Complicated Integration involving Bessel function and exponential function Sir,
while studying the quantum scattering for various cases, I have got the following integrals:
$$\int_{t_0=-\infty}^{t-R/c}\Big[\frac{\exp{(-i\omega t_0)}}{(t-t_0)}\Big]J_0\big(\sqrt{(t-t_0)^2-(R/c)^2}\big)\,dt_0$$ and similarly,
$$\int_{t_0=-\infty}^{t-R/c}\Big[-\frac{\exp{(-i\omega t_0)}}{(t-t_0)^2}\Big]J_0\big(\sqrt{(t-t_0)^2-(R/c)^2}\big)\,dt_0$$
where, $t$ is observation time, $t_0$ is the retarded time, $J_0$ is the Bessel function of first kind and $R$ is the distance between the observation point and the point on scatterer such that $R=\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$ , where, $(x,y,z)$ is the point of observation.$c$ is the speed of light and taken as constant. Further, the causality condition is followed such that $t-t_0\geq R/c$ or $t_0\leq t- R/c$. According to the physical resctriction imposed on the problem, it is also true that in the limit $t_0\rightarrow -\infty$, the integrand should be zero.
I am trying to find out the closed form answer of the integrals (if one can be solved, the other can be solved by the same technique).I searched through  the "Table of integrals" by Gradshteyn but somehow, I could not find out the exact match.
Would you kindly suggest me any method or any relevant texts from where I can get some help.

$$
\int_{-\infty}^{t-a}{\frac{e^{-i\omega u}}{t-u}J_0\left( \sqrt{\left( t-u \right) ^2-a^2} \right) du}
\\
u=t-r,=e^{-i\omega t}\int_a^{\infty}{\frac{e^{i\omega r}}{r}J_0\left( \sqrt{r^2-a^2} \right) dr}
\\
I\left( \omega \right) =\int_a^{\infty}{\frac{e^{i\omega r}}{r}J_0\left( \sqrt{r^2-a^2} \right) dr}
\\
I'\left( \omega \right) =i\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n}{\left( n! \right) ^2}}\left( \frac{1}{2} \right) ^{2n}\int_a^{\infty}{e^{i\omega r}\left( r^2-a^2 \right) ^ndr}
\\
r=ax,=i\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n}{\left( n! \right) ^2}}\left( \frac{a}{2} \right) ^{2n}\int_1^{\infty}{e^{i\omega ax}\left( x^2-1 \right) ^ndx}
\\
{H_{\nu}}^{\left( 1 \right)}\left( x \right) =-\frac{2i\left( \frac{x}{2} \right) ^{-\nu}}{\sqrt{\pi}\Gamma \left( \frac{1}{2}-\nu \right)}\int_1^{\infty}{\frac{e^{ixt}}{\left( t^2-1 \right) ^{\nu +\frac{1}{2}}}dt}
\\
{H_{-\frac{1}{2}-n}}^{\left( 1 \right)}\left( \omega a \right) =-\frac{2i\left( \frac{\omega a}{2} \right) ^{\frac{1}{2}+n}}{\sqrt{\pi}n!}\int_1^{\infty}{e^{i\omega at}\left( t^2-1 \right) ^ndt}
\\
I'\left( \omega \right) =i\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n}{\left( n! \right) ^2}}\left( \frac{a}{2} \right) ^{2n}\int_1^{\infty}{e^{i\omega ax}\left( x^2-1 \right) ^ndx}
\\
=-\frac{\sqrt{\pi}}{2\omega}\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n}{n!}}\left( \frac{a}{2\omega} \right) ^{n-\frac{1}{2}}{H_{-\frac{1}{2}-n}}^{\left( 1 \right)}\left( \omega a \right) 
