$$\sum_{m=1}^\infty\left[4m^2\ln(1-1/4m^2)+1\right]=-{1\over2}+
{4\over\pi^2}\int_0^{\pi/2}dz\,z^2\cot(z)\no$$ which means that $$\eqalign{\tp(2)&=-{4\over \pi^2}\zeta(3)
-{8\over\pi^2}\int_{0}^{\pi\over2}dz z^2\cot(z)\cr
\tm(2)&={3\over\pi^2}\zeta(3)+2\ln2\qquad {\rm and}\cr
\ln T(2)&=-{7\over\pi^2}\zeta(3)-2\ln(2)-
{8\over\pi^2}\int_0^{\pi/2}dz\,z^2\cot(z)}\no$$ The formula above for $T(2)$ can be pushed even further: By using it with Ray’s expression \[2\] for the torsion we can deduce that $$\eqalign{\ln T(p)&=-{4\over p}\sum_{j=1}^{(p-1)}\sum_{k=1}^p\cos({2jk\pi\over
p})\ln(2\sin({2k\pi\over p}))\exp[{2k\pi i\over  p}]\cr
&=-4\ln\left[2 \sin({\pi\over p})\right]\cr}\no$$ which, for $p=2$, becomes simply $$\ln T(2)=-4\ln(2)\no$$ Hence we straightaway have the identity $$-4\ln(2)=-{7\over\pi^2}\zeta(3)-2\ln(2)-
{8\over\pi^2}\int_0^{\pi/2}dz\,z^2\cot(z)\no$$ Or $$\zeta(3)={2\pi^2\over7}\ln(2)-{8\over7}\int_0^{\pi/2}dz\,z^2\cot(z)\no$$ In other words our computation of the torsion has given us a formula for the zeta function at its first odd argument. Equivalently we can use this relation to eliminate the integral and obtain quite simple expressions for the logarithms of the determinants of the Laplacians on zero and one forms respectively. We conclude this section by quoting these results

Noting first of all that the expressions for $\tp(2)$ and $\tm(2)$ in their simplest form now become $$\tp(2)= {3\over \pi^2}\zeta(3)-2\ln 2\qquad {\rm and }
\qquad\tm(2)={3\over \pi^2}\zeta(3)+2\ln 2\no$$ and from the definitions of these objects we have at once that $$\eqalign{
\ln \Det d^{*}d_{0} &= -{3\over 2\pi^2}\zeta(3) +\ln2\cr
\ln \Det d^{*}d_{1} &= -{3\over \pi^2}\zeta(3)-2 \ln2\cr
}\no$$ It is interesting to note the role that the Riemann zeta function $\zeta(3)$ plays in these expressions. Since these 