improper integrals - Closed form for $I(a)=int_0^infty lnleft(tanh(ax)right)dx$?

I have been messing around with this integral that has some particular special values $$I(a)=\int_0^\infty \ln\left(\tanh(ax)\right)dx$$

I found that $$I(1)=-\frac{\pi^2}{8}$$
$$I\left(\frac{1}{2}\right)=-\frac{\pi^2}{4}$$
$$I\left(\frac{1}{4}\right)=-\frac{\pi^2}{2}$$
$$...$$ and so on. It appears that $I(2^{-n})=-2^{n-3}\pi^2$. Can anyone explain how to derive a general closed form for $I(a)$ or at least why $I(2^{-n})$ takes on the particular values above?

Answer

We have $$I^\prime(a)=\int_0^\infty\frac{x\operatorname{sech}^2 ax dx}{\tanh ax}=\int_0^\infty\frac{2x dx}{\sinh 2ax}=\frac{1}{2a^2}\int_0^\infty\frac{y dy}{\sinh y},$$ so constants $A,\,B$ exist with $$I(a)=A-\frac{B}{a}.$$ From your results we can infer $A=0,\,B=\frac{\pi^2}{8}$.

Edit: a slicker way is to write $$\int_0^\infty\ln\frac{1-e^{-2ax}}{1+e^{-2ax}}dx=-2\int_0^\infty\sum_{n=0}^\infty\frac{1}{2n+1}e^{-(4n+2)ax} dx=-\frac{\pi^2}{8a}.$$