Scattering Greens function exactly at energy of bound state I have a small bit of confusion about the expansion I am seeing in literature for the Greens function in time independent scattering theory. For example here is an excerpt from Scattering Theory of Waves and Particles by R.G. Newton:

My question is quite simple. Why in equation 7.26 is there no $\pm i\epsilon$ in the denominator in the first term, i.e. the discrete sum over bound states. I understand that for most energies the $\epsilon$ in this term may be simply ignored, but what if $E$ is exactly equal to some $E_n$, then we have a pole at this energy right? So does this equation have some implicit assumption that we do not take $E$ exactly at the bound state energy? Or is there some other reason?
 A: The Green's function $G$ is defined as the solution to
\begin{align}
(E-H)G(E)=1.
\end{align}
You are correct that the Green's function is not well-defined when evaluated at the eigen-energies of the Hamiltonian. However, for a energies corresponding to the continuous part of the Hamiltonian ($E'$ in the textbook notation) we can hope to get around this by defining $G(E')$ via a limiting procedure: we could approach $E'$ from above (i.e., $E+i\epsilon$) or from below (i.e., $E-i \epsilon$), which is defined in Eq. (7.26). (Typically, this process can be understood as taking the analytic continuation of $G$ across the branch cut created by the continuous part of the spectrum.)
Your question relates more to the discrete eigenstates with energy $E_n$. For these, such a limiting procedure does not work and the divergence of the Green's function at these energies is unavoidable. So yes, we cannot "take $E$ exactly at the bound state energy". I suspect that this is why the author has left out the $\pm i \epsilon$. However, there are still benefits of leaving in the $\pm i \epsilon$, which other textbooks use. For example, the density of states can be written as
\begin{align}
N(E)=- \frac{1}{\pi} \Im \, \text{Tr}\, G(E+i\epsilon),
\end{align}
which will yield Dirac deltas at $E=E_n$ (corresponding to the contribution of a discrete state).
Hope this helps.
A: Two different situations are possible here, depending on whether one deals with eigenstates or a scattering problem. The former seems to be the case here - i.e., the spectrum of a Hamiltonian contains both bound and extended states. The examples of this a ubiquitous - a hydrogen atom has bound states for energies $E<0$ and extended for $E>0$. An easily solvable toy example is a negative delta-function potential. Since these are eigenstates of a Hamiltonian, there is no mixing between them (i.e. their eigenfunctions are mutually orthogonal). This would also hold, if a bound state lies within a continuum spectrum, although it is hard to come up with a realistic example.
The situation becomes more interesting in case of scattering - e.g., a particle scattered by an atom or an electron tunneling through or near the quantum dot. In such a scattering problem we start with the states of an isolated atom/quantum dot, but neither is a eigenstate of the joint Hamiltonian. The scattering amplitude in such cases will have resonances associated with the scattering near the bound states. In quantum dots this leads to distinct resonant tunneling and Fano resonance, depending on whether electron tunnels through the level or passes near it. Once interactions are incorporated it may result in even richer physics - e.g., Kondo effect can be viewed as a resonant tunneling/scattering of an electron pair.
