0
$\begingroup$

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 $2$.

$\endgroup$
  • 1
    $\begingroup$ $\cos(x)=\frac{e^{ix}+e^{-ix}} 2$ $\endgroup$ – Claude Leibovici May 3 '19 at 9:04
4
$\begingroup$

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}

$\endgroup$
  • 1
    $\begingroup$ LOL I feel so stupid. It's currently 2am and my brain is not working. You are completely right. Thanks for your help. $\endgroup$ – Steven May 3 '19 at 9:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.