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,

Li(x)

> 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).

Thanks.

Jean-Pierre Pratali

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