more specific information about integrals $J(g)\in$ . The following is true.

\[lem:jgbehave\] If $m = n$, and $n \geq N^*$ ($N^*$ as defined in ), $$\begin{aligned}
J(g) = 0 \quad \text{ if at least one } \quad g (\tilde{t}\,  ) = n, \label{eq:jequalzero} \\
\intertext{and}
J(g) = {\left(  \prod_{t = 1}^{j-1}2 (n - g_t) \right)}^{-1} \quad \text{otherwise} \label{eq:jequalnonzero}.\end{aligned}$$ If $m \neq n$, $$J(g) = 0 \,\text{ if }\,\# \tau(g) \neq 2, \, \, \text{where $\tau(g) = {\left\lbrace t: g_t = n \right\rbrace}$,} \label{eq:jnonequalzero}$$ and $$J(g) = \frac{1}{4(n-m)^2}{\left(  \prod_{t \not\in \tau(g)}^{j-1}2 (n - g_{t}) \right)}^{-1}  \text{ if } \# \tau(g) = 2. \label{eq:jnonequalnonzero}$$

The integrand of could have a pole inside of $\mathcal{D}_n$ only at $z_n = 2n + 1$. In the cases and the pole’s order $\geq 2$ or $F_g(z)$ is analytic on $\overline{\mathcal{D}_n}$, so $J(g) = 0$. In the cases , the pole’s order is one and $J(g)$ is the residue of $F_g(z)$ at $z_n$.

An odd potential v-super-o  {#subsect:oddpot}
---------------------------

As it is noticed in , $$\begin{gathered}
\begin{aligned}
v^0_{jk} &= {\left\langle ({\delta \left( x-b \right) } - {\delta \left( x + b \right) }) h_j, h_k \right\rangle} \\
 & = [1 - (-1)^{j + k}] a_j a_k
\end{aligned} 