Show that for large positive values of z, Jn(z) ∼ √2/πz cos (z - nπ/2 - π/4)

Question

Show that for large positive values of [latex]z[/latex],

[latex]J_{n}(z) \sim \sqrt{\frac{2}{\pi z}} \cos \left(z-\frac{n \pi}{2}-\frac{\pi}{4}\right)[/latex]

Accepted Answer

By Problem 6.33, we have

[latex]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 ight\}[/latex]

Let [latex]F(t)=i \sin t[/latex]. Then [latex]F^{\prime}(t)=i \cos t=0[/latex] where [latex]t=\pi / 2[/latex]. If we let [latex]t=\pi / 2+v[/latex], the integral in braces becomes

[latex] \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 ight)} 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} [/latex]

Let [latex]v^{2}=-2 i u^{2} / z[/latex] or [latex]v=(1-i) u / \sqrt{z}[/latex], i.e., [latex]u=\frac{1}{2}(1+i) \sqrt{z} v[/latex]. Then, the integral can be approximated by

[latex]\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[/latex]

or for large positive values of [latex]z[/latex],

[latex]\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}}[/latex]

and the real part is

[latex]\frac{1}{\sqrt{\pi z}}\left\{\cos \left(z-\frac{n \pi}{2} ight)+\sin \left(z-\frac{n \pi}{2} ight) ight\}=\sqrt{\frac{2}{\pi z}} \cos \left(z-\frac{n \pi}{2}-\frac{\pi}{4} ight)[/latex]

Higher-order terms can also be obtained [see Problem 10.162].

Question 10.47: Show that for large positive values of z, Jn(z) ∼ √2/πz cos ...

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