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:

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

| cite | improve this question | | | | |
  • $\begingroup$ I suppose a start could be to write down the definition of the Bessel functions. $\endgroup$ – 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?

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


Let $q=m\sinh u$ ,

Then $dq=m\cosh u$

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

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

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

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

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

| cite | improve this answer | | | | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.