$$\tau_{+}(2,s)=\sum_{m=1}^\infty{8 m^2\over(4m^2-1)^s},
\qquad\tau_{-}(2,s)=\sum_{m=0}^\infty{4m(2m+2)\over (2m+1)^{2s}}$$ and $$\tau(2,s)=\sum_{m=1}^\infty{8 m^2\over(4m^2-1)^s}
-\sum_{m=0}^\infty{2\over (2m+1)^{(2s-2)}}
+\sum_{m=0}^\infty{2\over (2m+1)^{2s}}\no$$ Now if we use the fact that $$\sum_{n=1,3,5,\dots}{1\over n^s}=(1-2^{-s})\zeta(s)\no$$ where $\zeta(s)$ is the usual Riemann zeta function then we get $$\tau_{-}(2,s)=2(1-2^{-(2s-2)})\zeta(2s-2)-2(1-2^{-2s})\zeta(2s)\no$$ and denoting $$A_{2}(m,0,s)={{(2m)}^2\over{\left\{{(2m)}^2-1\right\}}^s},
\qquad {\rm and}\qquad A_{2}(0,s)=\sum_{m=1}^{\infty}A_{2}(m,0,s)$$ (This notation is used to agree with the general case to be discussed in the next section. See also Appendix A.) Thus we have $$\tau_{+}(2,s)=2A_{2}(0,s)$$ and these combine to give $$\tau(2,s)=2A_{2}(0,s)-2(1-2^{-(2s-2)})\zeta(2s-2)+2(1-2^{-2s})\zeta(2s)\no$$ Since the terms involving the Riemann zeta function already have a well defined continuation it remains to continue $A_{2}(0,s)$. Now $$\eqalign{A_{2}(m,0,s)&={4m^2\over(4m^2-1)^s}={4m^2\over (4m^2)^s}
                      \left(1-{1\over4m^2}\right)^{-s}\cr
                {}&={1\over(4m^2)^{(s-
                                    1)}}\left\{1+{s\over4m^2}+\cdots\right\}\cr
                {}&={1\over(4m^2)^{(s-1)}}+{s\over(4m^2)^s}+R(m,s),
                                        \qquad(\hbox{def. of }R(m,s))\cr}\no$$ So that the remainder term $R(m,s)$ is given by $$\eqalign{R(m,s)&=A_{2}(0,m,s)-{1\over(4m^2)^{(s-1)}} -{s\over(4m^2)^s}\cr
                 &={4m^2\over(4m^2-1)^s}-{1\over(4m^2)^{(s-1)}}-
                                                  {s\over(4m^2)^s}\cr}\no$$

The definition of the remainder term is chosen to ensure that $$\left\vert R(m,s)\right\vert\le{(\ln m)^\alpha\over m^2}\no$$ and this has the vital consequence that the operations $d/ds$ (at $s=0$) and $\sum_m$ [*commute* ]{} when applied to $R(m,s)$.

Defining $$R(s)=\sum_{m=0}^\infty R(m,s)\no$$ allows us to tidy our expressions up somewhat. Collecting our regulated expressions we therefore have $$\eqalign{\tau_{+}(2,s)&={8\over{4}^{s}}\zeta(2s-2)
+{2s\over 4^s}\zeta(2s)+2 R(s)\qquad {\rm and}\cr
\tau_{-}(2,s)&=2(1-2^{-(2s-2)})\zeta(2s-2)-2(1-2^{-2s})\zeta(2s)\cr}\no$$ In fact the expression for $\tau(2,s)$ can be further tidied up to give $$\tau(2,s)=2\left\{{8\over4^{(s)}}-1\right\}\zeta(2s-2)
+2\left\{1+{(s-1)\over4^s}\right\}\zeta(2s)+2R(s)\no$$ The series for $R(s)$ is [*guaranteed*]{} to be convergent and the analytic continuation 