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


{\bf Question}

Consider the integral

$$\ds\int_0^{\infty} e^{-xt}e^{-\frac{1}{t}}\, dt,\ x\to+\infty$$

At first sight, it might seem that a direct application of
Watson's lemma should produce an asymptotic expansion for the
integral. Why is this not so? Use the change of variables
$t\sqrt{x}=s$ to show that

$$\ds\int_0^{\infty} e^{-xt}e^{-\frac{1}{t}}\,dt \sim
x^{-\frac{3}{4}}\sqrt{\pi}e^{-2\sqrt{x}}\ \ x\to+\infty$$

\vspace{.5in}

{\bf Answer}

$\ds\int_0^zinfty e^{-xt}e^{-\frac{1}{t}}\,dt\ \ x \to +\infty$

Watson's lemma requires $e^{-\frac{1}{t}}$ to be represented by an
asymptotic power series, with $e^{-\frac{1}{t}} \sim
a_0t^{\lambda_0}$ (where $\lambda_0>-1$) as $t\to 0^+$, say, as
the dominant term.

But $\lim_{t\to 0^+}
t^{-\lambda_0}e^{-\frac{1}{t}}=\lim_{\tau\to+\infty}\tau^{\lambda_0}e^{-\tau}=0
(\tau=\frac{1}{t})$ for all $\lambda_0$

Consequently no such expansion exists and we \un{cannot} use
Watson's lemma: $e^{-\frac{1}{t}}$ tends to zero faster than any
power of $t$ as $t\to 0^+$ and therefore has \un{no} power series
expansion about $t=0$.

Use change of variable given:$t\sqrt{x}=s$

$\Rightarrow  ds=\sqrt{x}\,dt$, so for $x>0$, the integral becomes

$\begin{array}{rcl} I & = & \ds\int e^{-xt}e^{-\frac{1}{t}}\,dt\\
& = & \df{1}{\sqrt{x}}\ds\int_0^\infty
e^{x\frac{s}{\sqrt{x}}}e^{-\frac{\sqrt{x}}{s}}\,ds\\ & = &
\df{1}{\sqrt{x}}\ds\int_0^\infty e^{-\sqrt{x}(s+\frac{1}{s})}\,ds
\end{array}$

We're now in the Laplace ball-park with large parameter $\sqrt{x}$

$$h(s)=s+\df{1}{s} \Rightarrow h'(s)=1-\df{1}{s^2} \Rightarrow
s=\pm 1\ \mbox{are min/max}$$

For $0 \leq s < \infty,\ s=+1$ is the critical point.

$h''(t)=+2$ so it's a min.

\newpage
Thus we use

$\begin{array}{rcl} u^2 & = & h(s)-h(1)=h(s)+2\\ 2u\,du & = &
h'(s)\,ds\\ u^2 & \approx & \df{h''(1)}{2}(s-1)^2=(s-1)^2
\end{array}$

so $u \approx s-1,\ h'(s)=h''(1)(s-1)=2(s-1) \approx 2u$

Therefore

\begin{eqnarray*} I & = &
\df{1}{\sqrt{x}}\ds\int_{-\sqrt{h(0)-h(1)}}^{+\infty}e^{-\sqrt{x}u^2-2\sqrt{x}}
\df{2u\, du}{h'(s(u))}\\ & \sim &
\df{2e^{-2\sqrt{x}}}{\sqrt{x}}\ds\int_{-\infty}^{+\infty}
\df{u\,du}{2u}\\ & \sim &
\df{e^{-2\sqrt{x}}}{\sqrt{x}}\ds\int_{-\infty}^{+\infty}e^{-\sqrt{x}u^2}\,du\\
& \sim & \df{e^{-2\sqrt{x}}}{\sqrt{x}}\sqrt{\df{\pi}{\sqrt{x}}}\\
& \sim & \df{e^{-2\sqrt{x}}\sqrt{\pi}}{x^{\frac{3}{4}}}
\end{eqnarray*}


\end{document}
