Stueckelberg Feynman propagator computation On page 35 of Itzykson-Zuber's textbook on quantum field theory, I am having trouble deriving equation (1-180):
$\displaystyle G_F(0,r) = \frac{i}{(2\pi)^2 r} \int_m^\infty dp \frac{p}{\sqrt{p^2-m^2}} e^{-pr}$
Here $G_F(0,r)$ is the Stueckelberg Feynman propogator $G_F(x) = \frac{-1}{(2\pi)^4} \int d^4 p e^{-i p\cdot x} \frac{1}{p^2 - m^2 + i\epsilon}$, when $x$ is of the form $(0,\vec{x})$ and $|\vec{x}| = r$. Presumably the $i$ in the exponent drops out because we are dealing with complex valued vector $\vec{x}$, but even with that assumption I cannot derive the formula. Any insight is appreciated! 
 A: Let $p = (p_0,{\bf p})$ and $x = (0,{\bf x})$. 
For convenience, define $\rho = |{\bf p}|$ and $r = |{\bf x}|$. 
Then 
$p\cdot x = -{\bf p}\cdot {\bf x} = - \rho r \cos\theta$ and 
$p^2 = p_0^2 - {\bf p}^2 = p_0^2 - \rho^2$. 
We find
$$\begin{eqnarray*}
G_F(0,r) = -\frac{1}{(2\pi)^4} \int d^3 p \, e^{i \rho r \cos\theta} 
\int_{-\infty}^\infty dp_0 \frac{1}{p_0^2 - E_\rho^2 + i\epsilon}
\end{eqnarray*}$$
where $E_\rho = \sqrt{\rho^2 + m^2}$. 
Close the contour in the $p_0$ integral above to pick up the poll at $p_0 = -E_\rho$. 
(Notice the pole at $E_\rho$ ($-E_\rho$) is slightly below (above) the $p_0$-axis. 
This is the Feynman prescription for handling the poles.)
We find 
$$\int_{-\infty}^\infty dp_0 \frac{1}{p_0^2 - E_\rho^2 + i\epsilon} 
= 2\pi i \frac{1}{-2E_\rho}
= -\frac{i \pi}{\sqrt{\rho^2 + m^2}}.$$
If we do the angular integral we arrive at 
$$\begin{eqnarray*}
G_F(0,r) &=& -\frac{1}{(2\pi)^4} \int_0^\infty d\rho\, \rho^2 \frac{-i\pi}{\sqrt{\rho^2 + m^2}}
\,2\pi \frac{e^{i\rho r} - e^{-i\rho r}}{i\rho r} \\
&=& \frac{i}{(2\pi)^2} \frac{1}{2r} \int_{-\infty}^\infty d\rho \frac{\rho e^{i\rho r}}{\sqrt{\rho^2+m^2}}.
\end{eqnarray*}$$
Move the contour to wrap around the cut in the upper half-plane. 
(The cut extending from $i m$ to $i\infty$.)
Define $p = -i \rho$. 
We find 
$$\begin{eqnarray*}
G_F(0,r) &=& \frac{i}{(2\pi)^2 r} \int_m^\infty dp\, \frac{p}{\sqrt{p^2 - m^2}} e^{-p r} \\
\end{eqnarray*}$$
as claimed by Itzykson and Zuber. 
This integral can be evaluated analytically. 
Let $p = m \cosh t$. 
We find 
$$G_F(0,r) = \frac{i m}{4\pi^2 r} \int_0^\infty dt\, \cosh t\, e^{-m r\cosh t}.$$
Recall the integral representation of the modified Bessel function of the second kind, 
$$K_n(z) = \int_0^\infty d t\, \cosh n t\ e^{-z \cosh t}.$$
Thus, 
$$G_F(0,r) = \frac{i m}{4\pi^2 r} K_1(m r).$$
For large $z$, $K_n(z) \sim \sqrt{\frac{\pi}{2z}}e^{-z}$. 
Therefore, for $r$ large we find
$$G_F(0,r) \sim \frac{i e^{-m r}}{(2\pi)^2 r^2} \left(\frac{\pi m r}{2}\right)^{1/2}$$
agreeing with the second half of Itzykson and Zuber's (1-180). 
A: Peskin details this integral. Excuse my laziness, but I'll just paste a photo of the relevant page here:

