# Talk:Logarithmic Integral as Non-Convergent Series

What the heck is $\ln 0$? --GFauxPas (talk) 23:25, 17 June 2013 (UTC)

Formally it should be $\lim_{x \to 0} \dfrac x {\log x} = 0$. Since $1/\log(t)$ isn't defined at 0, li$(z)$ is defined by:

$\ds \lim_{\epsilon \to 0} \int_\epsilon^z \frac {dt} t$

--Linus44 (talk) 11:19, 18 June 2013 (UTC)

For $z > 1$, it has to be understand as a Cauchy principal value:

$\ds {\rm li}\ z = \lim_{\epsilon\to 0}\left({\int_0^{1-\epsilon}\frac{\rd t}{\ln t} + \int_{1+\epsilon}^z\frac{\rd t}{\ln t}}\right)$

But all of this should be on a definition page. Aerendil97 (talk) 04:45, 22 June 2013 (UTC)

It all is. See Definition:Logarithmic Integral (as it is called on $\mathsf{Pr} \infty \mathsf{fWiki}$). --prime mover (talk) 23:19, 26 November 2015 (UTC)