MHB What is the Limit of the Bessel Function as x Goes to Infinity?

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Here is this week's POTW:

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Let $n$ be an integer. Show that as $x \to \infty$ on the positive real axis,

$$J_n(x) \sim \sqrt{\frac{2}{\pi x}}\left[\cos\left(x - \frac{n\pi}{2} - \frac{\pi}{4}\right)\right],$$

where $J_n(x)$ is the $n$th order Bessel function of the first kind.
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No one answered this week's problem. Here is my solution.

Spoiler

Using the integral representation

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

and Euler's formula, we have $J_n(x) = \frac{1}{2\pi}[I_1(x) + I_2(x)]$ where $I_1(x) = \int_0^\pi e^{ix\sin t - int}\, dt$ and $I_2(x) = \int_0^\pi e^{-ix\sin t + int}\, dt$. Now $I_1(x) = \int_0^\pi f_1(t)e^{ixg_1(t)}\, dt$ with $f_1(t) = e^{int}$ and $g_1(t) = \sin t$. The function $g$ is holomorphic around $[0,\pi]$, maps $\Bbb R$ to $\Bbb R$, and is $C^1[0,\pi]$. Further, the only critical point of $g_1$ is $\pi/2$. Since $g_1^{''}(\pi/2) = -1 < 0$, by the stationary phase theorem,

$$I_1(x) \sim \frac{e^{ixg_1(\pi/2)}\sqrt{2\pi}}{\sqrt{x}\sqrt{-g_1^{''}(\pi/2)}}e^{-\pi i/4}g_1(\pi/2) = \sqrt{\frac{2\pi}{x}}e^{i(x - n\pi/2 - \pi/4)}.$$

Consider $I_2(x) = \int_0^\pi f_2(t)e^{ixg_2(t)}\, dt$, with $f_2(t) = e^{-int}$ and $g_2(t) = -\sin t$. By a similar argument,

$$I_2(x) \sim \frac{e^{ixg_2(\pi/2)}\sqrt{2\pi}}{\sqrt{x}\sqrt{g_2^{''}(\pi/2)}}e^{\pi i/4}g_2(\pi/2) = \sqrt{\frac{2\pi}{x}}e^{-i(x - n\pi/2 - \pi/4)}.$$

Therefore

$$J_n(x) = I_1(x) + I_2(x) \sim \frac{1}{\pi}\cdot \sqrt{\frac{2\pi}{x}}\left[\cos\left(x - \frac{n\pi}{2} - \frac{\pi}{4}\right)\right] = \sqrt{\frac{2}{\pi x}}\left[\cos\left(x - \frac{n\pi}{2} - \frac{\pi}{4}\right)\right].$$