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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.

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    $\begingroup$ How do you extend $\ln(1-x)/x$ to $x=1$? $\endgroup$ – Eclipse Sun Jan 27 '19 at 21:11
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    $\begingroup$ In fact, $-\int_0^1 \frac{\ln (1-t)}{t}dt =\zeta(2)=\frac{\pi^2}{6}$. So you are right. $\endgroup$ – Myeonghyeon Song Jan 27 '19 at 21:13
  • $\begingroup$ @Song I don't think there would be an error on this worksheets. The next question is 'Show that F can be extended by continuity to the left of $x = 1$' which implies that x = 1 doesn't belong to the function's domain. I'm translating it, if someone doesn't understand just tell me. $\endgroup$ – Mário Belga Jan 27 '19 at 21:18
  • $\begingroup$ I think it is a matter of definition. If $-\int_0^x \frac{\ln(1-t)}{t}dt$ is defined as proper Riemann integral, then $x=1$ does not belong to the domain of $F$. (But if we allow improper Riemann integral, then $x=1$ belongs to the domain.) $\endgroup$ – Myeonghyeon Song Jan 27 '19 at 21:24
  • $\begingroup$ @EclipseSun Wait, you're right. It is divergent at $1$. Then how can F be extended by continuity in $x=1$(left)? $\endgroup$ – Mário Belga Jan 27 '19 at 21:24
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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 $$ \int_{x}^{1}\ln\left(t\right)\text{d}t \underset{x \rightarrow 0}{\rightarrow}-1 $$

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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.

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