, we aim to isolate the $\delta$-function contribution to the pole arising from the imaginary part. We define a pole-subtracted two-body scattering amplitude $\Delta \cM_2$ by
   
    \begin{align}
    \label{eq:delta-m-2}
    \Delta \cM_{2}(s_{2k'};\epsilon)
    &\equiv
    \cM_2(s_{sk'};\epsilon)
    - g^2\, i\pi \delta_\epsilon (s_{2k'}-s_b) \,,
    \end{align}
   
where we introduced an $\epsilon$-dependent delta function, which we denote $\delta_{\epsilon}$.
It is defined as
   
    \begin{align}
    \delta_\epsilon (s_{2k'}-s_b)&=\frac{\epsilon}{\pi \left((s_{2k'}-s_b)^2 + \epsilon^2\right)},
    \end{align}
   
and reproduces the Dirac delta function for $\epsilon \to 0$. It is straightforward to check that with this definition of $\delta_\epsilon$, $\Delta \cM_{2,\epsilon}$ is finite at the bound state pole and it contains the same branch cut as $\cM_2$. The $\epsilon$-regulated delta is necessary since $\cM_2$ only supports the Dirac delta function form for the imaginary part at the bound state pole.

With the definition of Eq.~\eqref{eq:delta-m-2}, we can isolate the pole in $\cM_2$ by writing it as
   
    \begin{align}
    \cM_2(s_{2k'}) 
    &= g^2\, i\pi \delta(s_{2k'} - s_b) + 
    \lim_{\epsilon\to 0}\Delta \cM_{2}(s_{2k'};\epsilon).
    \end{align}
   
Given that the pole is evaluated in terms of $s_{2k}$, we perform the change of variables in the integral with the pole,
   
    \begin{align}
    \label{eq:SA.jacobian}
    \int_{0}^{k_{\mathrm{max}}} \! \! \frac{\mathrm{d}k \, k^2}{(2 \pi)^2\, \omega_{k}}
    =
    \int_{0}^{(E-m)^2} \!\! \frac{ \mathrm{d} s_{2k'} }{16\pi^2 E^2} \,  \lambda^{1/2}(E^2, s_{2k'}, m^2) \,.
    \end{align}
   
With these identities, we proceed to modify the integral equation for $d$, Eq.~\eqref{eq:d_Sproj}, arriving to
   
    \begin{align}
    \label{eq:Mphib_eqv2}
    d^{(u,u)}_{S}(  p,   k; \epsilon) 
    &=
    - {G}_{S}( p,  k;\epsilon) - i g^2 \, {G}_{S}(  p,   q;\epsilon) \, \rho_{\varphi b} (E) \, d^{(u,u)}_{S}( q,  k; \epsilon) \nn\\[5pt] & \hspace{3cm} -\int \! \! \frac{\mathrm{d} k' \, k'^{2}}{(2 \pi)^2\, \omega_{k'}} \, {G}_{S}(  p,   k';\epsilon) \, { \Delta  {\cM}_2( k' ;\epsilon)} \, d^{(u,u)}_{S}( k',  k; \epsilon) \,,
    \end{align}
   
where in the last line we have used Eq,~\eqref{eq:qbs} and \eqref{eq:2p1.phase_space} after we integrated over the delta function with Eq.~\eqref{eq:SA.jacobian}. In the resulting integral equation, the remaining integral has at worst logarithmic singularities due to the exchange propagator.

Just as for the BF method, we introduce a uniform mesh to write this as,
   
    \begin{align}
    \label{eq:d_semi_Neps_eq}
    d^{(u,u)}_{S}(  p,   k;\epsilon, N) 
    &=
    - {G}_{S}( p,  k;\epsilon) 
    - i g^2 \, {G}_{S}(  p,   q;\epsilon)
    \, \rho_{\varphi b} (E)
    \, d^{(u,u)}_{S}( q,  k;\epsilon, N)
    \nn\\[5pt]
    & \hspace{3cm}
    - \sum^{N-1}_{n=0} \frac{ \Delta k' \,
    k_{n'}^2}{ (2\pi)^2 \, \omega_{k_{n'}}}
    \, G_{S}(  p, k'_n;\epsilon)
    \Delta \cM_2( k'_n ;\epsilon)
    \, d^{(u,u)}_{S}( k'_n, k;\epsilon, N) \,.
    \end{align}
   
This can then be solved in general by following a two-step process. First, solve for $d_S^{(u,u)}$ when $p=q$. Having this, one then readily insert this back into the second term of the right-hand side of the equation and solve for $d_S^{(u,u)}$ for arbitrary values of $p$. We proceed by introducing 
   
    \begin{align}
    \Delta B_{nn'}
    &=
    \delta_{k_{n}, k_{n'}}
    + \frac{ \Delta k' \,
    k_{n'}^2}{ (2\pi)^2 \, \omega_{k_{n'}}}  \, G_{S}(k_{n}, k_{n'}; \epsilon)
    \, \Delta \cM_2( k_{n'}; \epsilon) \,.
    \label{eq:dB}
    \end{align}
   
As before, this allows us to write Eq.~\eqref{eq:d_semi_Neps_eq} as a matrix equation,
   
    \begin{align}
    \label{eq:d_semi_mat}
    \left[\Delta B\,
    d^{(u,u)}_{S}\right]_{   nn'} 
    &=
    - {G}_{S, nn'}
    - i g^2 {G}_{S}(  k_n,   q;\epsilon)
    \, \rho_{\varphi b} (E) \,
    d^{(u,u)}_{S}( q,  k_{n'};\epsilon, N).
    \end{align}
   
By multiplying both sides by the inverse of $\Delta B$, proceeding to set $k_n=q$, we arrive at an algebraic equation for $d^{(u,u)}_{S}( q,  k_{n'},\epsilon, N)$, whose solution is
   
    \begin{align}
    \label{eq:eq:d_semi_qsol}
    g^2 d^{(u,u)}_{S}( q,  k_{n'};\epsilon, N)
    &=
    \frac{ \left[{\cK}_{\varphi b}\right]_{nn'}
    }{ 1- i  \left[ {\cK}_{\varphi b} \,  \rho_{\varphi b} \right]_{nn''} }
    \bigg|_{k_n=k_{n''}=q} \,
    \end{align}
   
where 
   
    \begin{align}
    {\cK}_{\varphi b,   nn'} 
    \equiv - g^2 \left[\Delta B^{-1} {G}_{S} \right]_{nn'},
    \end{align}
   
is an off-shell extension of the $K$ matrix of the $\varphi b$ system. In other words, when $k_n,k_{n'}=q$ this coincides with the physical $\varphi b$ $K$ matrix as defined in Eq.~\eqref{eq:2p1.Kmat}. In principle, $\cK_{\varphi b}$ must be a real function below the three-particle threshold. However, since we work with finite $N$, $\cK_{\varphi b}$ may be complex. The level of complexity is measured by the same $\Delta\rho_{\varphi b}$ as defined in Eq.~\eqref{eq:Deltarho}. As in the case with the BF method, one must then take the ordered, double limit defined Eq.~\eqref{eq:double_limit}. 


In Sec.~\ref{sec:numerics} we showed results for the $\varphi b$ amplitude as computed using the SA method. Both the BF and SA methods reliably recover the amplitude at the various $N$ we tested which were consistent with each other. We found that the SA method for most kinematics can yield results that have a unitarity deviation an order of magnitude lower than the BF method. Since