$\vep=0$.

Now we prove by induction on $k\geq 1$. Let $k=1$. Then $\vec s=s\in \ZZ$. The case when $s\leq 0$ follows from Eq. () with $i=1$. When $s>0$, we note that $Z'(\wvec{s}{r};\vep)=rZ(\wvec{s-1}{r};\vep)$. By Eq. (), $Z(\wvec{0}{r};\vep)$ is in $\CC\{\{\vep,\vep^{-1}\}[\ln(-\vep)]$ which is closed under integration (Lemma ). Thus $Z(\wvec{1}{r};\vep)$ is in $\CC\{\{\vep,\vep^{-1}\}[\ln(-\vep)]$ and, by an induction on $s$, the same holds for $Z(\wvec{s}{r};\vep)$ for any $s>0$.

Assume that the statements hold for $k\geq 1$ and prove for $Z(\wvec{\vec s}{\vec r}; \vep)$ with $\vec s=(s_1,\cdots,s_{k+1})$ in two cases.

[**Case 1.** ]{} Suppose $s_i\leq 0$ for some $1\leq i\leq k+1$. Then for fixed $n_{i-1}>n_{i+2}+1>0$, $$\begin{aligned}
\lefteqn{
\sum_{n_{i-1}>n_i>n_{i+1}>n_{i+2}}
\frac{e^{n_ir_i\vep+n_{i+1}r_{i+1}\vep}}{n_i^{s_i}n_{i+1}^{s_{i+1}}}}\\
&=& \sum_{n_{i-1}>n_i\geq n_{i+1}>n_{i+2}}
\frac{e^{n_ir_i\vep+n_{i+1}r_{i+1}\vep}}{n_i^{s_i}n_{i+1}^{s_{i+1}}}
-\sum_{n_{i-1}> n_i=n_{i+1}>n_{i+2}}
\frac{e^{n_ir_i\vep+n_{i+1}r_{i+1}\vep}}{n_i^{s_i}n_{i+1}^{s_{i+1}}}\\
&=& \sum_{n_{i-1}>n_{i+1}>n_{i+2}}
\frac{e^{n_{i+1}r_{i+1}\vep}}{n_{i+1}^{s_{i+1}}}
\sum_{n_i=n_{i+1}}^{n_{i-1}-1}\frac{e^{n_ir_i\vep}}{n_i^{s_i}}
-\sum_{n_{i-1}> n_{i+1}>n_{i+2}}
\frac{e^{n_{i+1}(r_i+r_{i+1})\vep}}{n_{i+1}^{s_i+s_{i+1}}}.\end{aligned}$$ Applying Eq. () to the inner sum of the first term, we have Thus Then we have The induction hypothesis applies to each term on the right hand side, completing the induction on $k$ in this case. In particular this completes the induction when $\vec{s}\in \ZZ^{k+1}_{\leq 0}$.

[**Case 2.**]{} Suppose $s_i>0$ for all $1\leq i\leq k+1$. We use induction on the sum $s:=\sum_{i=1}^{k+1} s_i$. Then $s\geq
k+1$. If $s=k+1$, then $s_i=1$ for $1\leq i\leq k+1$. Note that $$\frac {d}{d\vep } Z(\wvec{\vec 