<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\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}

Examine where the dominant contributions arises from, perform a
local expansion and Use Watson's lemma to show

\begin{description}
\item[(a)]
$\ds\int_0^\infty e^{-x(t^2+2t)}(1+t)^{\frac{5}{2}}\, dt \sim
\ds\frac{1}{2x},\ x\to+\infty$

\item[(b)]
$\ds\int_0^\infty e^{-x(t^2+2t)}\log(1+t)\, dt \sim \ds\frac{\log
2}{2x},\ x\to+\infty$

\item[(c)]
$\ds\int_0^\infty e^{-x(t^2+2t)}\log(1+t)\, dt \sim
\ds\frac{1}{4x^2},\ x\to+\infty$

\item[(d)]
$\ds\int_0^\infty e^{-x(t^2+2t)}(t+3t^2)^{-\frac{1}{2}}\, dt \sim
\sqrt{\ds\frac{\pi}{2x}},\ x\to+\infty$
\end{description}



\vspace{.5in}

{\bf Answer}

\begin{description}
\item[(a)]
$\ds\int_0^\infty e^{-x(t^2+2t)}(1+t)^{\frac{5}{2}}\, dt\ \ x \to
+\infty$

$h(t)=(t^2+2t) \Rightarrow \undb{h'(t)=2t+2} \Rightarrow h''(t)=2$

\hspace{.5in} min. at $t=-1$ which is \un{outside} our range of
integration

${}$

\setlength{\unitlength}{.5in}
\begin{picture}(4,4)
\put(0,2){\vector(1,0){4}}

\put(2,0){\vector(0,1){4}}

\put(1.5,2){\line(0,-1){.75}}

\put(2,2){\circle*{.125}}

\put(1.5,1.25){\circle*{.125}}

\put(3.8,1.5){\vector(-1,0){1.8}}

\put(3,1.25){\makebox(0,0){range of}}

\put(3,.75){\makebox(0,0){integration}}

\put(4,2){\makebox(0,0)[l]{$t$}}

\put(2,4){\makebox(0,0)[l]{$h(t)$}}
\end{picture}

Thus the minimum value of $e^{-x h(t)}$ occurs when $t=0,\
h(t)=0$. The dominant contribution will come from this \un{linear}
endpoint at $t=0$. This differs slightly from the examples in the
notes.

We could try an integration by parts but this looks messy. Try
instead a Watson type argument and Taylor expand about $t=0$.

$$h(t)-\undb{h(0)}=\undb{h'(0)}(t-0)+O(t=0)^2\ \ (1)$$

\hspace{1.5in} =0 \ \ \ \ \un{Not} zero here as it's a linear
endpoint $h'(0)=2$ (from above)

Therefore set $\begin{array}{rclc} u &amp; = &amp; h(t)-h(0) &amp; (2)\\ du &amp;
= &amp; h'(t)\,dt &amp; (3) \end{array}$

But $(1) \Rightarrow\ u \approx 2t$.

So in integral:

$\begin{array}{rcl} I &amp; = &amp; \ds\int_0^\infty
e^{-x(t^2+2t)}(1+t)^\frac{5}{2}\,dt\\ &amp; = &amp; e^{-x\ln(0)}
\ds\int_0^\infty
e^{-xu}\df{(1+t(u))^{\frac{5}{2}}}{h'(t(u))}\,du\\ &amp; \approx &amp;
\ds\int_0^\infty e^{-xu}\df{(1+\frac{u}{2})}{h'(t(u))}\, du
\end{array}$

$h'(t)=\undb{h'(0)}+\df{h''(0)}{2}(t-0)^2+\cdots=2$

\hspace{.5in} $\ne 0$ as it's a \un{linear} endpoint

from above, to leading order.

