one ingredient among those necessary for a calculation of the determinants and the resulting torsion: this is the construction of the analytic continuation of the series for $\tau(p,s)$. We shall construct this in the next section. The technique we shall use will be more easily followed if we first use it in a more simple case. Thus, to begin with, we set $p=2$ and then construct the continuation.

The series to be continued are $$\tau_{+}(p,s)=\sum_{n=1}^{\infty}
{2(n+1)\charav{n/2}\over\{n(n+2)\}^{s}}$$ $$\tau_{-}(p,s)=\sum_{n=1}^{\infty}
{n\charav{(n+1)/2}+(n+2)\charav{(n-1)/2}\over(n+1)^{2s}}$$ and their difference which leads to the torsion $$\tau(p,s)=\sum_n\left\{{2(n+1)\charav{n/2}\over \{n(n+2)\}^{s}}
-{n\charav{(n+1)/2}+(n+2)\charav{(n-1)/2}\over(n+1)^{2s}}\right\}
 \no$$ These already converges for $\hbox{Re}\, s>3/2$; however a calculation of the determinants and the torsion requires us to work at $s=0$, hence we see the need for, and the extent of, the analytic continuation.

Our interest in this section for illustrative purposes is in the case $p=2$ where we have $$\eqalign{\tau(2,s)&=\tau_+(2,s)-\tau_-(2,s)\cr
                  {}&=\sum_n\left\{{2(n+1)\left\langle\chi^{n/2}
\right\rangle_2\over \left\{n(n+2)\right\}^s}-
{n\left\langle\chi^{(n+1)/2}\right\rangle_2+
(n+2)\left\langle\chi^{(n-1)/2}\right\rangle_2
\over(n+1)^{2s}}\right\}\cr}\no$$ But using we find that $$\eqalign{\left\langle\chi^{(n+1)/2}\right\rangle_2&=
           \left\langle\chi^{(2k-j+1)/2}\right\rangle_2,\quad (n=2k-j)\cr
        \strut&{}\cr
        {}&=\cases{0,& $j=1$\cr
                    2k+2,& $j=0$\cr}
             \quad\equiv\cases{0,& $n$ odd\cr
                    2k+2,& $n$ even\cr}\cr}\no$$

Similarly $$\eqalign{\left\langle\chi^{(n-1)/2}\right\rangle_2&=
\left\langle\chi^{(2k-j)/2}\right\rangle_2,\quad (n=2k-j)\cr
        \strut&{}\cr
        {}&=\cases{2k,& $j=1$\cr
                    0,& $j=0$\cr}
             \quad\equiv\cases{(n+1),& $n$ odd\cr
                    0,& $n$ even\cr}\cr}\no$$ Thus $\tau(2,s)$ becomes $$\eqalign{\tau(2,s)&=\tau_+(2,s)-\tau_-(2,s)\cr
                    {}&=\sum_{n\;{\rm odd}}{
2(n+1)^2\over\left\{n(n+2)\right\}^s}
-\sum_{n\;{\rm even}}{2n(n+2)\over(n+1)^{2s}}\cr}\no$$ Setting $n=(2m-1)$ in $\tau_+(2,s)$ and $n=2m$ in $\tau_-(2,s)$ we have 