Logarithmic integral function $\mathrm{li}(x)$ I was reading the book of Zorich "Mathematical Analysis" about improper integrals and came across interesting question. First of all let me give some preliminaries.

In the case when the integrand is not bounded in a neighborhood of one
of the interior points $\omega$ of the closed interval $[a,b]$, we set
 $$\int \limits_{a}^{b}f(x)dx:=\int \limits_{a}^{\omega}f(x)dx+\int
 \limits_{\omega}^{b}f(x)dx,\quad \quad (6.81)$$ requiring that both of
the integrals on the right-hand side exist.
Besides (6.81), there is a second convention about computing the
integral of a function that is unbounded in a neighborhood of an
interior point $\omega$ of a closed interval of integration. To be
specific, we set $$\text{PV}\int \limits_{a}^{b}f(x)dx:=\lim
 \limits_{\delta\to 0+}\left(\int
 \limits_{a}^{\omega-\delta}f(x)dx+\int
 \limits_{\omega+\delta}^{b}f(x)dx\right),\quad \quad (6.82)$$ if the
limit on the right-hand side exists.
Example 21. The precise definition of the logarithmic integral can now be written
as  $$\mathrm{li}(x) =
     \begin{cases}
     \int \limits_{0}^{x}\frac{dt}{\ln t}, & \text{if }0<x<1, \\
     \mathrm{PV} \int \limits_{0}^{x}\frac{dt}{\ln t}, & \text{if }1<x.
     \end{cases}$$
    In the last case the symbol PV refers to the only interior singularity on the interval $(0,x]$, which is located at $1$. We
remark that in the sense of the definition in formula $(6.81)$ this
integral is not convergent.

I would like to clarify some questions related with the $\mathrm{li}\ (x)$.

*

*So as far as I inderstand $\mathrm{li}(x)$ is a real-valuable function defined on $(0,1)\cup (1,+\infty)$, right?


*For $x\in (0,1)$ we define $\mathrm{li}(x)=\int\limits_{0}^{x}\frac{dt}{\ln t}$. Do we consider this integral as improper or definite? The integrand $\frac{1}{\ln t}$ is not defined at $0$ but since $\lim \limits_{t\to 0+}\frac{1}{\ln t}=0$, then we can define $\frac{1}{\ln 0}=0$ to make it continuous on $[0,x]$. So is it improper or definite integral?


*Why $\mathrm{li}(x)$ is not defined at 1? What if we set $\mathrm{li}(1):=\int \limits_{0}^{1}\frac{dt}{\ln t}$? But I realised that this integral is divergent since $\ln t\sim t-1$ as $t\to 0$. Is that argument correct?


*I understand that if $x>1$, then $\int \limits_{0}^{x}\frac{dt}{\ln t}$ is divergent in the sense of formula $(6.81)$ since integral $\int \limits_{1}^{x}\frac{dt}{\ln t}$ is divergent since $\ln t\sim t-1$ as $t\to 0$. But why does $ \mathrm{PV} \int \limits_{0}^{x}\frac{dt}{\ln t}:=\lim \limits_{\epsilon\to 0+}\bigg(\int \limits_{0}^{1-\epsilon}\frac{dt}{\ln t}+\int \limits_{1+\epsilon}^{x}\frac{dt}{\ln t}\bigg)$ exist?
I'd be very grateful if someone can answer my questions!
 A: As per request we need to show that for $x>1$:
$$\mathrm{PV} \int \limits_{0}^{x}\frac{dt}{\ln t}:=\lim \limits_{\epsilon\to 0+}\bigg(\int \limits_{0}^{1-\epsilon}\frac{dt}{\ln t}+\int \limits_{1+\epsilon}^{x}\frac{dt}{\ln t}\bigg)$$ exists
First noting that $\int \limits_{0}^{c}\frac{dt}{\ln t}+\int \limits_{1+d}^{x}\frac{dt}{\ln t}$ exist for all fixed $0<c<1, 1<1+d<x$, it is enough to show that
$$\mathrm{PV} \int \limits_{c}^{1+d}\frac{dt}{\ln t}:=\lim \limits_{\epsilon\to 0+}\bigg(\int \limits_{c}^{1-\epsilon}\frac{dt}{\ln t}+\int \limits_{1+\epsilon}^{1+d}\frac{dt}{\ln t}\bigg)$$ exists with convenient $c,d$ as above (we will take $c=1/2$ and $d=1/2$ if $x \ge 3/2$, while $d=x-1<1/2$ will do otherwise)
Now with $t-1=a, |a|\le 1/2$ one has $|\log (1+a)-a|\le 2a^2$ say and also $|\log (1+a)| \ge |a|/2$ so $|\frac{1}{\log (1+a)}-\frac{1}{a}| \le 4$ hence $g(a)=\frac{1}{\log (1+a)}-\frac{1}{a}$ is integrable on $[-1/2,1/2]$
(it is actually continuous at $0$ but we do not need that - however if L'Hopital is handier than Taylor series estimates, one can easily show that $\frac{a-\log (1+a)}{a\log (1+a)}$ has a finite limit at zero so $g$ is continuous there when we define it by that limit)
In particular, this means that $\frac{1}{\log (t)}-\frac{1}{t-1}$ is integrable on $[1/2, 3/2]$ so $$\lim \limits_{\epsilon\to 0+}\bigg(\int \limits_{c}^{1-\epsilon}(\frac{dt}{\ln t}-\frac{dt}{t-1})+\int \limits_{1+\epsilon}^{1+d}(\frac{dt}{\ln t}-\frac{dt}{t-1})\bigg)= \int \limits_{c}^{1+d}(\frac{dt}{\ln t}-\frac{dt}{t-1})$$ hence it exists as we take $c=1/2, d \le 1/2$ by the above
So we just need to show that $$\mathrm{PV} \int \limits_{c}^{1+d}\frac{dt}{t-1}:=\lim \limits_{\epsilon\to 0+}\bigg(\int \limits_{c}^{1-\epsilon}\frac{dt}{t-1}+\int \limits_{1+\epsilon}^{1+d}\frac{dt}{t-1}\bigg)$$ exists and by a direct computation the sum of the two integrals is $(\log \epsilon - \log 1/2)+(\log d-\log \epsilon)=\log d-\log 1/2$ so the limit as $\epsilon \to 0$ obviously exists and we are done!
