Show that $\int_0^1\ln(-\ln{x})\cdot{\mathrm dx\over 1+x^2}=-\sum\limits_{n=0}^\infty{1\over 2n+1}\cdot{2\pi\over e^{\pi(2n+1)}+1}$ and evaluate it 
Considering this integral and sum are equal, how can we show that and evaluate its closed form?
$$\int_{0}^{1}\ln{(-\ln{x})}\cdot{\mathrm dx\over 1+x^2}=-\sum_{n=0}^{\infty}{1\over 2n+1}\cdot{2\pi\over e^{\pi(2n+1)}+1}=I_S\tag1$$

Note:
$$\int_{0}^{1}{\mathrm dx\over 1+x^2}=\sum_{n=0}^{\infty}{(-1)^n\over 2n+1}\tag2$$
An attempt:
$u=-\ln{x}$ then $xdu=-dx$
$(1)$ becomes
$$\int_{0}^{\infty}\ln{u}\cdot{\mathrm du\over e^u+e^{-u}}={1\over 2}\int_{0}^{\infty}\ln{u}\cdot{\mathrm du\over \cosh{u}}\tag3$$
$(3)$, seem complicate.
 A: The following is based on theory of theta functions and elliptic integrals. Let $0 < q < 1$ and consider the function $$a(q) = \sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 + q^{n})}\tag{1}$$ and $$b(q) = \sum_{n \text{ odd}}^{\infty}\frac{q^{n}}{n(1 + q^{n})}\tag{2}$$ We can see that $$b(q) = a(q) - \frac{a(q^{2})}{2}\tag{3}$$ Now we have
\begin{align}
a(q) &= \sum_{n = 1}^{\infty}\frac{1}{n}\sum_{m = 1}^{\infty}(-1)^{m - 1}q^{mn}\notag\\
&= \sum_{m = 1}^{\infty}(-1)^{m - 1}\sum_{n = 1}^{\infty}\frac{q^{mn}}{n}\notag\\
&= \sum_{m = 1}^{\infty}(-1)^{m}\log(1 - q^{m})\notag\\
&= \log\prod_{n = 1}^{\infty}\frac{1 - q^{2n}}{1 - q^{2n - 1}}\notag\\
&= \log\prod_{n = 1}^{\infty}\frac{1 - q^{n}}{(1 - q^{2n - 1})^{2}}\notag
\end{align}
Therefore
\begin{align}
2b(q) &= 2a(q) - a(q^{2})\notag\\
&= \log\prod_{n = 1}^{\infty}\frac{(1 - q^{n})^{2}}{(1 - q^{2n - 1})^{4}}\cdot\frac{(1 - q^{4n - 2})^{2}}{(1 - q^{2n})}\notag\\
&= \log\prod_{n = 1}^{\infty}\frac{(1 - q^{2n})^{2}}{(1 - q^{2n - 1})^{2}}\cdot\frac{(1 - q^{4n - 2})^{2}}{(1 - q^{2n})}\notag\\
&= \log\prod_{n = 1}^{\infty}(1 - q^{2n})(1 + q^{2n - 1})^{2}\notag\\
&= \log\vartheta_{3}(q)\notag
\end{align}
and thus $$b(q) = \frac{1}{4}\log\frac{2K}{\pi}\tag{4}$$ where $K$ is complete elliptic integral for modulus $k$ corresponding to nome $q = e^{-\pi K'/K}$.
The value of integral as claimed here is $$-2\pi b(e^{-\pi})$$ This corresponds to $q = e^{-\pi}, k = 1/\sqrt{2}, K = \Gamma^{2}(1/4)/4\sqrt{\pi}$ and hence the integral in question is claimed to be $$-\frac{\pi}{2}\log\frac{1}{\pi}\frac{\Gamma^{2}(1/4)}{2\sqrt{\pi}} = \frac{\pi}{2}\log \left(\sqrt{2\pi} \Gamma\left(\frac{3}{4}\right) / \Gamma\left(\frac{1}{4}\right)\right)$$ and this is evaluated in the question linked by "nospoon" in his comment.
It is however desirable to have a direct proof that the integral is equal to the infinite series in question without making use of the answers of the linked question. 
