Does anyone know an expression (in terms of simpler functions) for the following Epstein Zeta function:


$\sum \frac{1}{(m^2+m n+n^2)^s}$


I know an expression (in terms of the Dirichlet Beta function) exist for $\sum \frac{1}{(m^2+n^2)^s}$. I did look hard into the literature (being not an expert) but couldn't find anything.


Thank you very much!


fernando



Epstein zeta functions
I had thought the question meant to refer to Dedekind *zeta*, not *Beta*, since that Epstein zeta is ($4\times$) the Dedekind zeta function of the Gaussian integers. The case of the question gives ($6\times$) the Dedekind zeta function of the "Eisenstein integers", namely, $\mathbb Z[\rho]$, where $\rho$ is a cube root of unity. To see this, one uses the fact that that ring is Euclidean, and that $N(m+n\rho)=m^2+mn+n^2$.


Edit/addition: Ah, sorry, didn't realize the reference to Dedekind "beta"... Yes, by a special case of quadratic reciprocity, the Dedekind zeta function of the Gaussian integers factors as "ordinary" zeta times (what many might call) the Dirichlet $L$-function $L(s,\chi)=\sum\_n \chi(n)/n^s$ attached to the character $\chi(n)$ which is $0$ for $n=2\mod 4$, is $+1$ for $n=1\mod 4$, and is $-1$ for $n=3\mod 4$. This gives the Dedekind "beta function".


As @HenryCohn notes, in the case of $m^2+mn+n^2$, because of quadratic reciprocity for $-3$, the Epstein zeta factors again as $\zeta(s)\cdot L(s,\chi)$, now with $\chi(n)=+1$ for $n=1\mod 3$, $\chi(n)=-1$ for $n=2\mod 3$, and $\chi(n)=0$ for $n=0\mod 3$. 


Those factorizations due to quadratic reciprocity were known to Dirichlet for quadratic extensions of the rationals.


Further-edit: I had forgotten: although I think it's not the most important or revealing relation to other things, yes, such Dirichlet $L$-functions are indeed expressible in terms of the Hurwitz zeta: for example, for the $m^2+mn+n^2$ case, after the factor of $\zeta(s)$ is removed, the remaining $L(s,\chi)=\sum\_n\chi(n)/n^s$ is a linear combination of $\sum\_n 1/(n+{1\over 3})^s$ and $\sum\_n 1/(n+{2\over 3})^s$, and similarly in general.


