In the Gradshteyn and Ryzhik Table of Integrals, the following integral appears (3.876.1, page 486 in the 8th edition): \begin{equation} \int_0^{\infty} \frac{\sin (p \sqrt{x^2 + a^2})}{\sqrt{x^2 + a^2}} \cos (bx) dx = \begin{cases} \frac{\pi}{2} J_0 \left( a \sqrt{p^2 - b^2} \right) & 0 < b < p \\ 0 & b > p > 0 \end{cases} \end{equation} for $a > 0$, where $J_0$ is the Bessel function.
I am interested how this result could be derived (not necessarily rigorously proved), especially the range of $p$ such that the value is $0$.
Thank you.
$$I(a,b,p)\equiv\int_0^{\infty}\frac{\sin\left(p\sqrt{x^2+a^2}\right)}{\sqrt{x^2+a^2}}\cos(bx)\,dx\\\\ $$
Enforcing the substitution $x\to a\sinh x$ and assuming that $a>0$ yields
$$\begin{align} I(a,b,p)&=\int_0^{\infty}\sin\left(pa\cosh x\right)\cos(ba\sinh x)\,dx\\\\ &=\frac12\int_0^{\infty}\left(\sin\left(pa\cosh x+ba\sinh x\right)+\sin\left(pa\cosh x-ba\sinh x\right)\right)\,dx\\\\ &=\frac12\int_{-\infty}^{\infty}\sin\left(pa\cosh x+ba\sinh x\right)\,dx \end{align}$$
Recalling that
$$A\cosh x+B\sinh x= \begin{cases} \sqrt{A^2-B^2}\cosh(x-\text{artanh}(B/A))&,0<B<A\\\\ \sqrt{B^2-A^2}\sinh(x-\text{artanh}(A/B))&,B>A>0 \end{cases} $$
we have
$$I(a,b,p)= \begin{cases} \frac12\int_{-\infty}^{\infty}\sin\left(a\sqrt{p^2-b^2}\cosh x\right)\,dx&,0<b<p\\\\ \frac12\int_{-\infty}^{\infty}\sin\left(a\sqrt{b^2-p^2}\sinh x\right)\,dx&,b>p>0 \end{cases} $$
Noting that the integrand of the first integral is an even function of $x$, while the integrand of the second integral is an odd function of $x$ reveals
$$I(a,b,p)= \begin{cases} \int_{0}^{\infty}\sin\left(a\sqrt{p^2-b^2}\cosh x\right)\,dx&,0<b<p\\\\ 0&,b>p>0 \end{cases} $$
From Equation $(10.9.9)$ HERE, we see that
$$\int_{0}^{\infty}\sin\left(a\sqrt{p^2-b^2}\cosh x\right)\,dx=\frac{\pi}{2}J_0\left(a\sqrt{p^2-b^2}\right)$$
for $a\sqrt{p^2-b^2}>0$ and $p>b>0$
whereupon we have the final result
$$I(a,b,p)= \begin{cases} \frac{\pi}{2}J_0\left(a\sqrt{p^2-b^2}\right)\,dx&,0<b<p\\\\ 0&,b>p>0 \end{cases} $$
for $a>0$.