-\lambda B_{p}(\lambda,s)\right]\no$$ Let us further note that in the set $\{\lambda\}$ the non zero elements come in pairs of the form $\{\lambda,-\lambda\}$. Thus we can further write $$\tau_{+}(p,s)={2\over p}A_{p}(0,s)+{2\over p}\sum_{l=1}^{r-1}
\left[A_{p}^{+}(2l+1,s)-(2l+1)B_{p}^{-}(2l+1,s)\right]$$ where the $\mp$ superscripts refer to the symmetric or anti-symmetric combination with respect to the first argument: i.e. $A_{p}^{+}(x,s)=A_{p}(x,s)+A_{p}(-x,s)$ and $B_{p}^{-}(x,s)=B_{p}(x,s)-B_{p}(-x,s)$. We relegate the details of the computation of these functions and their analytic continuation to appendices A and B—the calculations are generalisations of those performed for $p=2$. Quoting here from appendices A and B we have that the relevant functions and their derivatives at $s=0$ are given by $$\eqalign{A_{p}(x,0)&=p^2\zeta(-2,1+{x\over p})
= -{x(x+p)(2x+p)\over 6p}\cr
B_{p}(x,0)&=p\zeta(-1,1+{x\over p})+{1\over 2p}=-{p\over
12}-{x\over p}+{x^2+1\over 2 p}\cr
H_{p}(x,0)&=-{px\over 12}-{x\over 2p}+{x^3\over 6p}\cr
}\no$$ Which immediately implies that $H_{p}^{+}(x,0)=0$ and we have our first result that for $p$ odd $$\tau_{+}(p,0)=0$$ The significance of this is that the analytic continuation of the scaling dimension of the determinant is zero.

Next we note again from appendices A and B that $$\eqalign{A_{p}^{+}(x)
=-{p^2\over\pi^2}\zeta(3)
&-{p^2\over\pi^2}\int_{0}^{(x+1)\pi\over p}
dz\, z(z-{2\pi x\over p})\cot(z)
-{p^2\over\pi^2}\int_{0}^{(x-1)\pi\over p}
dz\, z(z-{2\pi x\over p})\cot(z)\cr
&-x^2\ln\left[{2\sin\left({(x+1)\pi/ p}\right)\over x+1}\right]
-x^2\ln\left[{2\sin\left({(x-1)\pi/ p}\right)\over x-1}\right]\cr
}\no$$ and $$\eqalign{B_{p}^{-}(x)
&= {p\over\pi}\int_{0}^{(x+1)\pi\over p}dz\,z\cot(z)
+{p\over\pi}\int_{0}^{(x-1)\pi\over p}dz\, z\cot(z)\cr
&\qquad-x\ln\left[{2\sin\left({(x+1)\pi/ p}\right)\over x+1}\right]
-x\ln\left[{2\sin\left({(x-1)\pi/ p}\right)\over x-1}\right]\cr
}\no$$

Now combining our 