\documentclass[a4paper,12pt]{article}
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\begin{document}


{\bf Question}

Show that

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

and obtain one further term in the asymptotic expansion.

\vspace{.5in}

{\bf Answer}

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

$h(t)=\cos t$ has 3 minima on the range of integration at $t=0,\
\pm 2\pi$

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Now we see that $h(0)=h(\pm 2\pi)$ so it looks like all 3 will
give equally dominant contributions. In fact we have 2 quadratic
endpoints and one quadratic minimum. If we now consider the
periodicity of $h9t)$, we see

$$h(t\pm 2\pi)=-\cos(2\pm 2\pi)=-\cos t=h(t)$$

Thus the two endpoints add up to give one \un{full} contribution
identical to the one at $t=0$. Thus we focus on $t=0$ and just
double its contribution to get the full result.

Set $\begin{array}{rcl} u^2 & = & h(t)-h(0)\\ h(t) & = &
-1+\df{(t=0)^2}{2}+\cdots \end{array}$

$\begin{array}{rcl} h(t) & = & -\cos t\\ h'(t) & = & +\sin
t\\h''(t) & = & +\cos t \end{array}$

Therefore $u^2 \approx \df{t^2}{2},\ u \approx \df{t}{\sqrt{2}}$

and $2u\, du=h'(t)\,dt,\ h'(t) \approx h''(0)\, t$

Therefore

$\begin{array} {rcl} I & = &
2\ds\int_{-\sqrt{h(-2\pi)-h(0)}}^{\sqrt{h(2\pi)-h(0)}}\df{u(1+t(u))}{h'(t(u))}
e^{-xu^2-xh(0)}\,du\\ & \sim & 2e^x \ds\int_{-\infty}^{+\infty}
e^{-xu^2}\df{u(1+t(u))}{h'(t(u))}\,du\\ & \sim &
2e^x\ds\int_{-infty}^{+\infty}e^{-xu^2}\df{u}{h''(0)\sqrt{2}u}\,du\\
& \sim & \sqrt{2}e^x \ds\int_{-\infty}^{+\infty}e^{-xu^2}\,du\\ &
\sim & \sqrt{\df{2\pi}{x}}e^x\ \ x\to +\infty\end{array}$

Thus doubling up the contribution (including endpoints) we have
(as required)

$$I \sim 2e^x \sqrt{\df{2\pi}{x}}\ \ x\to + \infty$$

To get one further term in the expansion we need to retain more
terms in the expansion of

\begin{eqnarray*} \df{u(1+t)}{h'(t)} & = & \df{u(1+t)}{\sin t}\ \
\rm{about}\ t=0\\ & = & \df{u(1+t)}{(t-\frac{t^3}{3!}+\cdots)}\\ &
\approx & \df{u}{t}(1+t)\left(1-\df{t^2}{6}+\cdots\right)^{-1}\\ &
\approx & \df{u}{t}(1+t)\left(1+\df{t^2}{6}+\cdots\right)\\ & = &
\df{u}{t}\left(1+t+\df{t^2}{6}+O(t^3)\right)\ \ (A)
\end{eqnarray*}

Now also need $u$ to higher order in $t$:

$u^2=h(t)=h(0)=\df{h''(0)}{2!}(t-0)^2+\df{h'''(0)}{3!}(t-0)^3=\df{h^{iv}}{4!}(t-0)^4+\cdots$

so from above $u^2=\df{t^2}{2}-\df{t^4}{24}+\cdots$

so
$u=\df{t}{\sqrt{2}}\left(1-\df{t^2}{12}+\cdots\right)^{\frac{1}{2}}
\approx \df{t}{\sqrt{2}}\left(1-\df{t^2}{24}\right)\ \ (B)$

Therefore putting $(B)$ in $(A)$:

\begin{eqnarray*} \df{u(1+t)}{h'(t)} & \approx &
\df{t}{\sqrt{2}}\df{(1-\frac{t^2}{24})}{t}\left(1+t+\df{t^2}{6}+\cdots\right)\\
& \approx &
\df{1}{\sqrt{2}}\left(1-\df{t^2}{24}\right)\left(1+t+\df{t^2}{6}\right)\\
& = & \df{1}{\sqrt{2}}\left(1+t+\df{t^2}{8}\right)\
\end{eqnarray*}

Thus we have

$$\df{u(1+t)}{h'(t)} \approx
\df{1}{\sqrt{2}}\left(1+\sqrt{u}+\df{2u^2}{8}\right)$$

Substitution into original integral gives:

$$I \sim
2e^x\ds\int_{-\infty}^{+\infty}\df{e-xu^2}{\sqrt{2}}\left(1+\sqrt{2}+\df{2u^2}{8}\right)$$

First term gives $\sqrt{\df{2\pi}{x}} e^x$ as before.

Second term is ZERO: $\ds\int_{-\infty}^{+\infty}e^{-xu^2}\
u\,du=0$

\hspace{2.05in}$\stackrel{\uparrow}{\mbox{odd function}}$

Third term is
$\df{\sqrt{2}e^x}{4}\undb{\ds\int_{-\infty}^{+\infty}e^{-xu^2}\,u^2\,du}
=\df{e^x}{2\sqrt{2}}\df{1}{2x}\sqrt{\df{\pi}{x}}$

\hspace{1.4in} $=-\df{\pl}{\pl
x}\int_{-\infty}^{+\infty}e^{-xu^2}\,du$

Thus pulling everything together and multiplying by 2 for the 2
endpoint contributions, we have

$$I \sim
2e^x\sqrt{\df{2\pi}{x}}\left(1+\df{1}{8x}+O\left(\df{1}{x^2}\right)\right)$$

\un{Phew!!!}

\end{document}
