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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.

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Tito Piezas III
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2 Answers2

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The key here is a result which goes by the name of Kronecker's second limit formula. Using the formulation given in Wikipedia it can be proved that the desired sum in question is equal to $$-\frac{2\pi\log|2g^{4}(q)|} {\sqrt{163}} $$ where $q=\exp(\pi i\tau), 2\tau=1+i\sqrt{163}$ and $$g(q) =2^{-1/4}q^{-1/24}\prod_{n=1}^{\infty}(1-q^{2n-1})$$ With some manipulation it can be shown that the above sum is equal to $$-\frac{\pi\log(2G_{163}^{4})}{\sqrt{163}}$$ The value of $$G_{163}=\frac{6+\sqrt[3]{135-3\sqrt{489}}+\sqrt[3]{135+3\sqrt{489}}} {3\sqrt[4]{2}} $$ is (not so) well known and the calculations explained above can be performed with reasonable amount of labor to obtain a closed form for the sum in question.

Paramanand Singh
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  • I find that for $2\tau = -1+\sqrt{-163}$, then, $$\beta = \big|2g^4(q)\big| = \left(e^{-\pi i/24}\frac{\eta(\tau)}{\eta(2\tau)}\right)^2=x^2$$ where $x$ is the real root of the rather well-known cubic $x^3-6x^2+4x-2 = 0$. This then yields, $$ \sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + mn+ 41 n^2} = - \frac{2\pi \ln\big(x^2\big)}{\sqrt {163}} = -1.64493406\dots$$ But is the above eta quotient true for general $2\tau = -1+\sqrt{-d}$? – Tito Piezas III Oct 19 '17 at 03:48
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    @TitoPiezasIII: yes I think the relation between eta and Ramanujan $g$ should be valid. And you have finally managed to find the minimal polynomial for $\beta$. Great! – Paramanand Singh Oct 19 '17 at 04:15
  • I've added a supplementary answer to yours below. It seems the two forms can be expressed by $g_n$ and $G_n$ equally simply. – Tito Piezas III Oct 19 '17 at 06:00
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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,

$$e^{\pi\sqrt{163}}\approx x^{24}-24.00000000000000105\dots$$

Tito Piezas III
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