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

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. 
