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\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\df}{\ds\frac}
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\begin{document}


{\bf Question}

Calculate the leading order asymptotics of the following integrals
as $x\to+\infty$

\begin{description}
\item[(a)]
$\ds\int_0^1 \ds\frac{e^{ix^2}}{1+t}\, dt$

\item[(b)]
$\ds\int_0^\pi e^{ix(\sin
t+2)}t^{\frac{3}{2}}\cos(\sqrt{t+1})\,dt$

\item[(c)]
$\ds\int_{-1}^{+1}e^{ix\cosh t}\sqrt{1-t^2}\, dt$

\end{description}



\vspace{.5in}

{\bf Answer}

\begin{description}
\item[(a)]
$\ds\int_0^1 \df{e^{ixt^2}}{(1+t)}\,dt \sim \ds\int_0^{\infty
e^{i\frac{\pi}{4}}} \df{e^{ixt^2}}{(1+t)}\,dt$

Clearly $t=0$ is a stationary point.

Rotate contour to get exponential decay:

$Re(ixt^2)&lt;0,\ Im(ixt^2)=0$

PICTURE \vspace{1in}

Extend contour to $\infty e^{i\frac{\pi}{4}}$.

$\begin{array}{rcl} u &amp; = &amp; it^2\\ du &amp; = &amp; -2it\,dt \end{array}
\Rightarrow t^2=iu \Rightarrow t=e^{-\frac{\pi}{4}}\sqrt{u}$

Therefore

$\begin{array}{rcl} I &amp; \sim &amp; \ds\int_0^\infty
\df{e^{-xu}}{(1+e^{i\frac{\pi}{4}}\sqrt{u})}\df{du}{-2i \cdot
e^{i\frac{\pi}{4}}\sqrt{u}}\\ &amp; \sim &amp;
\df{1}{2e^{-i\frac{\pi}{4}}}\ds\int_0^\infty
\df{e^{-xu}}{\sqrt{u}}\,du\\ &amp; \sim &amp;
\df{e^{i\frac{\pi}{4}}}{2}\df{\Gamma(\frac{1}{2})}{x^\frac{1}{2}}\\
&amp; \sim &amp; \df{e^{i\frac{\pi}{4}}}{2} \sqrt{\df{\pi}{x}}
\end{array}$

${}$

cf. formula (3.100) $h(t)=-t^2,\ h'(t)=-2t,\ h''(t)=-2$ for all
$t$

$\theta=+\df{i\pi}{4}$ from diagram

$$\begin{array}{rcl} I &amp; \sim &amp; \sqrt{\df{\pi}{2|x \times
-2|}}e^{ix \cdot 0+\frac{i\pi}{4}} \times 1 \leftarrow f(0),\
f=\frac{1}{1+t}\\ &amp; \sim &amp;
\df{1}{2}\sqrt{\df{\pi}{x}}e^{i\frac{\pi}{4}}\ \
\surd\end{array}$$

\item[(b)]
$J=\ds\int_0^\pi e^{ix(\sin
t+t)}t^{\frac{3}{2}}\cos(\sqrt{t+1})\,dt$

$\left.\begin{array} {rcl}h(t) &amp; = &amp; -\sin t+t\\ h'(t) &amp; = &amp; -\cos
t+1\\ h''(t) &amp; = &amp; \sin t\\ h'''(t) &amp; = &amp; \cos t
\end{array}\right\} \Rightarrow t=0$ is stationary endpoint

$h(0)=0,\ h''(0)=0,\ h'''(0)=+1$

This ($h''(0)=0$) is a cubic endpoint!

Too difficult! Can MA415 students see how to do it?

\item[(c)]
$I=\ds\int_{-1}^{+1} e^{ix\cosh t}\sqrt{1-t^2}\,dt$

$\left.\begin{array} {rcl}h(t) &amp; = &amp; -\cosh t\\ h'(t) &amp; = &amp; -\sinh
t\\ h''(t) &amp; = &amp; -\cosh t\end{array}\right\} \Rightarrow t=0$ is a
stationary point

$h(0)=h''(0)=-1$

Set $\begin{array}{rcl} u^2 &amp; = &amp; i[-\cosh t+1]\\2u\,du &amp; = &amp;
-i\sinh t\, dt \end{array}$

Locally at stationary points:

$$u^2 \approx -\df{1}{2}(t-0)^2i=-\df{t^2i}{2}\Rightarrow u
\approx \df{t}{\sqrt{2}}e^{-i\frac{\pi}{4}}$$

Therefore locally rotate contour such that

$Re(ix[\cosh t-1])&lt;0$ and $Im(ix[\cosh t-1])=0$

$\Rightarrow \left.\begin{array}{l}
Re\left(\df{ixt^2}{2}\right)&lt;0\\
Im\left(\df{ixt^2}{2}\right)=0\end{array}\right\} \Rightarrow
\begin{array}{l} x \to +\infty\\ t = |t|e^{i\frac{\pi}{4}}\end{array}$

PICTURE \vspace{1in}

Extend contour to $\infty$.

Therefore

$\begin{array}{rcl} I &amp; \sim &amp; \ds\int_{R e^{+i\frac{\pi}{4}}}^{R
e^{i\frac{\pi}{4}}} e^{ix \cosh t}\sqrt{1-t^2}\,dt\\ &amp; \sim &amp;
e^{ix}
\ds\int_{-\infty}^{+\infty}e^{-xu^2}\df{2u\sqrt{1-t^2}}{-i\sinh
t}\,du\ \ x\to +\infty\\ &amp; \sim &amp;
e^{ix}\ds\int_{-\infty}^{+\infty}
e^{-xu^2}\df{2u\sqrt{1-2iu^2}}{-i\sqrt{2}e^{i\frac{\pi}{4}}u}\,du\
\ x\to +\infty\\ &amp; \sim &amp;
e^{ix}\ds\int_{-\infty}^{+\infty}e^{ixu^2}\,du
\sqrt{2}e^{i\frac{\pi}{4}}\ \ x\to + \infty\\ &amp; \sim &amp;
e^{ix+i\frac{\pi}{4}}\sqrt{\df{\pi}{x}}\sqrt{2}\\ &amp; \sim &amp;
\sqrt{\df{2\pi}{x}}e^{ix+i\frac{\pi}{4}}\ \ x\to + \infty
\end{array}$

Check with (3.99)

$I \sim \sqrt{\df{2\pi}{|xh''(0)|}}e^{-ixh(0)+i\theta}f(0)\ \ x
\to +\infty$

$h(0)=-1,\ h''(0)=-1,\ f(0)=\sqrt{1}=1,\ \theta=+\frac{\pi}{4}$
from diagram

$\sim \sqrt{\df{2\pi}{x}}e^{ix+i\frac{\pi}{4}}\ \ \surd$
\end{description}


\end{document}
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