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.
About the asymptotic formula of Bessel function
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2See [this](http://books.google.com/books?hl=en&id=Mlk3FrNoEVoC&pg=PA194). – 2012-07-31
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0In 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! – 2012-07-31
1 Answers
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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2It 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). – 2014-08-31
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0This is an old answer, but I needed it, so I tried to work through it. I ran into a number of obstacles. Each can be overcome, but as it was a bit of work, I list them here, with fixes: First, between the third and fourth line, you need to symmetrise the integrand. Easy enough: Substitute $-t$ for $t$, and replace the integral by the average of the original and the reversed version. So $e^{-i\nu t}$ gets replaced by $\cos(\nu t)$. You have clearly *done* it, but not written it up. (to be continued …) – 2017-10-20
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0(… continued) Second, in the $\sim$ line, there is no apparent exponential decay to justify moving the limits to $\pm\infty$. So just leave them at $\pm2^{-1/2}$. Later, when you change from integrating along the real line to integrating along the diagonal, the new limits of integration will be $\pm(1-i)$. To close the curve so you can use Cauchy's integral theorem, you add two vertical segments. But thanks to the symmetry of the integrand, the corresponding integrals cancel out. (to be continued …) – 2017-10-20
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0(… continued) Third, now that we are working on the diagonal $\{t-it\colon t\in\mathbb{R}\}$, we *do* have exponential decay, so moving the integration limits off to $\pm(1-i)\infty$ is justified, at last, and the calculation can proceed. (I haven't checked the detailed answer, but am convinced there are no more obstacles.) – 2017-10-20