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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: $$\lim_{x\to1^-}\ln x \int^\infty_2\ln(1-x^t)dt =\frac{\pi^2}6$$

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: $$\lim_{x\to1^-}\ln x \int^\infty_2\ln(1-x^t)dt =\frac{\pi^2}6$$

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 $x$:

$$\int^\infty_2\ln(1-\sqrt 2^{-t})dt=\frac{\pi^2}{6\ln 2}-\ln 2$$ $$\int^\infty_2\ln(1-\sqrt\phi^{-t})dt=-\frac{\pi^2}{10\ln\phi}-\ln\phi$$ $$\int^\infty_2\ln(1-{(\phi)}^{-t})dt=\frac{\pi^2}{30\ln\phi}-\frac{\ln\phi}2$$

Also, there is an interesting limit: $$\lim_{x\to1^-}\ln x \int^\infty_2\ln(1-x^t)dt =\frac{\pi^2}6$$

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: $$\lim_{x\to1^-}\ln x \int^\infty_2\ln(1-x^t)dt =\frac{\pi^2}6$$

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 at $x=\frac1{\sqrt2}$:

$$\int^\infty_2\ln(1-\sqrt 2^{-t})dt=\frac{\pi^2}{6\ln 2}-\ln 2$$

Also, there is an interesting limit: $$\lim_{x\to1^-}\ln x \int^\infty_2\ln(1-x^t)dt =\frac{\pi^2}6$$

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 $x$:

$$\int^\infty_2\ln(1-\sqrt 2^{-t})dt=\frac{\pi^2}{6\ln 2}-\ln 2$$ $$\int^\infty_2\ln(1-\sqrt\phi^{-t})dt=-\frac{\pi^2}{10\ln\phi}-\ln\phi$$ $$\int^\infty_2\ln(1-{(\phi)}^{-t})dt=\frac{\pi^2}{30\ln\phi}-\frac{\ln\phi}2$$

Also, there is an interesting limit: $$\lim_{x\to1^-}\ln x \int^\infty_2\ln(1-x^t)dt =\frac{\pi^2}6$$