I would need some help to work with the following integral:
$$f(x) = \int_2^\infty \log (1-x^t) dt ,\ \ \ \ \ \ \ \ \ |x|<1$$
I would like to get a closed form or something similar (which seems to be impossible), but any other type of exact equivalent expression to work with would be great.
Making a change of variables seems not to help much. I also tried to evaluate it as a complex integral, but the path of integration $[2, \infty)$ is not the easiest to work with.
Any idea will be welcomed.
By the substitution $u=x^t$, $$\begin{align} \int \ln(1-x^t)dt & =\frac1{\ln x}\int \frac{\ln(1-u)}{u}du\\ & =\frac1{\ln x}\int \frac{-\text{Li}_1(u)}udu\\ &=-\frac{\text{Li}_2(u)}{\ln x}+C\\ &=-\frac{\text{Li}_2x^t}{\ln x}+C\\ \end{align} $$
Applying the limits, one obtains $$\color{red}{\int^\infty_2\ln(1-x^t)dt=\frac{\operatorname{Li}_2 x^2}{\ln x}}$$
The expression has a value with a nice closed form for some special $x$:
$$\int^\infty_2\ln(1-\sqrt 2^{-t})dt=-\frac{\pi^2}{6\ln 2}+\ln 2$$ $$\int^\infty_2\ln(1-(\sqrt\phi^{-1})^t)dt=-\frac{\pi^2}{5\ln\phi}+2\ln\phi$$ $$\int^\infty_2\ln(1-{\phi}^{-t})dt=-\frac{\pi^2}{15\ln\phi}+\ln\phi$$
Also, there is an interesting limit: $$\lim_{x\to1^-}\ln x \int^\infty_2\ln(1-x^t)dt =\frac{\pi^2}6$$
