Asymptotic Expansion of Bessel Functions For Large Argument

This note is based on Appendix B.7 and B.8 of (Grafakos, 2008). First, we show that Bessel function can be written into the following form

$$\begin{equation} J_v(r) = \frac{(r/2)^v}{\Gamma(v+\frac12)\Gamma(\frac12)}\left[ie^{-ir} \int_0^\infty e^{-rt}(t^2+2it)^{v-\frac12}dt \\ -ie^{ir}\int_0^\infty e^{-rt}(t^2-2it)^{v-\frac12}dt\right]\,,\tag{*} \end{equation}$$

for $r>0$ and $Re(v)>-1/2$. This is done by considering the line integral of $e^{irz}(1-z^2)^{v-1/2}$ over the curves shown in the illustration (arrows should be flipped!) . First, the function is analytic in the region bounded by the curves. To see that, note that the complement region of the non-positive real line is a holomorphic branch of $\ln(z)$. For $\ln(1+z)$ this region is the complement of $(-\infty,-1)$. And to guarantee $-z^2$ stays in this region, we allow $z$ in the upper half plane together with the interval $[-1+\delta,1-\delta]$. Then by rewriting

$$\begin{equation} (1-z^2)^{v-1/2} = \exp((v-1/2)\ln(1-z^2))\,, \end{equation}$$

we are done.

Then by Cauchy’s theorem, we know the line integral of such a function will be $0$. Denote that

$$\begin{align} I_1 & = \int_{C_3 + C_5}e^{irz}(1-z^2)^{v-1/2}dz\\ & = ie^{-ir}\int_R^\delta e^{-rt}(t^2+2it)^{v-\frac12}dt + ie^{ir}\int_\delta^R e^{-rt}(t^2-2it)^{v-\frac12}dt\\ I_2 & = \int_{C_1} e^{irz}(1-z^2)^{v-1/2}dz \\ & = \int_{-1+\delta}^{1-\delta} e^{irt}(1-t^2)^{v-\frac12}dt\\ I_3 & = \int_{C_6+C_2} e^{irz}(1-z^2)^{v-1/2}dz \\ & = \int_{\pi/2}^0 i\delta e^{i\delta\theta} e^{ir(-1+e^{i\delta\theta})}(2e^{i\delta\theta}+e^{2i\delta\theta}) ^{v-1/2}d\theta \\ & + \int_{0}^{\pi/2} i\delta e^{i\delta\theta} e^{ir(1-e^{i\delta\theta})}(2e^{i\delta\theta}+e^{2i\delta\theta}) ^{v-1/2}d\theta \\ I_4 & = \int_{C_4} e^{irz}(1-z^2)^{v-1/2}dz \\ & = \int_{1}^{-1} e^{ir(t+iR)}(1-t^2+R^2-2iRt)^{v-1/2}dt\,. \end{align}$$

Clearly, when $\delta\to 0,R\to\infty$, $-I_1$ gives the right hand side of (*), and $I_2$ gives us the original definition of $J_v(r)$. For $I_3$ and $I_4$, we only need to recall that

$$\begin{equation} \vert(\vert z\vert e^{i\theta})^{a+bi}\vert = \vert z\vert^a e^{-\theta b}\,. \end{equation}$$

Then we get that $\vert I_3\vert = O(\delta)$ and $\vert I_4\vert = O(e^{-rR}R^{2Re(v)-1})$. So both $I_3$ and $I_4$ goes to $0$.

Using this formula, we can show that

$$\begin{equation} J_v(r) = \sqrt{\frac{2}{\pi r}}\cos\left(r-\frac{\pi v}{2} -\frac{\pi}{4}\right) + R_v(r)\,, \end{equation}$$

where $\vert R_v(r)\vert \le C_vr^{-3/2}$. From this asymptotics, we can see that for $r\gg v$, $J_v(r)$ is close to the decaying $\cos$.