Editing Logarithmic integral function (section)

== Asymptotic expansion ==
The asymptotic behavior for <math>x\to\infty</math> is
: <math> \operatorname{li}(x) = O \left( \frac{x }{\ln x} \right) . </math>
where <math>O</math> is the [[big O notation]]. The full [[asymptotic expansion]] is
: <math> \operatorname{li}(x) \sim \frac{x}{\ln x} \sum_{k=0}^\infty \frac{k!}{(\ln x)^k} </math>
or
: <math> \frac{\operatorname{li}(x)}{x/\ln x}  \sim  1 + \frac{1}{\ln x} + \frac{2}{(\ln x)^2} + \frac{6}{(\ln x)^3} + \cdots. </math>

This gives the following more accurate asymptotic behaviour:
: <math> \operatorname{li}(x) - \frac{x}{ \ln x} = O \left( \frac{x}{(\ln x)^2} \right) . </math>

As an asymptotic expansion, this series is [[divergent series|not convergent]]: it is a reasonable approximation only if the series is truncated at a finite number of terms, and only large values of ''x'' are employed. This expansion follows directly from the asymptotic expansion for the [[exponential integral]].

This implies e.g. that we can bracket li as:
: <math> 1+\frac{1}{\ln x} < \operatorname{li}(x) \frac{\ln x}{x} < 1+\frac{1}{\ln x}+\frac{3}{(\ln x)^2} </math>
for all <math>\ln x \ge 11</math>.