I have an integral of $\ R_{1} =-i\int_{t}^{\infty }\frac{ds}{s}J_{1}\left( 2s\right) e^{i\nu s}$,
where $J_{1}\left( 2s\right)$ is the Bessel function of the first kind. and $t$ is a positive number.
I try to solve it in the following method:
$\frac{dR_{1}}{d\nu }=\int_{t}^{\infty }dsJ_{1}\left( 2s\right) e^{i\nu s} $
$\ J_{n}\left( x\right) =\frac{1}{2\pi }\int_{-\pi }^{\pi }e^{i\left( x\sin \theta -n\theta \right) }d\theta$
$\frac{dR_{1}}{d\nu }=\frac{1}{2\pi }\int_{-\pi }^{\pi }d\theta \int_{t}^{\infty }dse^{i\left( 2s\sin \theta -\theta \right) }e^{i\nu s}=\frac{1}{2\pi }\int_{-\pi }^{\pi }d\theta \frac{ -e^{-i\left[ \theta -t\left( 2\sin \theta +\nu \right) \right] }}{i\left( 2\sin \theta +\nu \right)} $
Using the Jacobi-Anger expansion $\ e^{ix\sin \theta }=\sum_{n=-\infty }^{\infty }J_{n}\left( x\right) e^{in\theta } $,
$\frac{dR_{1}}{d\nu } =\frac{1}{2\pi }e^{it\nu }\sum_{n=-\infty }^{\infty }J_{n}\left( 2t\right) \int_{-\pi }^{\pi }d\theta \frac{-e^{-i\theta }e^{in\theta }}{i\left( 2\sin \theta +\nu \right) }$
Taking $\ z=e^{i\theta }$ , $dz=ie^{i\theta }d\theta =izd\theta$, and $\sin \theta =\frac{e^{i\theta }-e^{-i\theta }}{2i}=\frac{z-z^{-1}}{2i}$, I get
$\frac{dR_{1}}{d\nu }=-e^{it\nu }\sum_{n=-\infty }^{\infty }J_{n}\left( 2t\right) \frac{1}{2\pi i}\oint dz\frac{z^{n-1}}{z^{2}+i\nu z-1}$
According to the residue theorem, if $n>1$, then suppose there is only a single pole with $\left\vert z_{+}\right\vert <1$, while $n\leq 0$ there is also a pole at $z=0$.
$\frac{1}{z^{2}+i\nu z-1}=\frac{1}{\left( z-z_{+}\right) \left( z-z_{-}\right) }$, $z_{\pm }=\frac{-i\nu \pm \sqrt{\left( i\nu \right) ^{2}+4}}{2}$
$\frac{1}{\left( z-z_{+}\right) \left( z-z_{-}\right) }=\frac{1}{z_{+}-z_{-}}% \left[ \frac{1}{z-z_{+}}-\frac{1}{z-z_{-}}\right]$
For $z$ near $0$, $\frac{1}{z-z_{+}} =\frac{-1}{z_{+}}\frac{1}{1-\frac{z}{z_{+}}} =\left( \frac{-1}{z_{+}}\right) \sum_{j=0}^{\infty }\left( \frac{z}{z_{+}}% \right) ^{j}$, so
$\frac{dR_{1}}{d\nu }=-e^{it\nu }\sum_{n=-\infty }^{\infty }J_{n}\left( 2t\right) \frac{1}{2\pi i}\oint dz\frac{z^{n-1}}{z^{2}+i\nu z-1}$
$=-e^{it\nu }\left( \sum_{n=1}^{\infty }J_{n}\left( 2t\right) I_{n}+\sum_{n=-\infty }^{0}J_{n}\left( 2t\right) I_{n}\right) $
$=-e^{it\nu }\left\{ \sum_{n=1}^{\infty }J_{n}\left( 2t\right) \frac{\left( z_{+}\right) ^{n-1}}{z_{+}-z_{-}}\right. +\left. \sum_{n=-\infty }^{0}J_{n}\left( 2t\right) \left[ \frac{-1}{z_{+}-z_{-}}\left( \frac{1}{% \left( z_{+}\right) ^{1-n}}-\frac{1}{\left( z_{-}\right) ^{1-n}}\right) +% \frac{\left( z_{+}\right) ^{n-1}}{z_{+}-z_{-}}\right] \right\}$.
Because I seperate the sum into two parts from $-\infty$ to $0$ and from $1$ to $+\infty$, how can I combine them again to obtain the results? Do I make any mistakes during the calculations? I have been struggling for a couple of weeks. Also, do you have some better methods to solve this integral? Many thanks!