# Domain of Indefinite Integral - Fundamental Theorem of Calculus

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.

• How do you extend $\ln(1-x)/x$ to $x=1$? – Eclipse Sun Jan 27 '19 at 21:11
• In fact, $-\int_0^1 \frac{\ln (1-t)}{t}dt =\zeta(2)=\frac{\pi^2}{6}$. So you are right. – Myeonghyeon Song Jan 27 '19 at 21:13
• @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. – Mário Belga Jan 27 '19 at 21:18
• 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.) – Myeonghyeon Song Jan 27 '19 at 21:24
• @EclipseSun Wait, you're right. It is divergent at $1$. Then how can F be extended by continuity in $x=1$(left)? – Mário Belga Jan 27 '19 at 21:24

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 $$\int_{x}^{1}\ln\left(t\right)\text{d}t \underset{x \rightarrow 0}{\rightarrow}-1$$
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.