
> 
> **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](https://en.wikipedia.org/wiki/Riemann_zeta_function) and [Dirichlet beta](https://en.wikipedia.org/wiki/Dirichlet_beta_function) functions, respectively. 



What can be said about this double sum?
I don't know how "classical" you find these values, but here's perhaps something.


Define $E=\sum\_{m,n\in\mathbb{Z}}q^{m^2+n^2}$, which is known to be a weight 1 level 4 modular form. In fact $E$ is an eigenform for the Hecke operators, and if we write $E=\sum\_{r\geq0}a\_rq^r$ then $L(E,s)=\sum\_{r\geq1}a\_r/r^s$ equals $4\zeta(s)L(\chi,s)$ with $\chi$ the Dirichlet character of conductor 4. The factor of 4 is because $E$ is normalised so that $a\_1=4$. Note that $a\_r$ equals the number of ways $r$ can be written as $m^2+n^2$ (where we allow zero and negative values for $m$ and $n$ at this point). Note also that the $L$-function doesn't see the troublesome $m=n=0$ term.


Now note that setting $s=3/2$ (where everything converges) we get $L(E,3/2)$ is nearly what you want. In fact if your constant is $c$ then (allowing for signs) we get 


$$4c+4\sum\_{n\geq1}(1/n^2)^{3/2}=L(E,3/2)$$


($4c$ for the signs, and the other term for the $m=0$ and $n=0$ terms we missed out) and all the 4s cancel miraculously giving


$c=\zeta(3/2)L(\chi,3/2)-\zeta(3)$


or equivalently


$c=\zeta\_{\mathbb{Q}(i)}(3/2)-\zeta(3)$


with $\zeta\_{\mathbb{Q}(i)}$ the Dedekind zeta function of the number field. Let's check with pari-gp:



```
L=lfuncreate(x^2+1);
lfun(L,3/2)-zeta(3)

%2 = 1.0563485176156432910328906583178146441

```

which looks good to me.


Note finally that now I've done the calculation I realise that one could avoid the modular forms side of things and just consider the zeta function of $\mathbb{Q}(i)$ directly because we're summing some function of norms of elements; the final equation for $c$ follows essentially from the definition of the Dedekind $\zeta$ function.


Hmm, and now note finally finally that while I was typing this, Noam Elkies said exactly the same thing but rather more succinctly :-)


