I don't understand how the following simplification is done:
$\frac{1}{(2\pi)^3}\int d^3p e^{-it\sqrt{\vec{p}^2+m^2}}e^{i\vec{p}(\vec{x}-\vec{x}_0)}=\frac{1}{2\pi^2|\vec{x}-\vec{x}_0|}\int_0^{\infty}dp p \sin(p|\vec{x}-\vec{x}_0|)e^{-it\sqrt{p^2+m^2}}$
My first step is to use spherical coordinates:
$\frac{1}{(2\pi)^3}\int_0^{\infty} p^2 dp \int_0^\pi \sin\theta d\theta\int_0^{2\pi}d\phi e^{-it\sqrt{p^2+m^2}}e^{i\vec{p}(\vec{x}-\vec{x}_0)}=\underbrace{\frac{4\pi}{(2\pi)^3}}_{=1/(2\pi^2)}\int_0^{\infty} p^2 dpe^{-it\sqrt{p^2+m^2}}e^{i\vec{p}(\vec{x}-\vec{x}_0)}$
But now I don't know how to continue. The solution would suggest the relation $e^{ip|\vec{x}-\vec{x}_0|}=\frac{\sin(p|\vec{x}-\vec{x}_0|)}{p|\vec{x}-\vec{x}_0|}$
but I have never seen this relation before. The only relation I know with sinus and exponential function is: $e^{ix}=\cos x+i\sin x$. What am I missing?
Thank you for your answers in advance!
The problem is that $\mathbf{p}\cdot(\mathbf{x}-\mathbf{x}_0)$ is not independent of spherical angle, so you can't separate out the angular integrals. The correct way to do it is to make the polar axis parallel to $\mathbf{x}-\mathbf{x}_0$ in your spherical coordinate system so that $\mathbf{p}\cdot(\mathbf{x}-\mathbf{x}_0) = p|\mathbf{x}-\mathbf{x}_0|\cos\theta$ and integrate as \begin{multline} \frac{1}{(2\pi)^3}\int \exp\left(-it\sqrt{p^2 + m^2}\right)\exp[i\mathbf{p}\cdot(\mathbf{x}-\mathbf{x}_0)]d^3\mathbf{p} \\= \frac{1}{(2\pi)^3}\int_0^{2\pi}\!\!\!\!\int_0^\pi\!\!\!\!\int_0^\infty \exp\left(-it\sqrt{p^2 + m^2}\right)\exp(ip|\mathbf{x}-\mathbf{x}_0|\cos\theta)p^2\sin\theta dp\,d\theta\, d\phi \\ = \frac{1}{(2\pi)^3}\left[\int_0^{2\pi}\!\!d\phi\right]\int_0^\infty \!\!p^2\exp\left(-it\sqrt{p^2 + m^2}\right)\left[\int_{-1}^1\!\exp(ip|\mathbf{x}-\mathbf{x}_0|\xi)d\xi\right]dp, \end{multline} where I have substituted $\xi = \cos\theta$. You should be able to take it from here.