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.
 A: 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 $J_\nu(x)$.
