\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}

\begin{description}
\item[(a)]
$\ds\int_0^1 e^{ixt^3}\, dt \sim
\ds\frac{\Gamma(\frac{1}{3})e^{\frac{i\pi}{6}}}{3x^{\frac{1}{3}}},\
x\to+\infty$

\item[(b)]
$\ds\int_0^{\infty} \ds\frac{e^{ixt^3}}{\sqrt{t}}\, dt \sim
\ds\frac{\Gamma(\frac{1}{6})e^{\frac{i\pi}{12}}}{3x^{\frac{1}{6}}},\
x\to+\infty$
\end{description}
\vspace{.5in}

{\bf Answer}



%Question 18%

\begin{description}
\item[(a)]
$I+\ds\int_0^1 e^{-xt^3}\, dt\ \ x\to +\infty$

Clearly

$\begin{array}{rcl} h(t) & = & -t^3\\ h'(t) & = & -3t^2
\Rightarrow t=0\ \mbox{is a stationary point}\\ h''(t) & = & -6t\\
h'''(t) & = & -6 \end{array}$

Locally the $h(t)$ is obviously \un{cubic} rather than quadratic.
This will alter the twist angle.

\un{Twist}: $h(t=0)=0$

${}$

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\put(0.2,3.4){\makebox(0,0){V}}

\put(3.8,3.4){\makebox(0,0){V}}
\end{picture} Need $\left\{\begin{array}{l}Re(-ixt^3)<0\ \ x\to +\infty\\ t=re^{i\theta}\\ Re(ixr^3e^{3i\theta})<0\\
\Rightarrow -\sin 3\theta <0\\ \sin 3\theta>0\end{array}\right.$

Now there are \un{3} valleys around $t=0$ (only 2 for a quadratic
case).

Angle of twist from

\begin{eqnarray*} Im(-xt^3) & = & 0\\ \Rightarrow \cos 3\theta &
= & 0\\ \Rightarrow 3\theta & = & \df{\pi}{2}\\ \theta & = &
\df{\pi}{6}\ \mbox{to be in the correct valley} \end{eqnarray*}

Thus we have the twist:

PICTURE \vspace{1in}

$I \sim \ds\int_0^\infty e^{i\frac{\pi}{6}} e^{ixt^3}\,dt$

set $\begin{array}{rcl} u & = & -it^3\\ du & = & -3it^2\,dt
\end{array}$

$\begin{array}{rcl} I & \sim & \ds\int_0^\infty
e^{-xu}\df{du}{-3it^2}\\ & = & \ds\int_0^\infty \df{e^{-xu}\,
du}{-3i(\frac{u}{-i})^\frac{2}{3}}\\ & \sim &
\df{i}{3}i^\frac{2}{3}\ds\int_0^\infty
\df{e^{-xu}}{u^\frac{2}{3}}\,du\\ & \sim &
\df{(e^{i\frac{\pi}{2}})^\frac{1}{3}}{3}\df{\Gamma(\frac{1}{3})}{x^\frac{1}{3}}\\
\mbox{by definition of $\Gamma$-function}:\\ & &  \ds\int_0^\infty
e^{-xt}t^{\alpha-1}\,dt=\df{\Gamma(\alpha)}{x^\alpha}\\ & \sim &
\df{e^{\frac{i\pi}{6}}\Gamma(\frac{1}{3})}{3x^{\frac{1}{3}}}\ \
x\to +\infty \end{array}$

\newpage
\item[(b)]
$I=\ds\int_0^\infty \df{e^{ixt^3}}{\sqrt{t}}\,dt$

As above, the angle of twist is $+\df{\pi}{6}$, and we set
$u=-it^3$.

$\begin{array}{rcl} \Rightarrow I & \sim & \ds\int_0^\infty
\df{e^{-xu}}{\sqrt{t}}\df{du}{(-3it^2)}\\ & = & \ds\int_0^\infty
\df{e^{-xu}\,du}{-3i(\frac{u}{-i})^\frac{2}{3}\undb{(\frac{u}{-i})^\frac{1}{6}}}\\
& & \hspace{2in} =\sqrt{t}\\ & = & \df{1}{-3i}\ds\int_0^\infty
\df{e^{-xu}}{u^\frac{5}{6}}\,du (-i)^{\frac{2}{3}+\frac{1}{6}}\\ &
= & \df{i}{3(i)^\frac{5}{6}}\ds\int_0^\infty
\df{e^{-xu}}{u^\frac{5}{6}}\,du\\ & = &
\df{(e^{i\frac{\pi}{2}})}{3}\df{\Gamma(\frac{1}{6})}{x^\frac{1}{6}}\
\mbox{by $\Gamma$-function definition}\\ & = &
\df{e^{i\frac{\pi}{2}}}{3x^\frac{1}{6}}\Gamma\left(\df{1}{6}\right).
\end{array}$
\end{description}

\end{document}
