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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!

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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).

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Thank you for the detailed answer. – nbubis Apr 15 '12 at 7:29
@nbubis: Sure thing. Cheers! – user26872 Apr 15 '12 at 7:31

Peskin details this integral. Excuse my laziness, but I'll just paste a photo of the relevant page here: Peskin

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