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.

(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 $d=5,\,13,\,37$.
