for $p$ odd we get $$\eqalign{\tau_+(p,s)
=\sum_{k=1}^\infty\left[{2pk^2\over\left\{(pk-1)(pk+1)\right\}^s}\right.
              & +\sum_{l=1}^{r-1}{2(k-1)(pk-2l)\over\left\{(pk-2l-1)(pk-
                                                       2l+1)\right\}^s}\cr
             &+\sum_{l=0}^r\left.{2k(pk-2l+1)\over\left\{(pk-2l)(pk-
                                            2l+2)\right\}^s}\right]\cr}\no$$ To aid in marshalling the combinatorics of $\tau_+(p,s)$ we define $H_{p}(k,s,x)$ by $$H_{p}(k,s,x)={pk(pk+x)
\over\left\{(pk+x-1)(pk+x+1)\right\}^s}\no$$ The point being that each of the three summands in is of the form $H_{p}(k,s,\lambda)$ for appropriate $\lambda$. To see this we introduce precisely $p$ constants of the type $\lambda$ defined by $$\cases{\lambda_0=0&\cr
         \lambda_l=-2l+1,\;\,l=0,\ldots,r&\cr
          \bar\lambda_l=-2l+p,\;\,l=1,\ldots,(r-1)&\cr}\no$$ With this notation it can be checked that $\tau_+(p,s)$ is given by $$\tau_+(p,s)={2\over p}\sum_{k=1}^\infty\left[H_{p}(k,s,\lambda_0)
                          +\sum_{l=1}^{(r-1)}H_{p}(k,s,\bar\lambda_l)
                           +\sum_{l=0}^{r}H_{p}(k,s,\lambda_l)\right]\no$$ Also if we denote the entire set of $\lambda$’s by $\{\lambda\}$ i.e. $$\{\lambda\}\equiv\{\lambda_0,\lambda_l,\bar\lambda_l\}=\{-(p-2),
\ldots,-5,-3,-1,0,1,3,5\ldots(p-2)\}\no$$ then we have the even more concise expression $$\tau_+(p,s)={2\over p}\sum_{\{\lambda\}}\sum_{k=1}^\infty
H_{p}(k,\lambda,s)\no$$ The functions $H_{p}(\lambda,s)$ obtained by summing over $k$ then form a set of $p$ functions whose derivative at $s=0$ can be viewed as living on the appropriate space of sections for the Laplacian acting on $0$ forms on the lens space $L(p)$; taking the trace over these functions viewed as forming a matrix then gives the analytic continuation of the determinant of the Laplacian.

Next observing $$H_{p}(k,x,s)
={{(pk+x)}^2\over\{{(pk+x)}^2-1\}^s}
-x{(pk+x)\over\{{(pk+x)}^2-1\}^s}
\no$$ we see that there are therefore two additional functions of interest here i.e. $$A_{p}(k,x,s)={{(pk+x)}^2\over{\left\{{(pk+x)}^2-1\right\}}^s}\no$$ and $$B_{p}(k,x,s)={(pk+x)\over{\left\{{(pk+x)}^2-1\right\}}^s}\no$$ and we have $$H_{p}(k,x,s)=A_{p}(k,x,s)-xB_{p}(k,x,s)\no$$ and we can equally write $$\tau_{+}(p,s)
={2\over p}\sum_{\{\lambda\}}\left[A_{p}(\lambda,s)
