Show that $\sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 58 n^2} = - \frac{\pi \ln( 27 + 5 \sqrt {29})}{\sqrt {58}} $

Question

I wonder why this is true

$$ \sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 58 n^2} = - \frac{\pi \ln( 27 + 5 \sqrt {29})}{\sqrt {58}} $$

Where the sum omits the case $n = m = 0$ ofcourse.

Answer 1

Finally I managed to sum this series using Ramanujan's class invariants. We have the definition $$g(q) = 2^{-1/4}q^{-1/24}\prod_{n = 1}^{\infty}(1 - q^{2n - 1}), \, g_{p} = g(e^{-\pi\sqrt{p}})\tag{1}$$ Ramanujan established that if $p$ is a positive rational number then $g_{p}$ is an algebraic real number. Moreover we have the non-trivial identity $$g_{4/p} = 1/g_{p}\tag{2}$$ Another ingredient we need is the formula $$\sin \pi z = \pi z\prod_{n = 1}^{\infty}\left(1 - \frac{z^{2}}{n^{2}}\right)\tag{3}$$ Taking logs and differentiating with respect to $z$ we get $$\pi\cot\pi z = \frac{1}{z} -2z \sum_{n = 1}^{\infty}\frac{1}{n^{2} - z^{2}}$$ Replacing $z$ by $iz$ we get $$\pi\coth\pi z = \frac{1}{z} + 2z\sum_{n = 1}^{\infty}\frac{1}{n^{2} + z^{2}}$$ The above sum can be written as $$\sum_{n = 1}^{\infty}\frac{1}{n^{2} + z^{2}} = \frac{\pi}{2z} - \frac{1}{2z^{2}} + \frac{\pi e^{-2\pi z}}{z(1 - e^{-2\pi z})}\tag{4}$$ Consider the sum in question \begin{align} S(p) &= \sum_{m, n\in\mathbb{Z}}'\frac{(-1)^{m}}{m^{2} + pn^{2}}\notag\\ &= 2\sum_{m = 1}^{\infty}\frac{(-1)^{m}}{m^{2}} + 2\sum_{n = 1}^{\infty}\sum_{m\in\mathbb{Z}}\frac{(-1)^{m}}{m^{2} + pn^{2}}\notag\\ &= 2\sum_{m = 1}^{\infty}\frac{(-1)^{m}}{m^{2}} + \frac{2}{p}\sum_{n = 1}^{\infty}\frac{1}{n^{2}}+ 4\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}\frac{(-1)^{m}}{m^{2} + pn^{2}}\notag\\ &=\frac{(2- p)\pi^{2}}{6p} + 4\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}\frac{(-1)^{m}}{m^{2} + pn^{2}}\notag\\ &=\frac{(2- p)\pi^{2}}{6p} + 4\sum_{m = 1}^{\infty}(-1)^{m}\sum_{n = 1}^{\infty}\frac{1}{m^{2} + pn^{2}}\notag\\ &= \frac{(2- p)\pi^{2}}{6p} + \frac{4}{p}\sum_{m = 1}^{\infty}(-1)^{m}\sum_{n = 1}^{\infty}\frac{1}{m^{2}(1/p) + n^{2}}\notag\\ &= \frac{(2- p)\pi^{2}}{6p} + \frac{4}{p}\sum_{m = 1}^{\infty}(-1)^{m}\left(\frac{\pi\sqrt{p}}{2m} - \frac{p}{2m^{2}} + \frac{\pi\sqrt{p} e^{-2\pi m/\sqrt{p}}}{m(1 - e^{-2\pi m/\sqrt{p}})}\right)\text{ (using equation }(4)) \notag\\ &=\frac{\pi^{2}}{3p}-\frac{2\pi\log 2}{\sqrt{p}}+\frac{4\pi}{\sqrt{p}}\sum_{m=1}^{\infty} \frac{(-1)^{m}q^{m}}{m(1-q^{m})},\,q=e^{-2\pi/\sqrt{p}}\notag\\ &=\frac{\pi^{2}}{3p}-\frac{\pi\log 4}{\sqrt{p}}+\frac{4\pi}{\sqrt{p}}(a(q^{2})-a(q))\notag\\ &\, \, \, \, \,\,\,\,\text{where }a(q) =\sum_{n=1}^{\infty} \frac{q^{n}} {n(1-q^{n})}=-\log\prod_{n=1}^{\infty}(1-q^{n})\notag\\ &=\frac{\pi^{2}}{3p}-\frac{\pi\log 4} {\sqrt{p}}+\frac{4\pi}{\sqrt{p}}\log\prod_{n=1}^{\infty}(1-q^{2n-1}) \notag\\ &=\frac{\pi^{2}}{3p}-\frac{\pi}{\sqrt {p}} \log\frac{2}{q^{1/6}g^{4}(q)}\text{ (via equation }(1)) \notag\\ &= -\frac{\pi\log(2/g_{4/p}^{4})}{\sqrt{p}}\notag\\ &= -\frac{\pi\log (2g_{p}^{4})}{\sqrt{p}}\text{ (using equation }(2))\notag \end{align} It is well known that $g_{58} = \sqrt{(5 + \sqrt{29})/2}$ and this gives the desired closed form for $S(58)$. The above technique can also be used (with some more effort) to prove the Kronecker's second limit formula.

