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 $\cosh$:
$$\sinh (\pi x)=\pi x \prod_{n=1}^\infty \left(1+\frac{x^2}{n^2} \right)$$
$$\cosh (\pi x)=\prod_{n=1}^\infty \left(1+\frac{x^2}{(n-1/2)^2} \right)$$