Derivation of Gradshteyn and Ryzhik integral 3.876.1 (in question)

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.

Solutions Collecting From Web of "Derivation of Gradshteyn and Ryzhik integral 3.876.1 (in question)"

$$I(a,b,p)\equiv\int_0^{\infty}\frac{\sin\left(p\sqrt{x^2+a^2}\right)}{\sqrt{x^2+a^2}}\cos(bx)\,dx\\\\
$$

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 $a>0$.