H = H_V + v \delta(x)$$ where $$\label{example2}
H_V = \frac{p^2}{2} + V(x).$$ It is well-known that the Euclidean propagator $G[x_1, x_2; t]$ for $H$ obeys the following integral equation $$\label{inteq}
G[x_1, x_2; t] = G_V[x_1, x_2: t] - v \int_0^t ds \int dx
G_V[x_1, x; t-s] \delta(x) G[x, x_2; s]$$ where $G_V[x_1, x_2: t]$ is an Euclidean propagator for $H_V$. The Euclidean propagator $G[x_1, x_2; t]$ is related to the usual Feynman propagator(or Kernel) $K[x_1, x_2; t]$ as follows; $$\label{kerneldef}
K[x_1, x_2; t] = G[x_1, x_2; it].$$ Taking a Laplace transform $$\label{lapdef}
\hat{f} \equiv {\cal L} f(t) \equiv \int_0^{\infty} dt e^{-Et} f(t)$$ to both sides of Eq.(\[inteq\]) yields $$\label{example3}
\hat{G}[x_1, x_2; E] = \hat{G}_V[x_1, x_2; E] - v \hat{G}_V[x_1, 0; E]
\hat{G}[0, x_2; E]$$ which supports a solution $$\label{example4}
\hat{G}[0, x_2; E] = \frac{\hat{G}_V[0, x_2; E]}{1 + v \hat{G}_V[0, 0, E]}.$$ Inserting Eq.(\[example4\]) into Eq.(\[example3\]) again completes Schulman’s procedure; $$\label{schulman}
\hat{G}[x_1, x_2; E] = \hat{G}_V[x_1, x_2; E] - 
\frac{\hat{G}_V[x_1, 0; E] \hat{G}_V[0, x_2; E]}
     {\frac{1}{v} + \hat{G}_V[0, 0; E]}.$$ The usual energy-dependent Green’s function $\hat{K}[x_1, x_2; E]$ which is a Fourier transform of $K[x_1, x_2; t] \theta(t)$, where $\theta(t)$ is a step function, is also evaluated from the corresponding 