Logarithmic integral function: Difference between revisions

Revision as of 02:39, 27 March 2021

Revision as of 15:42, 21 April 2021

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The asymptotic behavior for ''x'' → ∞ 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

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This gives the following more accurate asymptotic behaviour:

:<math> \operatorname{li}(x) - \frac{x\over}{ \ln x} = O \left( \frac{x\over }{\ln^2 x} \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]].

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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>.