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Question. Can this number be expressed in terms of classical values? $$\sum_{n,m=1}^{\infty}\frac1{(n^2+m^2)^{\frac32}}=1.056348517615643291\dots$$

UPDATE. I'm encouraged by Noam, Kevin and Igor's directional replies. To spice things up, I ask: is this true? $$\sum_{n,m=1}^{\infty}\frac1{(n^2+m^2)^s}=\zeta(s)\beta(s)-\zeta(2s),$$ wherever convergence occurs. Here, $\zeta(s)$ and $\beta(s)$ are the Riemann zeta and Dirichlet beta functions, respectively. I also suspect that the function $r_2(k)$ (the number of ways of writing $k$ as a sum of integer squares) might come into the picture, if we were to avoid path through modular function.

Question. Can this number be expressed in terms of classical values? $$\sum_{n,m=1}^{\infty}\frac1{(n^2+m^2)^{\frac32}}=1.056348517615643291\dots$$

UPDATE. I'm encouraged by Noam, Kevin and Igor's directional replies. To spice things up, I ask: is this true? $$\sum_{n,m=1}^{\infty}\frac1{(n^2+m^2)^s}=\zeta(s)\beta(s)-\zeta(2s),$$ wherever convergence occurs. Here, $\zeta(s)$ and $\beta(s)$ are the Riemann zeta and Dirichlet beta functions, respectively.

Question. Can this number be expressed in terms of classical values? $$\sum_{n,m=1}^{\infty}\frac1{(n^2+m^2)^{\frac32}}=1.056348517615643291\dots$$

UPDATE. I'm encouraged by Noam, Kevin and Igor's directional replies. To spice things up, I ask if this is true? $$\sum_{n,m=1}^{\infty}\frac1{(n^2+m^2)^s}=\zeta(s)\beta(s)-\zeta(2s),$$ wherever convergence occurs. Here, $\zeta(s)$ and $\beta(s)$ are the Riemann zeta and Dirichlet beta functions, respectively. I also suspect that the function $r_2(k)$ (the number of ways of writing $k$ as a sum of integer squares) might come into the picture, if we were to avoid path through modular function.

Question. Can this number be expressed in terms of classical values? $$\sum_{n,m=1}^{\infty}\frac1{(n^2+m^2)^{\frac32}}=1.056348517615643291\dots$$

UPDATE. I'm encouraged by Noam, Kevin and Igor's directional replies. To spice things up, I ask: is this true? $$\sum_{n,m=1}^{\infty}\frac1{(n^2+m^2)^s}=\zeta(s)\beta(s)-\zeta(2s),$$ wherever convergence occurs. Here, $\zeta(s)$ and $\beta(s)$ are the Riemann zeta and Dirichlet beta functions, respectively. I also suspect that the function $r_2(k)$ (the number of ways of writing $k$ as a sum of integer squares) might come into the picture, if we were to avoid path through modular function.