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{\bf Question}

The Bessel function of order $n$ is defined by

$$J_n(x)=\ds\frac{1}{\pi}ds\int_0^\pi \cos(nt-x\sin t)\, dt$$

Use the method of stationary phase to show that

$$J_n(x)\sim \sqrt{\ds\frac{2}{\pi
x}}\cos\left(x-\ds\frac{n\pi}{2}-\ds\frac{\pi}{4}\right),\
x\to+\infty.$$

\vspace{.5in}

{\bf Answer}

${}$

$\begin{array}{rcl} J_n(x) &amp; = &amp; \df{1}{\pi}\ds\int_0^\pi
\cos(nt-x\sin t)\,dt\\ &amp; = &amp; \df{1}{2\pi}\ds\int_0^\pi
\left(e^{int-ix\sin t}+e^{-int+ix\sin t}\right)\,dt\\ &amp; = &amp;
\df{1}{2\pi}\sum_\pm\ds\int_0^\pi e^{\pm int\mp ix\sin t}\,dt\\ &amp;
= &amp; \sum_\pm J_\pm \end{array}$

Focus on $+\int-ix\sin t$ first.

We want asymptotes as $x\to +\infty,\ n$ fixed and finite

$\Rightarrow f^{(+)}(t)=e^{int},\ h^{(+)}(t)=\sin t$

$\Rightarrow h^{(+)}(t)=\cos t,\ {h^{(+)}}''(t)=-\sin t$

$t=\df{\pi}{2}$ is the stationary point (mid-range)

$h^{(+)}\left(\df{\pi}{2}\right)=+1,\
{h^{(+)}}''\left(\df{\pi}{2}\right)=-1$

Therefore

$\begin{array}{rcl}u^2 &amp; = &amp;
i\left[h^{(+)}(t)-h^{(+)}\left(\ds\frac{\pi}{2}\right)\right]=i(\sin
t-1)\\ 2u\,du &amp; = &amp; i\cos t, dt \end{array}$

Locally at stationary point

$u^2 \approx i \times
\df{(-1)}{2!}\left(t-\df{\pi}{2}\right)^2+\cdots \approx
-\df{i}{2}\left(t-\df{\pi}{2}\right)^2$

\newpage
\un{Twists}:

Locally require

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\put(0,4){\makebox(0,0){(t)}}

\put(2.3,1.7){\makebox(0,0){$\frac{\pi}{2}$}}

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\put(1,1.2){\makebox(0,0){V}}

\put(3,3.2){\makebox(0,0){V}}
\end{picture} $\left\{ \Rightarrow \begin{array}{l}Re(-ix[\sin
t-1])&lt;0\ \ x\to +\infty\\ Re(i[\sin t-1])&gt;0\\
Re\left(-\df{i}{2}\left(t-\df{\pi}{2}\right)^2\right)&gt;0\\
\left(t-\df{\pi}{2}\right)=re^{i\theta_+}\\
Re\left(-\df{i}{2}r^2e^{2i\theta_+}\right)&gt;0\\
Re(-ie^{2i\theta_+})&gt;0\\ Re(\sin 2\theta_+)&gt;0 \end{array}\right.$

Angle of twist from

\begin{eqnarray*}
Im\left\{i\left[h^{(+)}(t)-h^{(+)}\left(\df{\pi}{2}\right)\right]\right\}
&amp; = &amp; 0\\ \Rightarrow Im(-ie^{2i\theta_+}) &amp; = &amp; 0\\ \cos
2\theta_+ &amp; = &amp; 0\\ \Rightarrow \theta_+ &amp; = &amp; \un{\df{\pi}{4}}
\end{eqnarray*}

PICTURE \vspace{2in}

\newpage
Can use equation (3.99)

\begin{eqnarray*} J+ &amp; = &amp; \df{1}{2\pi} \ds\int_0^\pi
e^{int-ix\sin t}\, dt\\ &amp; \sim &amp; \df{1}{2\pi}
\sqrt{\df{2\pi}{|x{h^{(+)}}''(\frac{\pi}{2})|}}e^{-ixh^{(+)}(\frac{\pi}{2})
+i\theta+f^{(+)}(\frac{\pi}{2})}\\ &amp; \sim &amp; \df{1}{2\pi}
\sqrt{\df{\pi}{|x \times
-1|}}e^{-ix+i\frac{\pi}{4}}e^{in\frac{\pi}{2}}\\ &amp; \sim &amp;
\df{1}{2\pi}\sqrt{\df{2\pi}{x}}e^{-i(x-\frac{n\pi}{2}-\frac{\pi}{4})}
\end{eqnarray*}

For $J_{-}=\df{1}{2\pi}\ds\int_0^\pi e^{-int+ix\sin t}\,dt$ the
twist is in the \un{opposite} direction (due to the $+/-$
difference in $h(t)=+ix\sin t$).

Hence:

$$h^(-)\left(\df{\pi}{2}\right)=-1,\
{h^(-)}''\left(\df{\pi}{2}\right)=+1,\ \theta^{-}=-\df{\pi}{4}.$$

Hence using (3.99)

$\begin{array}{rcl} J_{-} &amp; \sim &amp; \df{1}{2\pi}\sqrt{\df{2\pi}{|(x
\times -1)|}}e^{+ix-\frac{i\pi}{4}}e^{-in\frac{\pi}{2}}\\ &amp; \sim &amp;
\df{1}{2\pi}
\sqrt{\df{2\pi}{x}}e^{+i(x-\frac{n\pi}{2}-\frac{\pi}{4})}\end{array}$

To get the final answer, we add together:

$J=J_+ + J_{-} \sim
\df{1}{2\pi}\sqrt{\df{2\pi}{x}}\left[e^{+i(x-\frac{n\pi}{2}-\frac{\pi}{4})}
+e^{-i(x-\frac{n\pi}{2}-\frac{\pi}{4})}\right]$

$J \sim \sqrt{\df{2}{\pi
x}}\cos\left(x-\df{n\pi}{2}-\df{\pi}{4}\right).$ as required.



\end{document}
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