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


{\bf Question}

Use Stirling's approximation to show that

$$\ds\frac{\Gamma(x+a)}{\Gamma(x+b)}\sim x^{a-b},\ x\to+\infty$$

\vspace{.5in}

{\bf Answer}

Stirling's formula is $\Gamma(x) \sim
\sqrt{2\pi}x^{x-\frac{1}{2}}e^{-x}\ \ x\to + \infty$

so substitute directly:

\begin{eqnarray*} \df{\Gamma(x+a)}{\Gamma(x+b)} & \sim &
\df{\sqrt{2\pi}(x+a)^{(x+a-\frac{1}{2})}e^{-x-a}}{\sqrt{2\pi}
(x+b)^{(x+b-\frac{1}{2})}e^{-x-b}}\ x \to +\infty\\ & \sim &
\left(\df{x+a}{a+b}\right)^xe^{b-a}\df{(x+a)^{a-\frac{1}{2}}}{(x+b)^{b-\frac{1}{2}}}\\
& \sim &
\df{(1+\frac{a}{x})^x}{(1+\frac{b}{x})^x}e^{b-a}x^{a-b}\df{(1+\frac{a}{x})
^{a-\frac{1}{2}}}{(1+\frac{b}{x})^{b-\frac{1}{2}}}
\end{eqnarray*}

Now $\left(1+\df{a}{x}\right)^x \to e^a$ as $x \to + \infty$ (by
definition of $e^a$).

Likewise $\left(1+\df{b}{x}\right)^x\to e^b$

Therefore $\begin{array}{rcl} \df{\Gamma(x+a)}{\Gamma(x+b)} & \sim
& \df{e^a}{e^b} \cdot \df{e^b}{e^a}
x^{a-b}\undb{\df{(1+\frac{a}{x})^{a-\frac{1}{2}}}{(1+\frac{b}{x})^{b-\frac{1}{2}}}}\\
& & \ \ \ \ \ \ =\df{1^{a-\frac{1}{2}}}{1^{b-\frac{1}{2}}}\
\rm{as}\ x \to +\infty\\ & \sim & x^{a-b}\ \ \rm{as}\ x\to +
\infty\end{array}$

Hence $\df{\Gamma(x+a)}{\Gamma(x+b)} \sim x^{a-b}\ \ \rm{as}\ x
\to +\ infty$ as required.

\end{document}