Logarithmic integral function: Difference between revisions

Revision as of 03:19, 26 March 2018

Revision as of 04:06, 23 August 2018

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==Integral representation==

The logarithmic integral has an integral representation defined for all positive [[real number]]s {{mvar|x}} ≠ 1 by the [[integral|definite integral]]

:<math> {\rm operatorname{li} (x) = \int_0^x \frac{dt}{\ln t}~. </math>

Here, {{math|ln}} denotes the [[natural logarithm]]. The function {{math|1/ln(''t'')}} has a [[mathematical singularity|singularity]] at {{mvar|t}}&nbsp;=&nbsp;1, and the integral for {{mvar|x}}&nbsp;>&nbsp;1 has to be interpreted as a ''[[Cauchy principal value]]'',

:<math> {\rm operatorname{li} (x) = \lim_{\varepsilon \to 0+} \left( \int_0^{1-\varepsilon} \frac{dt}{\ln t} + \int_{1+\varepsilon}^x \frac{dt}{\ln t} \right)~.</math>

==Offset logarithmic integral==

The '''offset logarithmic integral''' or '''Eulerian logarithmic integral''' is defined as

:<math> \operatorname{\rm Li}(x) = \operatorname{\rm li}(x) - \operatorname{\rm li}(2) \, </math>

or, integrally represented

:<math> {\rm operatorname{Li} (x) = \int_2^x \frac{dt}{\ln t} \, </math>

As such, the integral representation has the advantage of avoiding the singularity in the ___domain of integration.

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which is valid for ''x''&nbsp;>&nbsp;0. This identity provides a series representation of li(''x'') as

:<math> {\rm operatorname{li} (e^u) = \hbox{Ei}(u) =

\gamma + \ln |u| + \sum_{n=1}^\infty {u^{n}\over n \cdot n!}

\quad \text{ for } u \ne 0 \; , </math>

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:<math>

{\rm operatorname{li} (x) =

\gamma

+ \ln \ln x

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The asymptotic behavior for ''x''&nbsp;→&nbsp;∞ is

:<math> {\rm operatorname{li} (x) = O \left( {x\over \ln x} \right) \; . </math>

where <math>O</math> is the [[big O notation]]. The full [[asymptotic expansion]] is

:<math> {\rm operatorname{li} (x) \sim \frac{x}{\ln x} \sum_{k=0}^\infty \frac{k!}{(\ln x)^k} </math>

or

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