\\
=-\sqrt{\frac{\pi}{2\omega a}}\sum_{n=0}^{\infty}{\frac{1}{n!}}\left( -\frac{a}{2\omega} \right) ^n\left( J_{-\frac{1}{2}-n}\left( \omega a \right) +iY_{-\frac{1}{2}-n}\left( \omega a \right) \right) 
\\
=-\sqrt{\frac{\pi}{2\omega a}}\sum_{n=0}^{\infty}{\frac{1}{n!}}\left( -\frac{a}{2\omega} \right) ^n\left( \left( -1 \right) ^{n-1}Y_{\frac{1}{2}+n}\left( \omega a \right) +i\left( -1 \right) ^nJ_{\frac{1}{2}+n}\left( \omega a \right) \right) 
\\
=\sqrt{\frac{\pi}{2\omega a}}\sum_{n=0}^{\infty}{\frac{Y_{\frac{1}{2}+n}\left( \omega a \right)}{n!}}\left( \frac{a}{2\omega} \right) ^n-i\sqrt{\frac{\pi}{2\omega a}}\sum_{n=0}^{\infty}{\frac{J_{\frac{1}{2}+n}\left( \omega a \right)}{n!}}\left( \frac{a}{2\omega} \right) ^n
$$
The spherical Bessel functions have the generating functions
$$
\frac{\cos \left( \sqrt{z^2-2zt} \right)}{z}=\sum_{n=0}^{\infty}{\frac{t^n}{n!}}\sqrt{\frac{\pi}{2z}}J_{n-1/2}(z)
\\
\frac{\sin \left( \sqrt{z^2-2zt} \right)}{z}=\sum_{n=0}^{\infty}{\frac{t^n}{n!}}\sqrt{\frac{\pi}{2z}}Y_{n-1/2}(z)
\\
\therefore \frac{\sin \left( \sqrt{z^2-2zt} \right)}{\sqrt{z^2-2zt}}=\sum_{n=0}^{\infty}{\frac{t^n}{n!}}\sqrt{\frac{\pi}{2z}}J_{n+1/2}(z)
\\
-\frac{\cos \left( \sqrt{z^2-2zt} \right)}{\sqrt{z^2-2zt}}=\sum_{n=0}^{\infty}{\frac{t^n}{n!}}\sqrt{\frac{\pi}{2z}}Y_{n+1/2}(z)
\\
\therefore \frac{\sin \left( a\sqrt{\omega ^2-1} \right)}{a\sqrt{\omega ^2-1}}=\sum_{n=0}^{\infty}{\frac{\left( \frac{a}{2\omega} \right) ^n}{n!}}\sqrt{\frac{\pi}{2\omega a}}J_{n+1/2}(\omega a)
\\
-\frac{\cos \left( a\sqrt{\omega ^2-1} \right)}{a\sqrt{\omega ^2-1}}=\sum_{n=0}^{\infty}{\frac{\left( \frac{a}{2\omega} \right) ^n}{n!}}\sqrt{\frac{\pi}{2\omega a}}Y_{n+1/2}(\omega a)
\\
I'\left( \omega \right) =\sqrt{\frac{\pi}{2\omega a}}\sum_{n=0}^{\infty}{\frac{Y_{\frac{1}{2}+n}\left( \omega a \right)}{n!}}\left( \frac{a}{2\omega} \right) ^n-i\sqrt{\frac{\pi}{2\omega a}}\sum_{n=0}^{\infty}{\frac{J_{\frac{1}{2}+n}\left( \omega a \right)}{n!}}\left( \frac{a}{2\omega} \right) ^n
\\
=-\frac{\cos \left( a\sqrt{\omega ^2-1} \right)}{a\sqrt{\omega ^2-1}}-i\frac{\sin \left( a\sqrt{\omega ^2-1} \right)}{a\sqrt{\omega ^2-1}}
\\
I\left( \omega \right) =-\frac{1}{a}\int_{\infty}^{\omega}{\left( \frac{\cos \left( a\sqrt{t^2-1} \right)}{\sqrt{t^2-1}}+i\frac{\sin \left( a\sqrt{t^2-1} \right)}{\sqrt{t^2-1}} \right) dt}
\\
t=\cosh u,=\frac{1}{a}\int_{\mathrm{arc}\cosh \omega}^{\infty}{e^{ia\sinh u}du}
\\
\int_{-\infty}^{t-a}{\frac{e^{-i\omega u}}{t-u}J_0\left( \sqrt{\left( t-u \right) ^2-a^2} \right) du}=e^{-i\omega t}I\left( \omega \right) 
\\
=\frac{1}{a}e^{-i\omega t}\int_{\mathrm{arc}\cosh \omega}^{\infty}{e^{ia\sinh u}du}
$$
I think this integral may not have a closed-form solution.
