= Solution
The delta function in the specified <convolution> kernel collapses the $\mathbf k'$ integral to $-i\gamma G$. Set $\lambda=e\gamma$ and $\omega_0^2=k^2+m^2$. Then
$$
G(\mathbf k;\omega)=\frac1{-\omega^2-2i\lambda\omega+\omega_0^2},\qquad G(\mathbf k;\tau)=\int_{-\infty}^{\infty}\frac{d\omega}{2\pi}\frac{e^{-i\omega\tau}}{-\omega^2-2i\lambda\omega+\omega_0^2}.
$$
There is a change of assumptions here: the specified kernel is the transform of \b[$V=-i\gamma$], which is not real. These later parts define a formal <analytic continuation> to an effective damped equation, $[\partial_\tau^2+2\lambda\partial_\tau+\omega_0^2]G=\delta(\tau)$; they cannot simultaneously describe the earlier real external potential. The following <pole> analysis uses that continued operator.
Completing the square gives $D(\omega)=-(\omega+i\lambda)^2+(\omega_0^2-\lambda^2)$, so
$$
\boxed{\omega_\pm=-i\lambda\pm\sqrt{k^2+m^2-\lambda^2}.}
$$
For $e>0$, hence $\lambda>0$, the underdamped regime has two <poles> at $\pm\Omega-i\lambda$, with $\Omega=\sqrt{k^2+m^2-\lambda^2}>0$. Both lie below the <real axis>. In the overdamped regime they lie on the <imaginary axis> at $-i(\lambda\mp\kappa)$, where $\kappa=\sqrt{\lambda^2-k^2-m^2}<\lambda$ for $k^2+m^2>0$; both are again below the <real axis>. At equality they coalesce into a <double pole> at $-i\lambda$. If $m<|\lambda|$, the crossover occurs at $k_c=\sqrt{\lambda^2-m^2}$; increasing $k$ moves the two imaginary <poles> together and then apart horizontally. If $m>|\lambda|$, every $k$ is underdamped; if equal, only $k=0$ is critical.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2016/paper-1-green-function-poles.png]
{title=Underdamped, critically damped and overdamped pole positions for positive damping}
The sign of $e$ is not specified: for $e<0$ all nonzero-frequency <poles> are instead in the <upper half-plane>, describing <antidamping>; for $e=0$ they lie on the <real axis> and a causal prescription is required. When $m=k=0$, one <pole> is at zero even for nonzero $\lambda$ and a real-axis prescription is again needed. The figure uses illustrative $\lambda=1$, not supplied numerical data.
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