Question 10.47: Show that for large positive values of z, Jn(z) ∼ √2/πz cos ...
Show that for large positive values of z,
J_{n}(z) \sim \sqrt{\frac{2}{\pi z}} \cos \left(z-\frac{n \pi}{2}-\frac{\pi}{4}\right)
By Problem 6.33, we have
J_{n}(z)=\frac{1}{\pi} \int\limits_{0}^{\pi} \cos (n t-z \sin t) d t=\operatorname{Re}\left\{\frac{1}{\pi} \int\limits_{0}^{\pi} e^{-i n t} e^{i z \sin t} d t\right\}
Let F(t)=i \sin t. Then F^{\prime}(t)=i \cos t=0 where t=\pi / 2. If we let t=\pi / 2+v, the integral in braces becomes
\begin{aligned} \frac{1}{\pi} \int\limits_{-\pi / 2}^{\pi / 2} e^{-i n(\pi / 2+v)} e^{i z \sin (\pi / 2+v)} d v & =\frac{e^{-i n \pi / 2}}{\pi} \int\limits_{-\pi / 2}^{\pi / 2} e^{-i n v} e^{i z \cos v} d v=\frac{e^{-i n \pi / 2}}{\pi} \int\limits_{-\pi / 2}^{\pi / 2} e^{-i n v} e^{i z\left(1-v^{2} / 2+v^{4} / 24-\cdots\right)} d v \\ & =\frac{e^{i(z-n \pi / 2)}}{\pi} \int\limits_{-\pi / 2}^{\pi / 2} e^{-i n v} e^{-i z v^{2} / 2+i z v^{4} / 24-\cdots} d v \end{aligned}
Let v^{2}=-2 i u^{2} / z or v=(1-i) u / \sqrt{z}, i.e., u=\frac{1}{2}(1+i) \sqrt{z} v. Then, the integral can be approximated by
\frac{(1-i) e^{i(z-n \pi / 2)}}{\pi \sqrt{z}} \int\limits_{-\infty}^{\infty} e^{-(1+i) n u / \sqrt{z}} e^{-u^{2}-i u^{4} / 6 z-\cdots} d u
or for large positive values of z,
\frac{(1-i) e^{i(z-n \pi / 2)}}{\pi \sqrt{z}} \int\limits_{-\infty}^{\infty} e^{-u^{2}} d u=\frac{(1-i) e^{i(z-n \pi / 2)}}{\sqrt{\pi z}}
and the real part is
\frac{1}{\sqrt{\pi z}}\left\{\cos \left(z-\frac{n \pi}{2}\right)+\sin \left(z-\frac{n \pi}{2}\right)\right\}=\sqrt{\frac{2}{\pi z}} \cos \left(z-\frac{n \pi}{2}-\frac{\pi}{4}\right)
Higher-order terms can also be obtained [see Problem 10.162].