we introduce the infinitesimal positive constant $\epsilon$ as follows; $$\begin{aligned}
\label{toy2ddepsi}
\hat{G}_0^{Box}[0, 0; E]&\rightarrow&\hat{G}_0^{Box}[\epsilon^-, \epsilon, E]
= 2 \epsilon - 2 \sqrt{2E} \coth \sqrt{2E} L \epsilon^2 + 
{\cal O} (\epsilon^3)
                                         \\   \nonumber
\hat{G}_0^{Box}[L, L; E]&\rightarrow&\hat{G}_0^{Box}[L-\epsilon^-, L-\epsilon, E]
= 2 \epsilon - 2 \sqrt{2E} \coth \sqrt{2E} L \epsilon^2 + 
{\cal O} (\epsilon^3)
                                         \\   \nonumber
\hat{G}_0^{Box}[0, L; E]&\rightarrow&\hat{G}_0^{Box}[\epsilon, L-\epsilon, E] =
\frac{2 \sqrt{2E}}{\sinh \sqrt{2E} L} \epsilon^2 + {\cal O} (\epsilon^3)
                                          \\   \nonumber
\hat{G}_0^{Box}[L, 0; E]&\rightarrow&\hat{G}_0^{Box}[L-\epsilon, \epsilon, E] =
\frac{2 \sqrt{2E}}{\sinh \sqrt{2E} L} \epsilon^2 + {\cal O} (\epsilon^3)
                                          \\   \nonumber
\hat{G}_0^{Box}[a, 0; E]&\rightarrow&\hat{G}_0^{Box}[a, \epsilon, E] =
\frac{2 \sinh \sqrt{2E} (L-a)}{\sinh \sqrt{2E} L} \epsilon + {\cal O} (\epsilon^3)
                                          \\   \nonumber
\hat{G}_0^{Box}[a, L; E]&\rightarrow&\hat{G}_0^{Box}[a, L-\epsilon, E] =
\frac{2 \sinh \sqrt{2E} a}{\sinh \sqrt{2E} L} \epsilon + {\cal O} (\epsilon^3)
                                          \\   \nonumber
\hat{G}_0^{Box}[0, b: E]&\rightarrow&\hat{G}_0^{Box}[\epsilon, b; E] =
\frac{2 \sinh \sqrt{2E} (L-b)}{\sinh \sqrt{2E} L} \epsilon + {\cal O} (\epsilon^3)
                                          \\   \nonumber
\hat{G}_0^{Box}[L, b; E]&\rightarrow&\hat{G}_0^{Box}[L-\epsilon, b; E] =
\frac{2 \sinh \sqrt{2E} b}{\sinh \sqrt{2E} L} \epsilon + {\cal O} (\epsilon^3).\end{aligned}$$ Inserting Eq.(\[toy2ddepsi\]) into Eq.(\[toy2schul2\]) and Eq.(\[toy2schul1\]), and defining the renormalized constants $$\begin{aligned}
\label{toy2rencoup}
v_1^{ren}&=&\frac{1}{2\epsilon^2}
\left( \frac{1}{v_1} - 2\epsilon \right)    \\   \nonumber
v_2^{ren}&=&\frac{1}{2\epsilon^2}
\left( \frac{1}{v_2} + 2\epsilon \right),\end{aligned}$$ one can arrive at the following long expression after tedius calculation; $$\begin{aligned}
