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: π(''n'') ~ Li(n) = ∫<sub>''2''</sub><sup>''n''</sup> 1/ ln ''t'' d''t''.
This integral is in a connection with ''integral exponential function'' such as that li(''x'') = Ei (ln ''x''). If we substitute ''x'' with e<sup>''u''</sup>, we get a
:li(e<sup>''u''</sup>) = γ + ln ''u'' + ''u'' + ''u''<sup>2</sup>/2 · 2! + ''u''<sup>3</sup>/3 · 3! + ''u''<sup>4</sup>/4 · 4! - ...,
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