# About the asymptotic formula of Bessel function

For $\nu \in \Bbb R$, I want to prove the well-known formula $$J_\nu (x) \sim \sqrt{\frac{2}{\pi x}} \cos \left( x - \frac{2 \nu +1}{4} \pi \right) + O \left( \frac{1}{x^{3/2}} \right) \;\;\;\;(x \to \infty)$$ where $J_\nu$ denotes the Bessel function. How can I show this? Or would you tell me the Internet site which proves this formula? I could not find the proof of this.

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See this. –  Ｊ. Ｍ. Jul 31 '12 at 10:54
In the Olden Days we would find things like this in textbooks (such as G. N. Watson). Nowadays everyone wants things to be available on-line! –  GEdgar Jul 31 '12 at 17:52

## 1 Answer

First, some preliminary series expansions. Consider the substitution $$\cos(t)=1-u^2/2\tag{1}$$ We get the power series for $u=2\sin(t/2)$: $$u=t-t^3/24+t^5/1920-t^7/322560+t^9/92897280+O(t^{11})\tag{2}$$ and the inverse series for $t$; $$t=u+u^3/24+3u^5/640+5u^7/7168+35u^9/294912+O(u^{11})\tag{3}$$

We need to concentrate on the stationary points at $t=\pi/2$ and $t=-\pi/2$, away from which the integral decays exponentially in $x$. The contribution at $-\pi/2$ is the conjugate of the contribution at $\pi/2$, so the whole contribution is twice the real part of the contribution at $\pi/2$. \begin{align} J_\nu(x) &=\frac1{2\pi}\int_{-\pi}^\pi e^{-i(\nu t-x\sin(t))}\,\mathrm{d}t\\ &=2\mathrm{Re}\left(\frac1{2\pi}\int_0^\pi e^{-i(\nu t-x\sin(t))}\,\mathrm{d}t\right)\\ &=2\mathrm{Re}\left(e^{-i\nu\pi/2}\frac1{2\pi}\int_{-\pi/2}^{\pi/2} e^{-i\nu t}e^{ix\cos(t)}\,\mathrm{d}t\right)\\ &\sim2\mathrm{Re}\left(e^{-i\nu\pi/2}\frac1{2\pi}\int_{-\infty}^\infty\left(1-\nu^2u^2/2+O(u^4)\right)e^{ix(1-u^2/2)}\left(1+u^2/8+O(u^4)\right)\,\mathrm{d}u\right)\\ &=2\mathrm{Re}\left(e^{i(x-\nu\pi/2)}\frac1{2\pi}\int_{-\infty}^\infty\left(1-(4\nu^2-1)u^2/8+O(u^4)\right)e^{-ixu^2/2}\,\mathrm{d}u\right)\\ &=2\mathrm{Re}\left(e^{i(x-(2\nu+1)\pi/4)}\frac1{2\pi}\int_{-\infty}^\infty\left(1+i(4\nu^2-1)v^2/8+O(v^4)\right)e^{-xv^2/2}\,\mathrm{d}v\right)\\ &=2\mathrm{Re}\left(e^{i(x-(2\nu+1)\pi/4)}\frac1{2\pi}\left(\sqrt{\frac{2\pi}{x}}+i\frac{4\nu^2-1}{8}\frac1{2\pi}\sqrt{\frac{2\pi}{x}}^3+O\left(x^{-5/2}\right)\right)\right)\\ &=\cos\left(x-\frac{2\nu+1}{4}\pi\right)\sqrt{\frac{2}{\pi x}}-\sin\left(x-\frac{2\nu+1}{4}\pi\right)\frac{4\nu^2-1}{8}\sqrt{\frac{2}{\pi x^3}}+O\left(x^{-5/2}\right) \end{align} In the $\sim$ step, we make the $u$ substitution whose series is given above, and integrate over $(-\infty,\infty)$ instead of $[-\sqrt{2},\sqrt{2}]$ since the part outside the compact interval decays exponentially.

We also use the substitution $u=e^{-i\pi/4}v$, so that $u^2=-iv^2$, and change the path of integration, which is allowed since there are no singularities.

In the end, we get the first two terms of the asymptotic expansion of $J_\nu(x)$.

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(+1) Love the saddle point method. –  Jack D'Aurizio Aug 31 at 2:58
It is a very powerful tool. I have used it in a lot of asymptotic expansions (my graduate advisor used it a lot in dealing with singular integrals). –  robjohn Aug 31 at 3:19