Domain of Indefinite Integral - Fundamental Theorem of Calculus

Question

What's the domain of this function?

$F(x) = -\int _0^x\:\frac{\ln\left(1-t\right)}{t}dt$

In the 'Answers' Section it says $(-\infty,1)$ but I think it is $(-\infty,1]$ since $\frac{\ln\left(1-t\right)}{t}$ is defined and continuous in $\left(-\infty \:,\:0\right)\cup \left(0,\:1 \:\right)$ and continuous by extension in $x=0$ and $x=1$.

I don't know why I should treat $1$ differently from $0$, thank you for your time. Edit: I miss calculated $\lim _{t\to 1}\:\frac{log\left(1-t\right)}{t}$ and it happens to be divergent. Now I'm trying to show that $F$ is continuous by extension in $x=1$ (to the left) but I don't even know how that since the limit happens to be divergent.

Answer 1

The thing is for the point $1$ you have letting $x=1-h$ $$ \frac{\ln\left(1-x\right)}{x}=\frac{\ln\left(h\right)}{1-h}\underset{(0)}{\sim}\ln\left(h\right) $$

And $\displaystyle \left|\ln\left(h\right)\right|\underset{(0)}{=}o\left(\frac{1}{\sqrt{h}}\right)$ and you know that $h \mapsto \frac{1}{\sqrt{h}}$ is integrable on $\left]0,1/2\right]$. With equivalence criteria, $\displaystyle x \mapsto \frac{\ln\left(1-x\right)}{x}$ is integrable on $\left[1/2,1\right[$.

It is not because the integrand diverges that the integral does not exist, for example $$ \int_{0}^{1}\ln\left(x\right)\text{d}x=-1 $$ while $\ln\left(x\right) \underset{x \rightarrow 0}{\rightarrow}-\infty$ This, just because $$ \int_{x}^{1}\ln\left(t\right)\text{d}t \underset{x \rightarrow 0}{\rightarrow}-1 $$

Answer 2

We have $\displaystyle \int \ln(t) dt = x \ln(x) - x$, then we can calculate integrale of $\ln(x)$ in local $0^+$

$\dfrac{\ln(1-t)}{t} \sim \ln(1-t)$ in local $1^-$

Then $F(x=1)$ is defined.