Cauchy's integral theorem in deriving green's function In deriving the form of Green's function I get an integral of the form:
$$\int_{-\infty}^{\infty}d\omega(\frac{1}{ck-\omega}-\frac{1}{-ck-\omega})e^{-i\omega\tau}$$
Now we want to treat the singularities by shifting the poles in the complex plane. If we demand that the integral is non-zero for $\tau>0$ we should shift the poles down so we get:
$$\int_{-\infty}^{\infty}d\omega(\frac{1}{ck- i\epsilon-\omega}-\frac{1}{-ck - i\epsilon -\omega})e^{-i\omega\tau}$$
Using cauchy's integral theorem we then get: $2\pi i (-e^{-i (ck-i\epsilon)\tau}+e^{i(ck+i\epsilon)\tau})$ 
And taking the limit $\epsilon\rightarrow 0$ we get $2\pi i (-e^{-i ck\tau}+e^{ick\tau})$ 
Now my problem is that if I were to shift the poles up I get:
$$\int_{-\infty}^{\infty}d\omega(\frac{1}{ck+i\epsilon-\omega}-\frac{1}{-ck + i\epsilon -\omega})e^{-i\omega\tau}$$
and then: $2\pi i (-e^{-i (ck+i\epsilon)\tau}+e^{i(ck -i\epsilon)\tau})$  giving the same result after taking again the limit $\epsilon\rightarrow 0$. What am I missing here?
 A: To use Cauchy's theorem, you need to consider the integral in the complex plane.  In doing that, you need to carefully consider the contour itself.  Really, what you are doing is evaluating an integral
$$\oint_C dz \,\left (\frac{1}{ck-z} + \frac{1}{c k+z}\right )e^{-i \tau z}$$
where $C$ is a semicircular contour with deformations about the poles on the real axis.  Let's consider the case $\tau \lt 0$ first.  In this case, we use a semicircle of radius $R$ in the upper half plane.  Then we get that the contour integral is
$$\int_{-R}^{-c k-\epsilon} d\omega\, \left (\frac{1}{ck-\omega} + \frac{1}{c k+\omega}\right )e^{-i \tau \omega} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi}\, \left (\frac{1}{2 ck-\epsilon e^{i \phi}} + \frac{1}{\epsilon e^{i \phi}}\right ) e^{-i \tau (-c k+\epsilon  e^{i \phi})} + \\ \int_{-c k+\epsilon}^{c k - \epsilon} d\omega\, \left (\frac{1}{ck-\omega} + \frac{1}{c k+\omega}\right )e^{-i \tau \omega}+ i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi}\, \left (\frac{1}{2 ck+\epsilon e^{i \phi}} - \frac{1}{\epsilon e^{i \phi}}\right ) e^{-i \tau (c k+\epsilon  e^{i \phi})}\\+ \int_{c k+\epsilon}^R d\omega\, \left (\frac{1}{ck-\omega} + \frac{1}{c k+\omega}\right )e^{-i \tau \omega} + i R \int_0^{\pi} d\theta \, e^{i \theta} \, \left (\frac{1}{ck-R e^{i \theta}} + \frac{1}{ck+R e^{i \theta}}\right ) e^{-i \tau R e^{i \theta}} $$
We consider the limits as $\epsilon \to 0$ and $R \to \infty$.  The first, third, and fifth integrals combine to form the Cauchy principal value of the original integral (which is really the value you seek).  The sixth integral is bounded by
$$\frac12 \int_0^{\pi} d\theta \, e^{R \tau \sin{\theta}} \le \int_0^{\pi/2} d\theta \, e^{2 R \tau \theta/\pi}\le \frac{\pi}{2 R (-\tau)}$$
(Recall that $\tau \lt 0$.)  Thus, the sixth integral vanishes as $R \to \infty$.  The second and fourth integrals have nonzero contributions that are easily evaluated as $\epsilon \to 0$.
By Cauchy's theorem, the contour integral is zero because there are no poles inside the contour $C$.  Therefore
$$PV \int_{-\infty}^{\infty} d\omega\, \left (\frac{1}{ck-\omega} + \frac{1}{c k+\omega}\right )e^{-i \tau \omega} = i \pi \left (e^{i c k \tau} - e^{-i c k \tau}  \right ) \quad (\tau \lt 0)$$
Now consider the case $\tau \gt 0$. Note that if we use the same contour as above, the sixth integral will not vanish and indeed does not converge.  Therefore, we use another contour $C'$ which is a deformed semicircle in the lower half plane.  To exclude the poles on the real axis from the interior of $C'$, we deform below the real axis at these poles.  Thus, the contour integral in this case becomes
$$\int_{-R}^{-c k-\epsilon} d\omega\, \left (\frac{1}{ck-\omega} + \frac{1}{c k+\omega}\right )e^{-i \tau \omega} + i \epsilon \int_{\pi}^{2 \pi} d\phi \, e^{i \phi}\, \left (\frac{1}{2 ck-\epsilon e^{i \phi}} + \frac{1}{\epsilon e^{i \phi}}\right ) e^{-i \tau (-c k+\epsilon  e^{i \phi})} + \\ \int_{-c k+\epsilon}^{c k - \epsilon} d\omega\, \left (\frac{1}{ck-\omega} + \frac{1}{c k+\omega}\right )e^{-i \tau \omega}+ i \epsilon \int_{\pi}^{2 \pi} d\phi \, e^{i \phi}\, \left (\frac{1}{2 ck+\epsilon e^{i \phi}} - \frac{1}{\epsilon e^{i \phi}}\right ) e^{-i \tau (c k+\epsilon  e^{i \phi})}\\+ \int_{c k+\epsilon}^R d\omega\, \left (\frac{1}{ck-\omega} + \frac{1}{c k+\omega}\right )e^{-i \tau \omega} + i R \int_{2 \pi}^{\pi} d\theta \, e^{i \theta} \, \left (\frac{1}{ck-R e^{i \theta}} + \frac{1}{ck+R e^{i \theta}}\right ) e^{-i \tau R e^{i \theta}} $$
Now, note that the sixth integral vanishes for $\tau \gt 0$ because $\sin{\theta} \lt 0$ over the integration interval.  Also note how the limits of integration have also changed for the second and fourth integrals, to reflect the change in direction of integration.  Going through the same motions as above, therefore, we find that
$$PV \int_{-\infty}^{\infty} d\omega\, \left (\frac{1}{ck-\omega} + \frac{1}{c k+\omega}\right )e^{-i \tau \omega} = -i \pi \left (e^{i c k \tau} - e^{-i c k \tau}  \right ) \quad (\tau \gt 0)$$
as I think you expected. 
