Thermal green function and complex integration

[mercc]

green function and complex integration

Homework Statement

By reading this paper http://arxiv.org/pdf/hep-ph/0610391v4 I cannot proof the following relation on page 9 equation (23) , by a suitable choice of a contour in the complex omega plane:

[tex]\int\frac{d^3 p}{(2\pi)^2} \coth(\frac{|\vec{p}|}{2T})\frac{\cos(|\vec{p}| t-\vec{p}\vec{x})}{|\vec{p}|}=
\int\frac{d^3 p}{(2\pi)^3} e^{i\vec{p}\vec{x}}\int\frac{d\omega}{2\pi} \coth(\frac{\omega}{2T})\frac{e^{-i\omega t}}{\omega^2-|\vec{p}|^2-i \epsilon}}[/tex]T stands for the temperature, therefore T>0.

Homework Equations

[tex]\cos(|\vec{p}| t-\vec{p}\vec{x})=\frac{1}{2} e^{i \vec{p}| t-i \vec{p}\vec{x}}+\frac{1}{2} e^{-i \vec{p}| t+i \vec{p}\vec{x}}[/tex]

The Attempt at a Solution

First of all I calculated the residues at [tex]\omega=-|{\vec{p}}|[/tex] and [tex]\omega=|{\vec{p}}|[/tex]. This gives me:

[tex]Res(\omega=-|{\vec{p}}|)=\frac{coth(-|{\vec{p}}|/2T)}{-2|{\vec{p}}|} e^{i |{\vec{p}}|t}=\frac{coth(|{\vec{p}}|/2T)}{2|{\vec{p}}|} e^{i |{\vec{p}}|t}[/tex] since the coth is an odd function.
[tex]Res(\omega=|{\vec{p}}|)=\frac{coth(|{\vec{p}}|/2T)}{2|{\vec{p}}|} e^{-i |{\vec{p}}|t}[/tex]

The -i epsilon term in the denominator says one should circumvent the pole at [tex]\omega=|{\vec{p}}|[/tex] anti-clockwise and at [tex]\omega=-|{\vec{p}}|[/tex] clockwise (Feynman prescription). By suming up these contributions i get almost the result (missing just a numerical factor [tex]-2\pi i[/tex]).

Then I investigated the behaviour of the integrand for [tex]|\omega|\rightarrow \infty[/tex] and found out that
[tex]coth(\omega /2T)\rightarrow 1, |\omega|\rightarrow \infty[/tex] and
[tex]e^{-i\omega t} \rightarrow 0, |\omega|\rightarrow \infty[/tex] only if t<0 in the upper omega halfplane and t>0 in the lower halfplane. Else the contributions at infinity to the contour integral would not vanish (there would not be an damping factor exp(-Im(w)t).

The Problem is that I do not really know how to choose the contour in the complex plane, since additionally the coth has poles on the imaginary axis at [tex]\omega=i 2\pi nT[/tex] for every integer n too. Especially coth has a simple singularity at [tex]\omega=0[/tex].
I asked my professor and he told me to try the following contour (look at the attachment). But even he does not know what to do with the pole at zero.
I am trying to figure this out now for more than a week, but no success.

I hope one of you guys can help me out. Thanks in advance!

 

[_coyote_]

A:You should be able to evaluate the integral over $\omega$ as follows. The poles of the integrand are at $\omega = \pm |\vec{p}|$ and $\omega = i 2\pi nT$ for $n \in \mathbb{Z}$. Note that for $T>0$, the poles at $\omega = i 2\pi nT$ will all lie in the lower half-plane.Draw a semicircular contour in the upper-half plane with radius $R$ and centered at the origin. The integral over this contour is$$\oint_C d\omega \; \frac{\coth(\omega/2T)}{\omega^2 - |\vec{p}|^2 \pm i \epsilon}.$$The integral over the semicircle vanishes in the limit $R \to \infty$ because the integrand decays as $1/R$ (assuming $\epsilon > 0$). Therefore$$\int_{-\infty}^{\infty} d\omega \; \frac{\coth(\omega/2T)}{\omega^2 - |\vec{p}|^2 \pm i \epsilon} = 2 \pi i \sum_{\text{poles}} \text{Res}\left[\frac{\coth(\omega/2T)}{\omega^2 - |\vec{p}|^2 \pm i \epsilon}, \omega\right].$$The poles of the integrand are at $\omega = \pm |\vec{p}|$ and $\omega = i 2\pi nT$ for $n \in \mathbb{Z}$. The residues at these poles are\begin{align}\text{Res}\left[\frac{\coth(\omega/2T)}{\omega^2 - |\vec{p}|^2 \pm i \epsilon}, \omega=\pm|\vec{p}|\right] &= \pm \frac{\coth(\pm |\vec{p}|/2T)}{2|\vec{p