# Rewrite integral in terms of Bessel functions

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:

$$\frac{1}{4\pi r} \frac{\partial}{\partial r} J_0(m\sqrt{t^2-r^2}) r>0$$

• I suppose a start could be to write down the definition of the Bessel functions. – mathreadler Dec 18 '19 at 10:21

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?

• Not sure where you found this definition? Could you maybe give some more info? – Linus Dec 18 '19 at 11:01
• @Linus I took the definition directly from wikipedia, maybe you have some slightly different one in your course? – mathreadler Dec 18 '19 at 11:04
• Yes I tried to use J defined in some integral form that I found on Wolfram could you give some more steps? – Linus Dec 18 '19 at 11:11

Hint:

Let $$q=m\sinh u$$ ,

Then $$dq=m\cosh u$$

$$\therefore\int\dfrac{q^2}{2\pi^2}\dfrac{\sin(\sqrt{q^2+m^2}t)}{\sqrt{q^2+m^2}}\dfrac{\sin (qr)}{qr}~dq$$

$$=\int\dfrac{m^2\sinh^2u}{2\pi^2}\dfrac{\sin\left(t\sqrt{m^2\sinh^2u+m^2}\right)}{\sqrt{m^2\sinh^2u+m^2}}\dfrac{\sin(rm\sinh u)}{rm\sinh u}~d(m\sinh u)$$

$$=\dfrac{m}{2\pi^2r}\int\sin(mr\sinh u)\sin(mt\cosh u)\sinh u~du$$

$$=\dfrac{m}{4\pi^2r}\int\cos(mr\sinh u-mt\cosh u)\sinh u~du-\dfrac{m}{4\pi^2r}\int\cos(mr\sinh u+mt\cosh u)\sinh u~du$$

$$=\dfrac{m}{8\pi^2r}\int e^u\cos\left(\dfrac{m(r-t)e^u}{2}-\dfrac{m(r+t)}{2e^u}\right)~du-\dfrac{m}{8\pi^2r}\int e^{-u}\cos\left(\dfrac{m(r-t)e^u}{2}-\dfrac{m(r+t)}{2e^u}\right)~du-\dfrac{m}{8\pi^2r}\int e^u\cos\left(\dfrac{m(r+t)e^u}{2}-\dfrac{m(r-t)}{2e^u}\right)~du+\dfrac{m}{8\pi^2r}\int e^{-u}\cos\left(\dfrac{m(r+t)e^u}{2}-\dfrac{m(r-t)}{2e^u}\right)~du$$