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