\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}

Given that $b>a,\ \lambda>-1,\ h(x)>h(a),\ h'(a)\neq 0$, show that
as $x\to+\infty$

$$I(x)=\ds\int_a^b(t-a)^{\lambda}e^{-xh(t)}\,dt \sim
\ds\frac{e^{-xh(a)}\Gamma(\lambda+1)}{[xh'(a)]^{\lambda+1}}.$$

\vspace{.5in}

{\bf Answer}

$I(x)=\ds\int_a^b (t-a)^\lambda e^{-x\ln (t)}\,dt$

For convergence at $t=a$.\ \ \ \ It's a linear endpoint.

$\undb{b>a},\ \overbrace{\lambda>-1},\ \undb{h(x)>h(a)},\
\overbrace{h'(a) \ne 0}$


Integral on real line.

\hspace{1in}Endpoint at $x=a$ will dominate for $x>a$.

This is the general case of question 6: a dominant linear endpoint
as $x \to+\infty$ where $f(t)$ vanishes there.

Proceed as above:

\begin{eqnarray*} u & = & h(t)-h(a)\\ & = & h'(a)(t-a)+O[(t-a)^2]
\end{eqnarray*}

$\Rightarrow u \approx h'(a)(t-a)\ \ (1)$

Therefore
$I(x)=e^{-x\ln(a)}\ds\int_0^{h(b)-h(a)}e^{-xu}\df{[t(u)-a]^\lambda}{h'(t(u))}\,du$

$(1) \Rightarrow t-a \approx \df{u}{h'(a)}$

Therefore to leading order we have: $h'(t(u))=h'(a)$

$I(x) \approx
e^{-x\ln(a)}\ds\int_0^{h(b)-h(a)}\df{e^{-xu}u^\lambda}{[h'(a)]^{\lambda+1}}\,du$

as $x\to +\infty$ for Poincar\'{e} the upper limit can be set to
$\infty$ (with exponentially small error).

\newpage
Hence \begin{eqnarray*} I(x) & \sim &
\df{e^{-x\ln(a)}}{[h'(a)]^{\lambda+1}}\ds\int_0^\infty e^{-xu}\
u^{\lambda}\,du\\ & & \mbox{this is a known integral}\\ & \sim &
\df{e^{-x\ln(a)}\Gamma(\lambda+1)}{[x h'(a)]^{\lambda+1}}
\end{eqnarray*}

\end{document}