Therefore $du \approx 2\,dt$

Therefore $I \approx \ds\int_0^\infty
e^{-xu}\df{(1+\frac{u}{2})^\frac{5}{2}}{2}\,du$

Now apply Laplace: contribution centred about $u=0$ as $x \to
+\infty$.

\begin{eqnarray*} I &amp; \sim &amp;
\undb{\df{(1+\frac{0}{2})^\frac{5}{2}}{2}}\ds\int_0^\infty
e^{-xu}\, du,\ \rm{as}\ x \to +\infty\\ &amp; &amp; \mbox{to leading order
this is a constant.}\\ &amp; &amp; \mbox{So take it outside the
integral}\\ &amp; \sim &amp; \df{1}{2x}\ \ x\to +\infty\ \ \mbox{as
required}\end{eqnarray*}

\newpage
\item[(b)]
The dominant contribution again comes from $t=0$ (same $h(t)$ as
above). The only difference is the value of $f9t)=\log(2+t)$ at
$t=0$. Thus the method goes through as for (a) with:

\begin{eqnarray*} I &amp; = &amp; \ds\int_0^\infty
e^{-x(t^2+2t)}\log(2+t)\,dt\\ &amp; \sim &amp;
\df{\log(2+0)}{2}\ds\int_0^\infty e^{-xu}\,du\ \rm{as}\ x\to
+\infty\\ &amp; \sim &amp; \df{\log 2}{2x}\ \ x\to + \infty
\end{eqnarray*}

\item[(c)]
Here the dominant contribution is again from $t=0$ ($h(t)=t^2+2t$
again). But now $f(t)=\log(1+t)$ which is 0 at $t=0$. This does
\un{not} necessarily mean that the contribution from $t=0$
vanishes. Instead we must go to higher order in the expansion of
$\df{f(t)}{h'(t)}$, keeping it \un{inside} the integral.

Proceed as above until:

$I=\ds\int_0^\infty
e^{-x(t^2+2t)}\log(1=t)\,dt-\ds\int_0^\infty\df{\log(1+t(u))}{h'(t(u))}\,du$

where $h'(t)\approx 2$ and $u \approx 2t$.

Now just expand the log inside the integral:

$I\sim\ds\int_0^\infty
\df{e^{-xu}(t(u)-\frac{t^2(u)}{2}+\cdots)}{2}\,du \sim
\ds\int_0^\infty e^{-xu}\df{u}{4}$ to leading order

${}$

$x\to + \infty$

Therefore $I \sim \df{1}{4}\ds\int_0^\infty e^{-xu}\, u \sim
\df{1}{4x^2}\ \ x\to +\infty$

\item[(d)]
As above the dominant contribution is from the linear $t=0$
endpoint ($h(t)=t^2+2t$).

Consider $f(t)$

$$f(t)=\df{1}{\sqrt{t(1+3t)}}=\df{1}{\sqrt{t}}+O(t^{\frac{1}{2}}),\
t \to o^+$$

The method proceeds as above, but we now retain the leading order
of $f(t)$ as $t\to 0^+$ in the integral.

$J=\ds\int_0^\infty
e^{-x(t^2+2t)}(t+3t^2)^{-\frac{1}{2}}\,dt=\ds\int_0^\infty
e^{-xu}\df{[t(u)+3t^2(u)]^{-\frac{1}{2}}}{h'(t(u))}\,du$

$h'(t) \approx 2$ to leading order and $u \approx 2t$

Thus $J \sim \ds\int_0^\infty \df{e^{-xu}}{2} \cdot
\df{1}{\sqrt{t(u)}} \approx \df{1}{2}\ds\int_0^\infty
\df{e^{-xu}}{\sqrt{\frac{u}{2}}}\,du=\df{1}{\sqrt{2}}\ds\int_0^\infty
\df{e^{-xu}}{u^{\frac{1}{2}}}\,du$

Remembering the definition of the $\Gamma$-function, this last
integral is $\df{\Gamma(\frac{1}{2})}{x^{\frac{1}{2}}}$.

Therefore $\ds\int_0^\infty e^{-x(t^2+2t)}(t+3t^2)\,dt \sim
\df{\Gamma(\frac{1}{2})}{\sqrt{2x}}=\sqrt{\df{\pi}{2x}}\ \ x \to
+\infty$
\end{description}




\end{document}
</pre></body></html>