---

The function $a(q) $ used above is related to work of Simon Plouffe. See this answer for details.

Answer 2

• Let $F=\mathbb{Q}(\sqrt{-58}),\mathcal{O}_F=\mathbb{Z}[\sqrt{-58}]$. Its ideal class group is $C_F= \{ (1),(2,\sqrt{-58})\}$

  thus the ideals with their norm are $N((n+\sqrt{-58}m))= n^2+58m^2$, $ N(\frac{2n+\sqrt{-58}m}{2}(2,\sqrt{-58}))= \frac12(4n^2+58m^2)$

  then your series is $\ \ 2 \ L(1,\psi)$

  where $\psi((n+\sqrt{-58}m))=1, \psi(\frac{2n+\sqrt{-58}m}{2}(2,\sqrt{-58}))=-1$ is the Hecke character induced by the non-trivial character of $C_F$.
  $$ L(s,\psi) = \sum_I \psi(I) N^{-s} =\frac{1}{|\mathcal{O}_F^\times|} \sum_{n,m \in \mathbb{Z}^2}' N((n+\sqrt{-58}m))^{-s}- 2^{s} N((2n+\sqrt{-58}m))^{-s}$$

• Using $|C_F| = 2$ and some class field theory, we find that $H = F(\sqrt{-2})$ is the Hilbert class field of $F$ and $$\zeta_H(s) = \zeta_F(s) L(s,\psi)$$

  Now by chance it happens that $H/\mathbb{Q}$ is itself an abelian extension. Thus we can write $\zeta_H$ as a product of Dirichlet L-functions

  $$\zeta_H(s) = \zeta(s)\ L(s,{\scriptstyle \left(\frac{-58}{.}\right)})\ L(s,{\scriptstyle \left(\frac{-2}{.}\right)})\ L(s,{\scriptstyle \left(\frac{29}{.}\right)})$$ Together with $\displaystyle\zeta_F(s) = \zeta(s)\ L(s,{\scriptstyle \left(\frac{-58}{.}\right)})$ it means $$L(1,\psi) = L(1,{\scriptstyle \left(\frac{-2}{.}\right)})\ L(1,{\scriptstyle \left(\frac{29}{.}\right)})$$ and we use quadratic reciprocity to write $(\frac{-d}{.}) = (\frac{.}{\Delta})$ and conclude.

Answer 3

(Too long for a comment.)

We have,

$$ \sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 10n^2} = - \frac{2\pi \ln( \sqrt2\; U_{5})}{\sqrt {10}} $$

$$ \sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 58 n^2} = - \frac{2\pi \ln( \sqrt2\; U_{29})}{\sqrt {58}} $$

with fundamental units $U_5 = \frac{1+\sqrt5}2$ and $U_{29} = \frac{5+\sqrt{29}}2$.

P.S. Presumably there might be a family for $d = 5,\,13,\,37$.

---

Added:

Courtesy of a comment by Paramanand Singh, we have the closed-form in terms of the Dedekind eta function $\eta(\tau)$ as,

$$\sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + pn^2} = - \frac{2\pi \ln(\sqrt2\,g_p^2)}{\sqrt {p}} =- \frac{\pi \ln(2\,g_p^4)}{\sqrt {p}} $$ where,

$$g_p = 2^{-1/4}\frac{\eta(\tfrac12\sqrt{-p})}{\eta(\sqrt{-p})}$$

In particular, $$\sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 6n^2} = - \frac{2\pi \ln\big(\sqrt2\,(1+\sqrt2)^{1/3}\big)}{\sqrt{6}}$$ $$\sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 22n^2} = - \frac{2\pi \ln\big(\sqrt2\,(1+\sqrt2)\big)}{\sqrt{22}}$$ and more complicated ones for $d=5,\,13,\,37$.