Proving double sum $\sum _{a=1}^{\infty }\sum _{b=1}^{\infty }\frac{\left(-1\right)^{a+b}}{\left(a^2+b^2\right)^s} $

Asked Dec 11 '22 at 19:50

Active Dec 11 '22 at 20:17

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I have see this formula in Wolframe Alpha Double series's documentation, but what is the procces of proving? $$\sum _{a=1}^{\infty }\sum _{b=1}^{\infty }\frac{\left(-1\right)^{a+b}}{\left(a^2+b^2\right)^s}=\eta \left(2s\right)-\eta \left(s\right)\beta \left(s\right) $$ Where $$ \beta \left(s\right)=\sum _{n=1}^{\infty }\frac{\left(-1\right)^{n-1}}{\left(2n-1\right)^s}$$

$$\eta \left(s\right)=\sum _{n=1}^{\infty }\frac{\left(-1\right)^{n-1}}{n^s}$$

edited Dec 11 '22 at 20:00

asked Dec 11 '22 at 19:50

asmatqatea

What is $\beta (s)$? – Kamal Saleh Dec 11 '22 at 19:53
@KamalSaleh, dirichlet beta function – asmatqatea Dec 11 '22 at 19:54
Thanks for clarifying :) – Kamal Saleh Dec 11 '22 at 19:54
1

$\beta(s)$ is better understood as $L(s,\chi_4)$. Did you prove that $\sum_{a,b\in \Bbb{Z}^2- (0,0)} (a^2+b^2)^{-s}=4 \zeta(s)L(s,\chi_4)$ ? – reuns Dec 11 '22 at 20:14
For the non-alternating version of the sum (as in the comment above), see (1), (2), (3), etc. Then you can follow any of these approaches to handle the present (alternating) case. (I would post an answer myself, but your question is about to be qualified as MCoOD.) – metamorphy Dec 12 '22 at 19:53

Proving double sum $\sum _{a=1}^{\infty }\sum _{b=1}^{\infty }\frac{\left(-1\right)^{a+b}}{\left(a^2+b^2\right)^s} $

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