parameter $\xi_1$ represents the type of BC at $x = 0$. Then, the fixed-energy amplitude $\hat{G}_0^{Box}$ can be computed by introducing an infinite barrier at $x = L$ to the half-line constraint system. As we commented in section 2 the infinite barrier is introduced by $\delta$- and $\delta^{\prime}$-functions potential at $x = L$ with assumption that the coupling constant is infinite. Thus the final form of $\hat{G}_0^{Box}$ is dependent on the two parameters as follows; $$\label{toy2box1}
\hat{G}_0^{Box}[a, b; E] = \xi_2 \hat{G}_0^{\xi_1, N}[a, b; E]
+ (1 - \xi_2) \hat{G}_0^{\xi_1, D}[a, b; E]$$ where $$\begin{aligned}
\label{toy2gdgn}
\hat{G}_0^{\xi_1, D}[a, b; E]&=&\hat{G}_F^{\xi_1}[a, b; E] - 
\frac{\hat{G}_F^{\xi_1}[a, L; E] \hat{G}_F^{\xi_1}[L, b; E]}
     {\hat{G}_F^{\xi_1}[L, L^-; E]}
                                        \\   \nonumber
\hat{G}_0^{\xi_1, N}[a, b; E]&=&\hat{G}_F^{\xi_1}[a, b; E] -
\frac{\hat{G}_{F,b}^{\xi_1}[a, L; E] \hat{G}_{F, a}^{\xi_1}[L, b; E]}
     {\hat{G}_{F, ab}^{\xi_1}[L, L^-; E]}.\end{aligned}$$ Of course the parameter $\xi_2$ in Eq.(\[toy2box1\]) parametrizes the various BCs at $x = L$.

Explicit calculation shows $$\begin{aligned}
\label{toy2explicit}
\hat{G}_0^{\xi_1, D}[a, b; E]&=&
\frac{1}
     {\sqrt{2E} [\xi_1 \cosh \sqrt{2E} L + (1 - \xi_1) \sinh \sqrt{2E} L]}
                                                         \\   \nonumber
& & \times
\Bigg[ \xi_1 \bigg\{\sinh \sqrt{2E} (L - |a - b|) - \sinh \sqrt{2E} ((a + b) 
                                                             - L) \bigg\}
                                                   \\  \nonumber
& & \hspace{1.0cm}
      + (1 - 