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\begin{document}


{\bf Question}

Use Laplace's method to show that

$$\ds\int_{-1}^1 dt,\ e^{-x(2t^2-t^4)}\sqrt{2+t} \sim
\sqrt{\ds\frac{\pi}{x}},\ x\to+\infty$$

and obtain the corresponding result for

$$\ds\int_0^3dt,\ e^{-x(2t^2-t^4)}\sqrt{2+t}\ \ x\to+\infty$$

\vspace{.5in}

{\bf Answer}

$I=\ds\int_{-1}^{+1}\,dt\,e^{-x(2t^2-t^4)}{\sqrt{2+t}}$

$h(t)=2t^2-t^4,\ h'(t)=4t-4t^3, \Rightarrow t=0\ \rm{or}\ \pm 1\
\rm{are\ min/max}$

$h''(t)=4-12t^2 \Rightarrow \begin{array}{l} t=0: h''(0)=4>0
\Rightarrow min\\ t=\pm 1: h''(\pm 1)=-8<0 \Rightarrow
max\end{array}$

Don't contribute at leading order of exponentials

Sketch of $h(t)$:

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\put(2,3){\makebox(0,0)[l]{\ 1}}

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\newpage
Thus we have a \un{full} contribution from the min. at $t=0$.

Set

\hspace{1.4in}=0

$\begin{array} {rcl} u^2 & = & h(t)-\overbrace{h(0)}=2t^2-2^4\\
2u\,du & = & h'(t)\,dt \end{array}$

Therefore $I=\ds\int_{-\sqrt{h(-1)-h(0)}}^{\sqrt{h(1)-h(0)}}
e^{-xu^2-xh(0)} \undb{\df{\sqrt{2+t(u)}}{h'(t(u))}}\cdot 2u,\ du$

\hspace{2in} Need to expand this about $t=0\ (u=0)$

$h'(t) approx h''(0)t\ \ t\to +\infty$ by Taylor

and $u^2=h(t)-h(0) \approx \df{h''(0)t^2}{2},\ t\to + \infty$

$\Rightarrow u=\pm \sqrt{\df{h''(0)}{2}}t$ (take $+\sqrt{}$ as $t$
increases when $u$ increases.)

Therefore

\begin{eqnarray*} I & \approx & \ds\int_{-\sqrt{h(1)-h(0)}}^{\sqrt{h(1)-h(0)}}
e^{-xu^2}\ 2u\
\df{\sqrt{2+t(u)}}{h''(0)u\sqrt{\frac{2}{h''(0)}}}\,du\\ & \approx
& 2\ds\int_{-\infty}^{+\infty}\df{e^{-xu^2}}{\sqrt{h''(0)}}\,du\ \
x\to +\infty\\ & & \mbox{errors are exponentially small as}\ x\to
+\infty\\ & = &
\df{2}{\sqrt{4}}\ds\int_{-\infty}^{+\infty}e^{-xu^2}\,du\\ & = &
\sqrt{\df{\pi}{x}}\ \ x\to + \infty\end{eqnarray*}

(By standard Gaussian integral)

KNOW: $\ds\int_{-\infty}^{+\infty}e^{-\alpha
x^2}\,dx=\sqrt{\df{\pi}{\alpha}}$

When we consider

$$\ds\int_0^3 dt\, e^{-x(2t^2-t^4)}\sqrt{2+t},\ x\to + \infty$$

Looking at the graph of $h(t)=2t^2-t^4$ as above, we see that
$t=0$ is an endpoint minimum but $t=3$ is an \un{overall} minimum
on the range of integration. Thus the \un{dominant} behaviour will
come from $t=3$, a \un{linear} endpoint. Thus we proceed as in
question 6:

\begin{eqnarray*} u & = & h(t)-\undb{h(3)}=h(t)+63\\ & & \hspace{.5in} 2
\times 9 - 81\\ du & = & h'(t)\,dt \end{eqnarray*}

\hspace{1.5in} NB endpoint is at RHS

$J=\ds\int_0^3 e^{-x(2t^2-t^4)}\sqrt{2+t}\,dt \sim
\ds\int_{h(0)-h(3)=+63}^0 e^{-xu+63x}
\df{\sqrt{2+t(u)}}{h'(t(u))}\,du$

\hspace{2in} NB

$\begin{array}{rcl} u = h(t)-h(3) & \approx & h'(3)(t-3) t\to 3\\
& = & (4\cdot 3-4\cdot 3^3)(t-3)\\ & = & -96(t-3) \end{array}$

\hspace{.5in} NB minus sign will reverse limits in integral

$\begin{array}{rcl} h'(t) & \approx & h'(3)+h''(3)(t-3)\ \rm{by\
Taylor}\\ & = & -96\ \rm{to leading order} \end{array}$

Also need expansion of $\sqrt{2+t}$ about $t=+3$:

$\sqrt{2+t}=\sqrt{5}+O(t-3)\ t\to 3\ \rm{by Taylor}$

Therefore

$\begin{array}{rcl} J & \sim & \ds\int_{63}^0
e^{-xu+63x}\df{\sqrt{5}}{-96}\,du\ \ x\to+\infty\\ & = &
+e^{63x}\df{\sqrt{5}}{96}\ds\int_0^{63}e^{-xu}\,du\
(-\int_{63}^0=+\int_0^{63})\\ & \sim &
e^{63x}\df{\sqrt{5}}{96}\ds\int_0^\infty e^{-xu}\,du\ \ x\to
+\infty\\ & \sim & e^{63x}\df{\sqrt{5}}{96x}\ \ x\to +\infty
\end{array}$

The two integrals above only differ by their range of integration.
However they have dramatically different behaviour as $x\to
+\infty$. The first tends to zero $\left(\df{1}{\sqrt{x}} \to 0\
\rm{as}\ x\to +\infty\right)$, the second $\to + \infty\
(e^{63x}\to \infty\ \rm{as}\ x \to +\infty)$. In fact, even when
$x=+1,\ e^{63} \approx 2.3 \times 10^{27}$.

\end{document}