\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}

Show that as $x\to\infty$

$$\ds\int_x^\infty e^{-t}t^{\lambda-1}\, dt\sim x^\lambda
e^{-x}\left[\ds\frac{1}{x}+\ds\frac{\lambda-1}{x}
+\ds\frac{(\lambda-1)(\lambda-2)}{x^2}+O\left(\ds\frac{1}{x^4}\right)\right]$$


\vspace{.5in}

{\bf Answer}

Use integration by parts

$I=\ds\int_x^\infty e^{-t}t^{\lambda-1}\,dt$

$\begin{array} {rclrcl} u & = & t^{\lambda-1} & dv & = & e^{-t}\\
du & = & (\lambda-1)t^{\lambda-1} & v & = & -e^{-t} \end{array}$

\begin{eqnarray*} I & = &
[-t^{\lambda-1}e^{-t}]_x^\infty+(\lambda-1)\ds\int_x^\infty
t^{\lambda-2{e^{-t}}}\,dt\\ & = &
x^{\lambda-1}e^{-x}+(\lambda-1)\left[[-t^{\lambda-1}e^{-t}]_x^\infty
+\ds\int_x^\infty(\lambda-2)t^{\lambda-3}e^{-t}\,dt\right]\\ & &
\begin{array} {rclrcl} u & = & t^{\lambda-2} & dv & = & e^{-t}\\
du & = & (\lambda-2)t^{\lambda-3} & v & = & -e^{-t}\end{array}\\ &
= & x^{\lambda-1}e^{-x}+(\lambda-1)x^{\lambda-2}e^{-x}+
(\lambda-1)(\lambda-2)\ds\int_x^\infty e^{-t}t^{\lambda-3}\,dt\\ &
& \begin{array} {rclrcl} u & = & t^{\lambda-3} & dv & = & e^{-t}\\
du & = & (\lambda-3)t^{\lambda-4} & v & = & -e^{-t}\end{array}\\ &
&  \vdots\\ & = &
x^{\lambda-1}e^{-x}+(\lambda-1)x^{\lambda-2}e^{-x}+
(\lambda-1)(\lambda-2)(\lambda-3)x^{\lambda-3}e^{-x}\\ & & +
(\lambda-1)(\lambda-2)(\lambda-3)\ds\int_x^\infty
t^{\lambda-4}e^{-t}\,dt\\ & = & x^\lambda
e^{-x}\left(\df{1}{x}+\df{(\lambda-1)}{x^2}
+\df{(\lambda-1)(\lambda-2)(\lambda-3)}{x^3}+\df{R}{x^\lambda
e^{-x}}\right) \end{eqnarray*}

\begin{eqnarray*}
R & = & (\lambda-1)(\lambda-2)(\lambda-3)\ds\int_x^\infty
t^{\lambda-4}e^{-t}\\ & \leq &
|(\lambda-1)(\lambda-2)(\lambda-3)|\left|\ds\int_x^\infty
t^{\lambda-4}e^{-t}\,dt\right|\\ & \leq &
(\lambda-1)(\lambda-2)(\lambda-3)\ds\int_x^\infty|t^{\lambda-4}|e^{-t}\,dt\\
& & x>0,\ \rm{so}\\& \leq &
(\lambda-1)(\lambda-2)(\lambda-3)\ds\int_x^\infty
t^{\lambda-4}e^{-t}\,dt\\ & = &
\df{(\lambda-1)(\lambda-2)(\lambda-3)}{x^{4-\lambda}}\ds\int_x^\infty
t^{\lambda-4}x^{4-\lambda}e^{-t}\,dt\\ & = &
\df{(\lambda-1)(\lambda-2)(\lambda-3)}{x^{4-\lambda}}\ds\int_x^\infty
\left(\df{x}{t}\right)^{4-\lambda} e^{-t}\,dt\\ & & \rm{But}\ t
\geq x.\ \rm{Therefore}\ \left(\df{x}{t}\right)\leq 1\\
\Rightarrow R & \leq &
\df{(\lambda-1)(\lambda-2)(\lambda-3)}{x^{4-\lambda}}\ds\int_x^\infty
e^{-t}\,dt\\ & = &
\df{(\lambda-1)(\lambda-2)(\lambda-3)}{x^{4-\lambda}}e^{-x}\\ & &
\mbox{which, when divided by}\ x^{-1}e^{-x}\ \rm{is}\
O\left(\df{1}{x^4}\right)\\ \rm{Therefore}\ I & = & x^\lambda
e^{-x}\left(\df{1}{X}+\df{(\lambda-1)}{x}\right.\\ & & \left.+
\df{(\lambda-1)(\lambda-2)(\lambda-3)}{x^3}+O\left(\df{1}{x^4}\right)
\right)\end{eqnarray*}

Implied constant $(\lambda-1)(\lambda-2)(\lambda-3)$.



\end{document}