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


{\bf Question}

Use Watson's lemma to calculate the full Poincar\'{e} asymptotic
expansion of

\begin{description}

\item[(a)]
$\ds\int_0^{\infty}e^{-xt}\log(1+\sqrt{t})\,dt,\ x\to+\infty$

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

\item[(c)]
$\ds\int_0^\infty \ds\frac{e^{-xt}}{\sqrt{t(2+t)}}\, dt,\
x\to+\infty$
\end{description}


\vspace{.5in}

{\bf Answer}

With Watson's lemma we must check existence of expansion of $f(t)$
about $t=0$ \un{and} $|f(t)|<Ae^{bt}$.

$\ds\int_0^\infty e^{-xt} f(t)\,dt$ for some finite $A,\ b$.

\begin{description}
\item[(a)]
$\ds\int_0^\infty e^{-xt}\log(1+\sqrt{t})$

Clearly $|log(1+\sqrt{t})|<e^t,\ \ t>0$

Also $\log(1+\sqrt{t})=\sum_{n=0}^\infty
\df{t^{\frac{n}{2}}}{n}(-1)^n$ by Maclaurin / Taylor series about
$t=0$.

Thus $\ds\int_0^\infty e^{-xt}\log(1+\sqrt{t})\,dt \sim
\sum_{n=0}^\infty \df{(-1)^n}{n}\ds\int_0^\infty
e^{-xt}t^{\frac{n}{2}}$.

$\left(\lambda_n=\df{n}{2},\
a_n=\df{(-1)^n}{n}\right)=\sum_{n=0}^\infty
\df{(-1)^n}{n}\df{\Gamma(\frac{n}{2}+1)}{x^{\frac{n}{2}}+1},\ \
x\to +\infty$ by Watson.

\newpage
\item[(b)]
$\ds\int_0^\infty
\df{e^{-xt}}{\sqrt{t(2+t)}}\,dt=\left[\ds\int_0^1+\ds\int_1^\infty\right]
\df{e^{-xt}}{\sqrt{t(2+t)}}\,dt$

As framed in notes, we only have Watson for
$\ds\int_0^\infty$-type integrals.

BUT $\ds\int_1^\infty$ is exponentially small as $x\to +\infty$:

\begin{eqnarray*} \left|\ds\int_1^\infty \df{e^{-xt}}{\sqrt{t(2+t)}}\,dt\right| &
\leq & \ds\int_1^\infty \df{e^{-xt}}{\sqrt{t(2+t)}}\,dt\\ & &
\sqrt{t(2+t)} \geq \sqrt{3}\ \rm{for\ all}\ t \geq 1\\& \leq &
\ds\int_1^\infty \df{e^{-xt}}{\sqrt{3}}\,dt\\ & = &
\df{e^{-x}}{x\sqrt{3}}\\ & = & O\left(\df{e^{-x}}{x}\right)
\end{eqnarray*}

So for Poincar\'{e} purposes,

$$\ds\int_0^\infty\,dt \sim \ds\int_0^1\,dt$$

Thus we apply Watson. Must check:

\begin{description}
\item[(i)]
$f(t)=\df{1}{\sqrt{t(2+t)}}<Ae^{b-t}$ for all $t>0$, for some $A,\
b>0$

This is clearly satisfied by $A=1,\ b=1,$ i.e.,

$$\df{1}{\sqrt{t(2+t)}}<e^t\ \ \surd$$

\item[(ii)]
$f(t)$ has an expansion about $t=0$.

$f(t)=\df{1}{\sqrt{2t}}\left(1+\df{t}{2}\right)^{-\frac{1}{2}}=\df{1}{\sqrt{2t}}
\sum_{s=0}^\infty
\left(\df{t}{2}\right)^s\df{(-1)^s}{s!}\df{\Gamma(s+\frac{1}{2})}{\Gamma(\frac{1}{2})}
\ \ \surd$

We can work this last fraction out by using factorial notation to
spot this as $\df{(s-\frac{1}{2})!}{(-\frac{1}{2})!}$ and
converting back to $\Gamma$ functions by adding one to top and
bottom.

Thus $\lambda_s-\df{1}{2}$

$d_s=\df{(-1)^s}{2^{s+\frac{1}{2}}}\df{\Gamma(s+\frac{1}{2})}{s!\Gamma(\frac{1}{2})}$

Therefore, from Watson:

$$\ds\int_0^1 \df{e^{-xt}}{\sqrt{t(2+t)}}\,dt \sim
\sum_{s=0}^\infty
\df{(-1)^s\Gamma(s+\frac{1}{2})}{2^{s+\frac{1}{2}}s!\Gamma(\frac{1}{2})}\df
{\Gamma(s+\frac{1}{2})}{x^{s+\frac{1}{2}}}\ \ x\to +\infty$$

Now $\Gamma(\frac{1}{2})=\sqrt{\pi}$: KNOW THIS FOR THE EXAM!

Therefore $\sim \sum_{s=0}^\infty \df{(-1)^s
[\Gamma(s+\frac{1}{2})]^2}{2^{s+\frac{1}{2}}s!\sqrt{\pi}x^{s+\frac{1}{2}}},\
\ x \to +\infty$
\end{description}

\item[(c)]
From above, for Poincar\'{e} expansion

$$\ds\int_0^1 \sim \ds\int_0^\infty$$

so $\ds\int_0^\infty \df{e^{-xt}}{\sqrt{t(2+t)}} \sim
\sum_{s=0}^\infty
\df{(-1)^s[\Gamma(1+\frac{1}{2})]^2}{2^{s+\frac{1}{2}}s!\sqrt{\pi}x^{s+\frac{1}{2}}}$

\end{description}


\end{document}