we can reduce to Ray’s expression. One may conclude that, by following two alternate derivations, we have arrived at what is a sequence of non-trivial identities. As we saw in the case of $p=2$ utilising these identities one can obtain non-trivial formulae for $\zeta(3)$ and also can be used to further simplify the expressions for the individual determinants. The resulting expressions for $\zeta(3)$ are $$\eqalign{\zeta(3)&={2\pi^2\over (p^3-1)}\left[
2(p-1)\ln\left[2\sin({\pi\over p})\right]
-2\sum_{l=1}^{(p-3)\over 2}
4l^2\ln\left[2\sin({2l\pi\over p})\right] \right.\cr
&\qquad-{p^2\over\pi^2}\int_{0}^{\pi\over p}dz\,
z(z+{(p-2)\pi\over p})\cot(z)
-{p^2\over\pi^2}\int_{0}^{(p-1)\pi\over p}dz\,
z(z-{(p-2)\pi\over p})\cot(z)\cr
&\qquad\left.-2\sum_{l=1}^{(p-3)\over 2}
{p^2\over\pi^2}\int_{0}^{2l\pi\over p}dz\, z(z-{4l\pi\over p})\cot(z)
\right]\cr
}\no$$

For the sake of illustration let us quote the implications of these formulae for the simplest odd case: $p=3$. On utilising all of the information at our disposal we find that $$\eqalign{\tp(3)&=-{\zeta(3)\over 3\pi^2}-{4\over3}\ln3
+{2\over\pi}\int_{0}^{\pi\over3}dz\,z \cot(z)\cr
\tm(3)&=-{\zeta(3)\over 3\pi^2}+{2\over3}\ln3
+{2\over\pi}\int_{0}^{\pi\over3}dz\,z \cot(z)\qquad {\rm and}\cr
\ln T(3)&=-2\ln3\cr
}$$ The relation between the expressions we have obtained which we expressed in terms of a formula for $\zeta(3)$ (the analog of for $p=2$) becomes $$\zeta(3)={2\pi^2\over13}\ln 3
-{9\over 13}\int_{0}^{\pi\over3}dz\,z(z+{\pi\over3})\cot(z)
-{9\over 13}\int_{0}^{2\pi\over3}dz\,z(z-{\pi\over3})\cot(z)$$ We will now turn to the case of even $p$. When $p$ is even we follow a slightly different route to that used in the previous section but we arrive at expressions of a similar general form 