where we have denoted by $\{\nu\}$ the set $$\{\nu\}\equiv\{\nu_0,\nu_l,\bar\nu_l\}=\{-(p-3),
\ldots,-6,-4,-2,-1,0,1,2,4,6\ldots(p-3)\}\no$$ The sums occuring in are naturally expressible in terms of Hurwitz zeta functions and this provides the natural analytic continuation of this expression. We also note that $$\sum_{\{\nu\}}\sum_{k=1}^{\infty}{1\over {(kp+\nu)}^{(2s-2)}}
=\zeta(2s-2)-1-\sum_{l=1}^{(p-3)\over2}{(2l)}^{(2-2s)}\no$$ On utilising these relations we find that $$\eqalign{\tau_{-}(p,s)
&={2\over p}
\left[\zeta(2s-2)-1-\sum_{l=1}^{(p-3)\over2}{(2l)}^{2-2s}\right]\cr
&\qquad (p-2){p}^{-2s}
\left[\zeta(2s-1,1+{1\over p})-\zeta(2s-1,1-{1\over p})\right]\cr
&\qquad-2{p}^{-2s}\sum_{l=-{(p-3)\over2}}^{(p-3)\over2}
(2l)\zeta(2s-1,1+{2l\over p})\cr
&\qquad-{p}^{-2s}
\left[\zeta(2s,1+{1\over p})+\zeta(2s,1-{1\over p})\right]
}\no$$ If we analytically continue the RHS of this expression to $s=0$, and observe that $\zeta(-1,1+x)-\zeta(-1,1-x)=-x$, then we obtain $$\tau_{-}(p,0)=-{2\over p}
\left[1+\sum_{l=1}^{(p-3)\over2}{(2l)}^2\right]
+(p-2)(-{1\over p})
-2\sum_{l=1}^{(p-3)\over2}(2l)(-{2l\over p})\no$$ which immediately gives $$\tau_{-}(p,0)=0\no$$ Thus again the scaling dimension of the associated determinant is zero.

Passing to $\tm(p)$ by taking the derivative at $s=0$ gives $$\eqalign{\tm(p)
&={4\over p}\left[\zeta '(-2)-\ln{p}
+\sum_{l=1}^{(p-3)\over2}{(2l)}^2\ln[{2l\over p}]\right]\cr
&\qquad+2(p-2)\left[\zeta '(-1,1+{1\over p})
-\zeta '(-1,1-{1\over p})\right]\cr
&\qquad-4\sum_{l=1}^{(p-3)\over 2}(2l)
\left[\zeta '(-1,1+{2l\over p})-\zeta '(-1,1-{2l\over p})\right]\cr
&\qquad-2
\left[\zeta '(0,1+{1\over p})+\zeta '(0,1-{1\over p})\right]\cr
}\no$$ A useful identity for Hurwitz zeta functions derived in appendix C is $$\zeta '(-1,1+x) -\zeta '(-1,1-x)=
-x\ln\left[{2\sin(\pi x)\over x}\right]+{1\over \pi}\int_{0}^{\pi
x}dz z\cot(z)$$ Using this and the expression $\zeta '(-2)=-{\zeta(3)/ 4{\pi}^2}$ we conclude that $$\eqalign{\tm(p)
&= -{1\over p{\pi}^2}\zeta(3)
+{4\over p}\ln\left[2\sin({\pi\over p})\right]
+{4\over p}\sum_{l=1}^{(p-3)\over 2}
{(2l)}^2\ln\left[2\sin({2\pi l\over p})\right]\cr
&\qquad+{2(p-2)\over \pi}\int_{0}^{\pi\over p}dz z\cot(z)
-{4\over\pi}\sum_{l=1}^{(p-3)\over 2}
(2l)\int_{0}^{2\pi l\over p}dz z\cot(z)\cr
}\no$$ But $-\tm(p)$ is 