Trying to rewrite this integral:
$$\int dq \frac{q^2}{2\pi^2} \frac{\sin(\sqrt{q^2+m^2}t)}{\sqrt{q^2+m^2}} \frac{\sin (qr)}{qr}$$
In terms of the Bessel function of the first kind, $J_0$ but have no idea how to since I'm not used to Bessel functions.
I know that the answer should be:
$$\frac{1}{4\pi r} \frac{\partial}{\partial r} J_0(m\sqrt{t^2-r^2}) r>0$$
Hint:
Let $q=m\sinh u$ ,
Then $dq=m\cosh u$
$\therefore\int\dfrac{q^2}{2\pi^2}\dfrac{\sin(\sqrt{q^2+m^2}t)}{\sqrt{q^2+m^2}}\dfrac{\sin (qr)}{qr}~dq$
$=\int\dfrac{m^2\sinh^2u}{2\pi^2}\dfrac{\sin\left(t\sqrt{m^2\sinh^2u+m^2}\right)}{\sqrt{m^2\sinh^2u+m^2}}\dfrac{\sin(rm\sinh u)}{rm\sinh u}~d(m\sinh u)$
$=\dfrac{m}{2\pi^2r}\int\sin(mr\sinh u)\sin(mt\cosh u)\sinh u~du$
$=\dfrac{m}{4\pi^2r}\int\cos(mr\sinh u-mt\cosh u)\sinh u~du-\dfrac{m}{4\pi^2r}\int\cos(mr\sinh u+mt\cosh u)\sinh u~du$
$=\dfrac{m}{8\pi^2r}\int e^u\cos\left(\dfrac{m(r-t)e^u}{2}-\dfrac{m(r+t)}{2e^u}\right)~du-\dfrac{m}{8\pi^2r}\int e^{-u}\cos\left(\dfrac{m(r-t)e^u}{2}-\dfrac{m(r+t)}{2e^u}\right)~du-\dfrac{m}{8\pi^2r}\int e^u\cos\left(\dfrac{m(r+t)e^u}{2}-\dfrac{m(r-t)}{2e^u}\right)~du+\dfrac{m}{8\pi^2r}\int e^{-u}\cos\left(\dfrac{m(r+t)e^u}{2}-\dfrac{m(r-t)}{2e^u}\right)~du$
Hint: One possible start that is a bit too long for a comment could be to start with definition for Bessel function according to wikipedia:
$$x^2\frac{\partial^2 y}{\partial x^2} + x\frac{\partial y}{\partial x}+(x^2-\alpha^2)y = 0$$
Subtract last term from both sides:
$$x^2\frac{\partial^2 y}{\partial x^2} + x\frac{\partial y}{\partial x} = -(x^2-\alpha^2)y$$
Assume $(x^2-\alpha) \neq 0$ and divide on both sides: $$\frac{x^2}{x^2-\alpha^2}\frac{\partial^2 y}{\partial x^2} + \frac{x}{x^2-\alpha^2}\frac{\partial y}{\partial x} = -y$$
Maybe you can continue from there?