Closed form for the series $\sum_{k=1}^\infty (-1)^k \ln \left( \tanh \frac{\pi k x}{2} \right)$ 
Is there a closed form for: $$f(x)=\sum_{k=1}^\infty (-1)^k \ln \left( \tanh \frac{\pi k x}{2} \right)=2\sum_{n=0}^\infty \frac{1}{2n+1}\frac{1}{e^{\pi (2n+1) x}+1}$$

This sum originated from a recent question, where we have:
$$f(1)= -\frac{1}{\pi}\int_0^1 \ln \left( \ln \frac{1}{x} \right) \frac{dx}{1+x^2}=\ln \frac{\Gamma (3/4)}{\pi^{1/4}}$$
If we differentiate w.r.t. $x$, we obtain:
$$f'(x)=\sum_{k=1}^\infty (-1)^k \frac{\pi k}{\sinh \pi k x}$$
There is again a closed form for $x=1$ (obtained numerically):
$$f'(1)=-\frac{1}{4}$$
So, is there a closed form or at least an integral definition for arbitrary $x>0$?

The series converges absolutely (numerically at least):
$$\sum_{k=1}^\infty \ln \left( \tanh \frac{\pi k x}{2} \right)< \infty$$
Thus, this series can also be expressed as a logarithm of an infinite product:
$$f(x)=\ln \prod_{k=1}^\infty \tanh (\pi k x) - \ln \prod_{k=1}^\infty \tanh  \left( \pi (k-1/2) x \right)$$
$$e^{f(x)}=  \prod_{k=1}^\infty \frac{\tanh (\pi k x)}{\tanh  \left( \pi (k-1/2) x \right)}$$
This by the way leads to:
$$\prod_{k=1}^\infty \frac{\tanh (\pi k)}{\tanh  \left( \pi (k-1/2) \right)}=\frac{\pi^{1/4}}{\Gamma(3/4)}$$
I feel like there is a way to use the infinite product form for $\sinh$ and $\cosh$:
$$\sinh (\pi x)=\pi x \prod_{n=1}^\infty \left(1+\frac{x^2}{n^2} \right)$$
$$\cosh (\pi x)=\prod_{n=1}^\infty \left(1+\frac{x^2}{(n-1/2)^2} \right)$$
 A: Let’s use $~\displaystyle\prod\limits_{k=1}^\infty (1+z^k)(1-z^{2k-1}) =1~$ . $\enspace$ (It's explained in a note below.)
For $~z:=q^2~$ and $~q:=e^{-\pi x}~$ with $~x>0~$ we get 
$\displaystyle e^{f(x)} = \prod\limits_{k=1}^\infty\frac{\tanh(k\pi x)}{\tanh((k-\frac{1}{2})\pi x)} = \prod\limits_{k=1}^\infty\frac{ \frac{q^{-k}-q^k}{q^{-k}+q^k} }{ \frac{q^{\frac{1}{2}-k}-q^{k-\frac{1}{2} }}{q^{\frac{1}{2}-k}+q^{k-\frac{1}{2}}} } = \prod\limits_{k=1}^\infty\frac{(1-q^{2k})(1+q^{2k-1})}{(1+q^{2k})(1-q^{2k-1})} =$
$\displaystyle = \prod\limits_{k=1}^\infty (1-q^{2k})(1+q^{2k-1})^2 = \sum\limits_{k=-\infty}^{+\infty} q^{k^2} = \vartheta(0;ix)$
The “closed form” for $\,f\,$ is:

$$f(x) = \ln\vartheta(0;ix)$$ 

Please see e.g. Theta function .

Note:
$\displaystyle\prod\limits_{k=1}^\infty (1+z^k)(1-z^{2k-1}) =1$
$\Leftrightarrow\hspace{2cm}$ (logarithm) 
$\displaystyle \sum\limits_{k=1}^\infty \sum\limits_{v=1}^\infty \frac{(-1)^{v-1}z^{kv}}{v} = \sum\limits_{k=1}^\infty \ln(1+z^k) = -\sum\limits_{k=1}^\infty \ln(1-z^{2k-1}) = \sum\limits_{k=1}^\infty \sum\limits_{v=1}^\infty \frac{z^{(2k-1)v}}{v}$
$\Leftrightarrow\hspace{2cm}$ (exchanging the sum symbols which is valid for $~|z|<1~$ 
$\hspace{2.7cm}$ and using $~\displaystyle\frac{x}{1-x}=\sum\limits_{k=1}^\infty x^k~$)
$\displaystyle\sum\limits_{v=1}^\infty \frac{(-1)^{v-1}}{v}\frac{z^v}{1-z^v} = \sum\limits_{v=1}^\infty \frac{1}{v}\frac{z^v}{1-z^v} - 2\sum\limits_{v=1}^\infty \frac{1}{2v}\frac{z^{2v}}{1-z^{2v}} = \sum\limits_{v=1}^\infty \frac{1}{v}\frac{z^v}{1-z^{2v}}$
