Ways to "milk" the integral $\int^{1}_{0}\ln\left(\phi^{x}-\phi^{-x}\right)\ln\left(\phi\right)dx=-\frac{\pi^2}{20}$

Question

After reading the MSE post on "Integral Milking", my first instinct was to try it out on one of my favorite integrals:

$$\int^{\ln{\phi}}_{0}\ln\left(e^{x}-e^{-x}\right)dx=-\frac{\pi^2}{20}$$

which is equivalent to

$$\int^{\ln{\phi}}_{0}\ln\left(2\sinh{x}\right)dx=-\frac{\pi^2}{20}$$

I personally prefer the first one.

After trying to milk the integral with various techniques for a while, I came up with

$$\int^{1}_{0}\ln\left(\phi^{x}-\phi^{-x}\right)\ln\left(\phi\right)dx=-\frac{\pi^2}{20}$$

I like this one a lot except that it is multiplied by the constant $\ln\left(\phi\right)$

My solution, which made use of the change of base formula was not as elegant as I had envisioned:

$$\int_{0}^{1}\log_{\sqrt[\ln\left(\phi\right)]{e}}\left(\phi^{x}-\phi^{-x}\right)dx=-\frac{\pi^2}{20}$$

My "milking" of this integral used very simple techniques. I am curious what can be done with more advanced ones so my question is this:

What are other ways to "milk" the integral $\int^{1}_{0}\ln\left(\phi^{x}-\phi^{-x}\right)\ln\left(\phi\right)dx=-\frac{\pi^2}{20}$?

Answer 1

Making the problem more general $$I=\int \log (a) \log \left(a^x-a^{-x}\right)\,dx$$ $$I=-\text{Li}_2\left(-a^x\right)+\text{Li}_2\left(1-a^x\right)+\frac {1}{2} \log \left(a^x\right) \log \left(a^{-x}+a^x-2\right)$$ $$\color{blue}{J=\int_0^1 \log (a) \log \left(a^x-a^{-x}\right)\,dx=}$$ $$\color{blue}{\text{Li}_2(1-a)-\text{Li}_2(-a)+\frac{1}{2} \log (a) \log \left(\frac{(a-1)^2}{a}\right)-\frac{\pi ^2}{12}}$$ and for $a=\phi$ the nice result.