Proof that the logarithmic integral $\int\limits_0^x \frac{\mathrm{d}t}{\log t}$ exists Considering the integral logarithm
$$\text{li}(x)=\int\limits_0^x\frac{\mathrm{d}t}{\log t}$$
it is not clear what happens around the value $t=1$, so this is defined as
$$\text{li}(x)=\lim\limits_{\epsilon\to0} \Bigg(\int\limits_0^{1-\epsilon}\frac{\mathrm{d}t}{\log t} + \int\limits_{1+\epsilon}^x \frac{\mathrm{d}t}{\log t}\Bigg)$$
or for sake of simplicity, we regard
$$\text{Li}(x)=\int\limits_2^x \frac{\mathrm{d}t}{\log t}$$
to avoid the value $t=1$.
However, I have never seen an actual proof why the above limit should exist and I dont think it is clear it does. Also the approach just using $\text{Li}(x)$ only works because
$$\text{Li}(x)=\text{li}(x)-\text{li}(2)$$
where $\text{li}(2)$ is a finite value, but here it is also not clear why this is the case.
Maybe anyone could provide a argument why any of the aboves statements is true.
 A: I just will answer the first question as the second follows trivially from the existence of the principal value. Making the change of variable $s=\frac1{t}$ we have that
$$
\int_{0}^{1-\epsilon }\frac1{\log t}\,d t=-\int_{\frac1{1-\epsilon }}^{+\infty }\frac1{s^2 \log s}\,d s
$$
Therefore for $y>1$ and $\epsilon \in(0,1)$ we have that
$$
\begin{align*}
\operatorname{P.V.}\int_{0}^y \frac{dt}{\log t}&:=\lim_{\epsilon \to 0^+}\left(\int_{0}^{1-\epsilon }\frac{dt}{\log t}+\int_{1+\epsilon }^{y}\frac{dt}{\log t}\right)\\
&=\lim_{\epsilon \to 0^+ }\left(\int_{1+\epsilon }^{y}\frac{dt}{\log t}-\int_{\frac1{1-\epsilon }}^{+\infty }\frac{dt}{t^2\log t}\right)\\
&=\lim_{\epsilon \to 0^+}\left(\int_{1+\epsilon }^{y}\frac{t^2 \,dt}{t^2\log t}-\int_{\frac1{1-\epsilon }}^{+\infty }\frac{dt}{t^2\log t}\right)\\
&=\lim_{\epsilon \to 0^+}\left(\int_{1+\epsilon }^{\frac1{1-\epsilon }}\frac{dt}{\log t}+\int_{\frac1{1-\epsilon }}^{y}\frac{t^2-1}{t^2\log t}d t-\int_{y}^{+\infty }\frac{dt}{t^2 \log t}\right)\\
&=\int_{1}^{y}\frac{t^2-1}{t^2\log t}d t-\int_{y}^{+\infty }\frac1{t^2 \log t}dt\tag{*}
\end{align*}
$$
where in the last equality I used the fact that when $\epsilon \in(0,1)$ we have that $1+\epsilon <\frac1{1-\epsilon }$ and consequently
$$
\lim_{\epsilon \to 0^+}\left| \int_{1+\epsilon }^{\frac1{1-\epsilon }}\frac{dt}{\log t} \right|\leqslant \lim_{\epsilon \to 0^+}\frac{\frac1{1-\epsilon }-(1+\epsilon )}{\log (1+\epsilon )}=0
$$
Then finally to see that the first integral in (*) is a finite number it is enough to observe that
$$
\lim_{t\to 1^+}\frac{t^2-1}{\log t}=\lim_{t\to 1^+}\frac{2t}{\frac1{t}}=2
$$
∎
