# Integral of $\log(1-x^t)$ with respect to $t$

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. Thank you in advantage.

• "which seems to be impossible"....right Aug 13 '18 at 0:16
• unless you want to do a complex integral you should ask that 0<x<1 instead of -1<x<1 as you currently have because say x=-1/2, then you have (-1/2)^(1/2) when t=1/2 Aug 13 '18 at 0:45
• @mathew Regarding the limits of integrations, $t \ge 2$ Aug 13 '18 at 0:48
• sorry, t=1/2 was a bad example, here is another, take t=5/2, this still gives the same problem if complex numbers appearing in the integrand Aug 13 '18 at 1:04

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: $$\lim_{x\to1^-}\ln x \int^\infty_2\ln(1-x^t)dt =\frac{\pi^2}6$$
• There is a typo in the backward substitution, $u \neq \log_x t$. Aug 13 '18 at 3:26