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## Main Question or Discussion Point

I think it is a good way to obtain the formula for Pi(x),where

Pi(x)={Sum over primes<x}of 1

We know that S(p)Exp(-sp)=Int(0,infinite)Pi(x)Exp(-sx)..where s is a parameter s>0 and S(p) means sum over all primes.

But Int(0,infinite)Pi(x)Exp(-sx) is just the Laplace transform (will be donoted by L) of the function Pi(x) so we would have.

S(p)Exp(-sp)=L(Pi(x)) or taking the inverse

L**(-1)Exp(-sp)=Pi(x). (1)

To get the sum S(p)Exp(-sp) we will use the formula

S(0,infinite){Pi(x)-Pi(x-1)+1}(-1)**nZ**n=2Z**2-S(p)Z**p+1/1+Z**p (2)

as you can check a(n)={Pi(n)-Pi(n-1)+1} we would have an alternate series so we could use an Euler transform to improve the convergence and from (2) we could obtain S(p)Exp(-sp) and substitute it into the formula given in (1) to obtain an aproach to formula Pi(x).

I submitted it by e-mail and letter to several universities in my country (spain) but i did not have any answer..i do not know if there is something wrong in my method or..if it is useles...or if perhaps will be only snobism..to accept that a non-mathematician person this important formula....

Pi(x)={Sum over primes<x}of 1

We know that S(p)Exp(-sp)=Int(0,infinite)Pi(x)Exp(-sx)..where s is a parameter s>0 and S(p) means sum over all primes.

But Int(0,infinite)Pi(x)Exp(-sx) is just the Laplace transform (will be donoted by L) of the function Pi(x) so we would have.

S(p)Exp(-sp)=L(Pi(x)) or taking the inverse

L**(-1)Exp(-sp)=Pi(x). (1)

To get the sum S(p)Exp(-sp) we will use the formula

S(0,infinite){Pi(x)-Pi(x-1)+1}(-1)**nZ**n=2Z**2-S(p)Z**p+1/1+Z**p (2)

as you can check a(n)={Pi(n)-Pi(n-1)+1} we would have an alternate series so we could use an Euler transform to improve the convergence and from (2) we could obtain S(p)Exp(-sp) and substitute it into the formula given in (1) to obtain an aproach to formula Pi(x).

I submitted it by e-mail and letter to several universities in my country (spain) but i did not have any answer..i do not know if there is something wrong in my method or..if it is useles...or if perhaps will be only snobism..to accept that a non-mathematician person this important formula....