In quantum mechanical scattering theory, we often use Green's functions which contain poles. For example, in Schroedinger quantum mechanics the free Green's function is given by
$$ G_0(\vec{p}) = \frac{1}{E-\frac{p^2}{2m}+i\epsilon} $$
in momentum space where the infinitesimal constant $\epsilon >0$ has been introduced to take care of the pole at $E = \frac{p^2}{2m}$. The imaginary part of $G_0$ is then
$$ \text{Im}G_0 = -\frac{\epsilon}{(E-\frac{p^2}{2m})^2+\epsilon^2} $$
and letting $\epsilon$ go to zero, we get (using the Sokhotski–Plemelj formula)
$$ \lim_{\epsilon\to0}\text{Im}\,G_0=-\pi\delta\left(E-\frac{p^2}{2m}\right). $$
The full Green's function is then given by the Dyson equation
$$ G = G_0 + G_0 VG_0 + G_0VG_0VG_0+\ldots $$
with the scattering potential $V$. Looking at the second term in the Dyson equation, we see that the free Green's function appears twice giving rise to an expression proportional to $\frac{1}{(E-p^2/(2m)+i\epsilon)^2}$. I wonder what the imaginary part of this expression is. Phyiscally, there should still be some delta function because the physical particle fulfills the relation $E=\frac{p^2}{2m}$ even after elastic scattering but I don't see how the delta enters the game. So, how can I get a delta function from the fraction
$$ \frac{1}{(E-\frac{p^2}{2m}+i\epsilon)^2}? $$