Is it possible to express the following sum in closed form? $$S(p)=\sum_{n=-\infty}^{\infty}\sum_{m=-\infty}^{\infty}{\left(n^2+m^2\right)^{\large-{p}}}$$ where the point $(n,m)=(0,0)$ is not taken into account in the sum (i.e. $(n,m)\in \mathbb{Z}^2/\{(0,0)\}$).
In the book Lattice Sums Then and Now by Borwein et al. it is stated (page 66, Appendix 1.7), that $S(p)=4\beta(p)\zeta(p)$, with $\beta$ and $\zeta$ the Dirichlet beta and Riemann zeta functions.
However I cannot find any derivation of this result. Any thoughts?
If $\chi_4$ denotes the unique non-trivial Dirichlet character modulo 4, then it is proved that for $k \geq 1$,
$$ r_2(k) := \#\{ (m,n)\in\mathbb{Z}^2 : m^2+n^2=k\} = 4 \sum_{d \mid k} \chi_4(d). $$
(I do not remember the proof of this fact, but the keyword sum of squares function may be useful for finding out a reference.)
Now utilizing this, we have
\begin{align*} S(p) = \sum_{k=1}^{\infty} k^{-p} r_2(k) &= 4 \sum_{k=1}^{\infty} \left( \sum_{d\mid k} \chi_4(d) \right) k^{-p} \\ &= 4 \left( \sum_{d=1}^{\infty} \chi_4(d) d^{-p} \right) \left( \sum_{l=1}^{\infty} l^{-p} \right) = 4 \beta(p)\zeta(p). \end{align*}