Spherical transformation in Peskin Book In the page 27 the authors calculate the propagator
$$D(x-y)=\int{\frac{\mathrm d^3p}{(2\pi)^3}\frac{1}{2E_{\textbf p}}e^{i\textbf{p}\cdot\textbf{r}}}$$
and it transforms the integral from Cartesian to spherical coordinates for $p$ and they found that
$$D(x-y)=\frac{2\pi}{(2\pi)^3}\int_{0}^{\infty}{~\mathrm dp~\frac{p^2}{2E_{\textbf p}}\frac{ e^{i pr}- e^{-i pr}}{ipr}}$$
I can not understand how $e^{i\textbf{p}\cdot\textbf{r}}$ became $\frac{ e^{i px}- e^{-i pr}}{2ipr}$. 
How is this is possible?
 A: Without loss of generality, we can create spherical coordinates $(p, \theta, \phi)$ such that $\mathbf{r}$ is along the $\theta=0$ axis.  This may seem a little backwards, since we usually have $\mathbf{r}$ as the position vector, which naturally varies.  But here we leave $\mathbf{r}$ constant, and vary in $\mathbf{p}$-space, since that's what we're integrating over.  In particular, variations in $\theta$ and $\phi$ correspond to variations in $\mathbf{p}$ rather than $\mathbf{r}$.  Then, we can transform the integral into those coordinates, perform the trivial integral over $\phi$, then change coordinates again using $\sin\theta\, d\theta = -d\cos\theta$, and finally perform that integral too, leaving only the integral over $p$.
\begin{align}
  D(x-y) &=
  \int_0^\infty \int_0^\pi \int_0^{2\pi} \frac{1}{(2\pi)^3}\,
    \frac{1}{2E_{\mathbf{p}}} e^{i\, p\, r\, \cos\theta}\, p^2\, \sin\theta\,
  d\phi\, d\theta\, dp \\
  &=
  \frac{1}{(2\pi)^2}\,
  \int_0^\infty \frac{p^2}{2E_{\mathbf{p}}} \left[\int_0^\pi
    e^{i\, p\, r\, \cos\theta}\, \sin\theta\,
  d\theta\right]\, dp \\
  &=
  \frac{1}{(2\pi)^2}\,
  \int_0^\infty \frac{p^2}{2E_{\mathbf{p}}} \left[\int_1^{-1}
    -e^{i\, p\, r\, x}\, dx\right]\, dp \\
  &=
  \frac{1}{(2\pi)^2}\,
  \int_0^\infty \frac{p^2}{2E_{\mathbf{p}}}
  \frac{-e^{-i\, p\, r} - -e^{i\, p\, r}}{i\, p\, r}\, dp.
\end{align}
Note that this assumes $E_{\mathbf{p}}$ is independent of $\theta$ and $\phi$, but that's fine because we usually have $E_{\mathbf{p}} = \sqrt{ \lvert \mathbf{p} \rvert^2 + m^2} = \sqrt{ p^2 + m^2}$.
