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Re: [PrimeNumbers] A mystery about the logarithmic integral function Li(x) ?>>>

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  • Ehsan Mohammadi
    *********************************************************************************** what legendre probably did is this : he knew that Pi(x) is approximately
    Message 1 of 2 , Feb 17, 2005
      what legendre probably did is this :
      he knew that Pi(x) is approximately " x/ln(x)"
      so :
      d(Pi(x)) 1 1
      ---------- = ------- - ------------
      d(x) ln(x) ln(x)^2

      know if we integrate again --->>> Pi(x)= Li(x) - int(1/ln(x)^2)

      he found that the second part is not valuable.so Li(x) was discoverd.


      Jean-Pierre Pratali <jpierre.pratali@...> wrote:

      ----- Original Message -----
      From: "Jud McCranie" <j.mccranie@...>
      To: "Jean-Pierre Pratali" <jpierre.pratali@...>
      Cc: <primenumbers@yahoogroups.com>
      Sent: Sunday, February 13, 2005 3:31 AM
      Subject: Re: [PrimeNumbers] A mystery about the logarithmic integral
      function Li(x) ?

      > I use formula (6) and (7) at
      > http://mathworld.wolfram.com/LogarithmicIntegral.html
      > Note: according to that reference, li(x) is the logarithmic integral,
      > is the
      > Polylogarithm.
      > We would like to see it - that could be important.
      Thanks. I now understand that formula (6) (Nielsen-Ramanujan) of MathWorld
      is the one used to compute li(x) (a short time ago it was written Li(x)).

      I'll try to set out briefly how I established my own formula:
      1. Taking li(x), I had the idea to integrate "by parts" :
      li(x)=x/logx+(Integral of 1/logx^2), and then to reiterate that at each time
      on the remaining integral :
      2. After n times : li(x)=x/logx(Sum of k!/logx^k, k from 0 to n-1)
      +n!(Integral of 1/logx^(n+1))
      3. I saw that, first that I could not get rid of an integral, and secondly
      that the series k!/logx^k is (very) divergent!
      4. So I had the thought to truncate it at its least term : using the
      logarithmic derivate of k!/logx^k and the simplified Stirling formula
      k(logk-1) for logk!, it's easy to show that this derivate is nil for k=logx.
      5. Finally I built the sum x/logx(Sum of k!/logx^k, k from 0 to |logx|),
      |logx| meaning "integer part" of logx (preferably "floor").
      That's all.
      Using Mathematica :
      n = ...;
      L := N[LogIntegral[n], 100]; N1 := Floor[L];
      S := 0;
      p = N[n/Log[n], 100];
      For [k = 0, k < Floor[Log[n]], k++, S = S + N[k!/Log[n]^k, 100]];
      S = p*S; N2 := Floor[S];
      Print[N1, " ", N2, " ", N2 - N1]

      it's easy to note that there is a nearly perfect sticking between N1 and N2.
      The absolute value of the difference is never greater than 1 (and probably
      for the little numbers only).


      Jean-Pierre Pratali

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