Renormalising $\Delta$ baryon mass in chiral effective field theory I have essentially no experience of quantum field theory, other than a superficial knowledge of some basic ideas - my apologies if I've phrased anything unusually or made any mistakes in my question here.
For context, I've been looking at matrix methods in quantum mechanics, and I'm trying to understand a paper on a matrix model for $\Delta\to N\pi$ decay that uses chiral effective field theory to construct the Hamiltonian:
https://arxiv.org/abs/1303.4157
In the paper, the bare mass $\Delta_0$ of the $\Delta$ is adjusted to account for the $\Delta\to N\pi$ interaction. This is done by:
$$\Delta_0=E_\text{res}-\Sigma_{\Delta N}(k_\text{res})$$
where $E_\text{res}$ is the phenomenological value of the $\Delta$ resonance, $k_\text{res}$ is the corresponding pion momentum $E_\text{res}=\sqrt{k_\text{res}^2+m_\pi^2}$, and $\Sigma_{\Delta N}$ is given by the integral:
$$\Sigma_{\Delta N}(k)=\int_0^\infty dk'\frac{k'^2g_{\Delta N}^2(k')}{\omega_\pi(k')\left[\omega_\pi(k)-\omega_\pi(k')-i\epsilon\right]}$$
where $\omega_\pi(k)=\sqrt{k^2+m_\pi^2}$ is the pion energy and $g_{\Delta N}^2$ is the $\Delta- N\pi$ coupling (which I understand to be a sort of potential energy). It is said to be related to this diagram:

My question is essentially: how does one construct these integrals from these diagrams? For example, I am in particular interested in how to construct the integral for the self-energy associated with $\Delta\to\Delta \pi$ interaction, with the diagram:

 A: Although the paper mostly deals with Quenched QCD, Young et al. (https://arxiv.org/abs/hep-lat/0205017) does have some discussion of the Sigma Commutator in full QCD which is relevant here. 
In the paper you referenced, they make use of a $\chi_{\Delta}$ in the $g_{\Delta N}^2$ term. Using the notation from Young et al., $\chi_{\Delta}$ from Hall et al. can be written as
$$\chi_{\Delta N} = \frac{3}{32\pi f_{\pi}^2}G_{\Delta N}\,,$$
where $G_{\Delta N} = \frac{2}{9}\mathcal C^2$ (Young uses $G_{\Delta N} = \frac{8}{25}(F+D)^2$, but that can be reduced to $\frac{2}{9}\mathcal C^2$ using the approximation $F = \frac{2}{3}D$). 
To find the form of the Sigma Commutator for the $\Delta\to\Delta+\pi$ interaction then, you just have to use $\chi_{\Delta\Delta}$ instead, with $G_{\Delta\Delta} = (F+D)^2$. This then gives you an expression for $g_{\Delta\Delta}^2$, which can be used to find $\Sigma_{\Delta\Delta}(k)$.
For further discussion on the Sigma Commutator in a bit more general terms, you may want to try https://arxiv.org/abs/nucl-th/0005003. 
