1}{2}e\log x\ \rm{as}\ x \to +\infty\ \ \mbox{as
required.}\end{array}$

\end{document}