# Proving $J_0(x)\approx\left(\frac{2}{\pi x}\right)^{1/2}\cos\left(x-\frac{\pi}{4}\right)$ as $x\rightarrow\infty$

I am trying to show that if $$J_0(x)=\frac{2}{\pi}\int_0^{\pi/2}\cos(x\cos(\theta)) \ d\theta, \ \ \text{then}$$ $$J_0(x)\approx\left(\frac{2}{\pi x}\right)^{1/2}\cos\left(x-\frac{\pi}{4}\right) \ \ \text{as x\rightarrow\infty.}$$

My attempt:

We use the method of stationary phase. Let $$h(\theta)=\cos(\theta)\implies h'(\theta)=-\sin(\theta)$$. We seek $$h'(\theta)=0\implies \theta=0$$. We will use the approximation $$h(\theta)=1-\frac{\theta^2}{2}.$$ \begin{align} J_0(x)&=\frac{2}{\pi}\int_0^{\pi/2}\cos\left(x\left(1-\frac{\theta^2}{2}\right)\right) \ d\theta \\ &=\frac{4}{\pi}\int_0^{\pi/2}\frac{e^{ix\left(1-\frac{\theta^2}{2}\right)}+e^{-ix\left(1-\frac{\theta^2}{2}\right)}}{2} \ d\theta\\ &=\frac{2}{\pi}\int_0^{\pi/2}e^{ix\left(1-\frac{\theta^2}{2}\right)} \ d\theta+\frac{2}{\pi}\int_0^{\pi/2}e^{-ix\left(1-\frac{\theta^2}{2}\right)} \ d\theta \\ &=\frac{e^{ix}}{\pi}\int_{-\pi/2}^{\pi/2}e^{-i\frac{x}{2}\theta^2} \ d\theta+\frac{2e^{-ix}}{\pi}\int_{-\pi/2}^{\pi/2}e^{i\frac{x}{2}\theta^2} \ d\theta \\ &\approx\frac{e^{ix}}{\pi}\int_{-\infty}^{\infty}e^{-i\frac{x}{2}\theta^2} \ d\theta+\frac{e^{-ix}}{\pi}\int_{-\infty}^{\infty}e^{i\frac{x}{2}\theta^2} \ d\theta \\ &=\frac{e^{ix}}{\pi}\left(e^{-i\pi/4}\sqrt{\frac{2\pi}{x}}\right)+\frac{e^{-ix}}{\pi}\left(e^{i\pi/4}\sqrt{\frac{2\pi}{x}}\right) \\ &=\frac{2}{\pi}\sqrt{\frac{2\pi}{x}}\left(\frac{\left(e^{i\left(x-\pi/4\right)}+e^{-i\left(x-\pi/4\right)}\right)}{2}\right) \\ &=2\left(\frac{2}{\pi x}\right)^{1/2}cos\left(x-\frac{\pi}{4}\right) \end{align} I seem to be out by a factor of $$2$$.

• $\cos(x)=\frac{e^{ix}+e^{-ix}} 2$ – Claude Leibovici May 3 '19 at 9:04

Your second $$=$$ sign incorrectly takes $$\cos y=e^{iy}+e^{-iy}$$; it's half that. The calculation should read
\begin{align} J_0(x)&=\frac{2}{\pi}\int_0^{\pi/2}\cos\left(x\left(1-\frac{\theta^2}{2}\right)\right) \ d\theta \\ &=\frac{2}{\pi}\int_0^{\pi/2}\frac{e^{ix\left(1-\frac{\theta^2}{2}\right)}+e^{-ix\left(1-\frac{\theta^2}{2}\right)}}{2} \ d\theta\\ &=\frac{1}{\pi}\int_0^{\pi/2}e^{ix\left(1-\frac{\theta^2}{2}\right)} \ d\theta+\frac{1}{\pi}\int_0^{\pi/2}e^{-ix\left(1-\frac{\theta^2}{2}\right)} \ d\theta \\ &=\frac{e^{ix}}{2\pi}\int_{-\pi/2}^{\pi/2}e^{-i\frac{x}{2}\theta^2} \ d\theta+\frac{e^{-ix}}{2\pi}\int_{-\pi/2}^{\pi/2}e^{i\frac{x}{2}\theta^2} \ d\theta \\ &\approx\frac{e^{ix}}{2\pi}\int_{-\infty}^{\infty}e^{-i\frac{x}{2}\theta^2} \ d\theta+\frac{e^{-ix}}{2\pi}\int_{-\infty}^{\infty}e^{i\frac{x}{2}\theta^2} \ d\theta \\ &=\frac{e^{ix}}{2\pi}\left(e^{-i\pi/4}\sqrt{\frac{2\pi}{x}}\right)+\frac{e^{-ix}}{2\pi}\left(e^{i\pi/4}\sqrt{\frac{2\pi}{x}}\right) \\ &=\frac{1}{\pi}\sqrt{\frac{2\pi}{x}}\left(\frac{\left(e^{i\left(x-\pi/4\right)}+e^{-i\left(x-\pi/4\right)}\right)}{2}\right) \\ &=\left(\frac{2}{\pi x}\right)^{1/2}\cos\left(x-\frac{\pi}{4}\right). \end{align}