expressions , and summing over $\{\lambda\}$ we obtain for $\tp(p)$ the expression $$\eqalign{\tp(p)
&=-{2\over p}\left[{p^3\over2\pi^2}\zeta(3)
+{p^2\over\pi^2}\int_{0}^{\pi\over p}dz\, z^2\cot(z)
+{p^2\over\pi^2}\int_{0}^{(p-1)\pi\over p}
dz\, z(z-{(p-2)\pi\over p})\cot(z)\right.\cr
&\qquad\left.+2\sum_{l=1}^{(p-3)\over2}
{p^2\over\pi^2}\int_{0}^{2l\pi\over p}dz\, z(z-{2l\pi\over p})\cot(z)
\right]\cr
}\no$$ for $p$ odd. But the expression $\tp(p)$ above is $2\ln\Det\dstar d_0 $: i.e. twice the logarithm of the Laplacian on 0-forms for the lens space $L(p)$. More precisely our analytic continuation has shown us that $$\eqalign{\ln\Det\dstar d_0
&={1\over p}\left[{p^3\over2\pi^2}\zeta(3)
+{p^2\over\pi^2}\int_{0}^{\pi\over p}dz z^2\cot(z)
+{p^2\over\pi^2}\int_{0}^{(p-1)\pi\over p}dz
z(z-{(p-2)\pi\over p})\cot(z)\right.\cr
&\qquad\left.+2\sum_{l=1}^{(p-3)\over2}
{p^2\over\pi^2}\int_{0}^{2l\pi\over p}dz z(z-{2l\pi\over p})\cot(z)
\right]\cr
}\no$$ Let us now turn to $\tau_{-}(p,s)$ for $p$ odd. $$\tau_{-}(p,s)=\sum_{n=1}^{\infty}
{n\left<\chi^{(n+1)\over2}\right>_{p}
+(n+2)\left<\chi^{(n-1)\over2}\right>_{p}\over {(n+1)}^{2s}}\no$$ which on decomposing $n$ over the conjugacy classes mod $p$ $n=pk-j,\quad j=0,\dots,(p-1)$ as for $\tau_{+}(p,s)$, distinguishing the two parities as in yields $$\eqalign{\tau_{-}(p,s)
&=\sum_{k=1}^{\infty}{kp{\left\langle\chi^{kp+1\over2}\right\rangle}_{p}
+(kp+2){\left\langle\chi^{kp-1\over2}\right\rangle}_{p}\over{(kp+1)}^{2s}}
+\sum_{k=1}^{\infty}{(kp-2){\left\langle\chi^{kp-1\over2}\right\rangle}_{p}
+kp{\left\langle\chi^{kp-3\over2}\right\rangle}_{p}\over{(kp-1)}^{2s}}\cr
&\qquad+\sum_{k=1}^{\infty}\sum_{l=2}^{(p-3)\over2}
{(kp-2l){\left\langle\chi^{kp-2l+1\over2}\right\rangle}_{p}
%% FOLLOWING LINE CANNOT BE BROKEN BEFORE 80 CHAR
+(kp-2l+2){\left\langle\chi^{kp-2l-1\over2}\right\rangle}_{p}\over{(kp-2l+1)}^{2s}}\cr
%% FOLLOWING LINE CANNOT BE BROKEN BEFORE 80 CHAR
&\qquad+\sum_{k=1}^{\infty}{(kp-p+1){\left\langle\chi^{kp-p\over2}\right\rangle}_{p}
%% FOLLOWING LINE CANNOT BE BROKEN BEFORE 80 CHAR
+(kp-p+3){\left\langle\chi^{kp-p-2\over2}\right\rangle}_{p}\over{(kp-p+2)}^{2s}}\cr
&\qquad+\sum_{k=1}^{\infty}\sum_{l=0}^{r-1}
{(kp-2l-1){\left\langle\chi^{kp-2l\over2}\right\rangle}_{p}
%% FOLLOWING LINE CANNOT BE BROKEN BEFORE 80 CHAR
+(kp-2l+1){\left\langle\chi^{kp-2l-2\over2}\right\rangle}_{p}\over{(kp-2l)}^{2s}}\cr
}\no$$ Using our degeneracy formula we have with a little re-arrangement $$\eqalign{\tau_{-}(p,s)
&=\sum_{k=1}^{\infty}{2(kp+1)k+kp\over{(kp+1)}^{2s}}
+\sum_{k=1}^{\infty}{2(kp-1)k-kp\over{(kp-1)}^{2s}}\cr
&\qquad+\sum_{l=-{(p-3)\over2}}^{(p-3)\over2}
\sum_{k=1}^{\infty}{2(kp+2l)k\over{(kp+2l)}^{2s}}\cr
}\no$$ We now observe that this is of the form $$\eqalign{\tau_{-}(p,s)
&={2\over p}\sum_{\{\nu\}}\sum_{k=1}^{\infty}
{kp(kp+\nu)\over{(kp+\nu)}^{2s}}\cr
&+\sum_{k=1}^{\infty}{kp\over{(kp+1)}^{2s}}
-\sum_{k=1}^{\infty}{kp\over{(kp-1)}^{2s}}\cr
}\no$$ 