TITLE: What is the closed-form for $\displaystyle\sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + mn+41n^2}$?
QUESTION [12 upvotes]: Omitting the case $m = n = 0$, if,
$$ \sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 58 n^2} = - \frac{2\pi \ln\big(\tfrac{5 + \sqrt {29}}{\sqrt2}\big)}{\sqrt {58}} $$
as in this post, then is,
$$ \sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + mn+ 41 n^2} = - \frac{2\pi \ln\big(\beta\big)}{\sqrt {163}} $$
for some algebraic number $\beta$? If yes, then what is $\beta$?

P.S. Incidentally, we have the nice approximation,
$$e^{\pi\sqrt{58}} \approx \Big(\tfrac{5 + \sqrt {29}}{\sqrt2}\Big)^{12} +23.999999988776\dots$$
and the "excess" near $24$ has to do with the Dedekind eta function.

REPLY [3 votes]: It seems we can get an aesthetic summary using Ramanujan's $g_n$ and $G_n$ functions (also via the Weber modular functions) as,
$$ \sum_{p,q = - \infty}^{\infty}\, \frac{(-1)^p}{p^2 + n q^2} = - \frac{\pi \ln{\big(2\,g^4_n\big)}}{\sqrt {n}} $$
$$ \sum_{p,q = - \infty}^{\infty}\, \frac{(-1)^p}{p^2 + pq+ k q^2} = - \frac{\pi \ln{\big(2\,G^4_n\big)}}{\sqrt {n}} $$
where $n=4k-1$ for $G_n$ and,
$$g_n =2^{-1/4}\,\frac{\eta\big(\tfrac12\sqrt{-n}\big)}{\eta\big(\sqrt{-n}\big)}$$
and,
$$G_n = 2^{-1/4}\,\frac{\eta^2\big(\sqrt{-n}\big)}{\eta\big(\tfrac12\sqrt{-n}\big)\,\eta\big(2\sqrt{-n}\big)}$$
with Dedekind eta function $\eta(\tau)$. 

For example,
$$g_{58} = \sqrt{\frac{5+\sqrt{29}}2} =2.27872\dots $$
$$G_{163} = 2^{-1/4}x =4.47241\dots$$
with $x$ as the real root of $x^3-6x^2+x-2=0$. Incidentally,
$$e^{\pi\sqrt{163}}\approx x^{24}-24.00000000000000105\dots$$