- #1

- 492

- 0

where 8 means infinity, and the `means derivative:

now we set in (1) f(x)=x^-s so we would have that:

Sum(p)p^-s=g(s)=-sInt(1,8)Pi(x)x^-(s+1) (2)

(2) has an inverse transform of the form:

-Int(C)g(s)x^s/2pi.is=Pi(x) C=is the real line conecting c-i8 and c+i8

Another useful formula is that:

Sum(p)p^-s=Sum(1,8)mu(n)LnR(ns)/n with R(s)=1+2^-s+3^-s+.......

On the other hand we have that

Sum(1,8)mu(n)/n^(4-q)=1/R(4-q)=M[w(x)mu(x)/x^3]

where M is the Mellin Transform and w(x)=sum(1,8)d(x-n) with d the delta function,reagruping all we have that:

PI(x)=(1/4Pi^2)Int(0,8)In(d-i8,d+i8)Int(c-i8,c+i8)(n^-q+2)x^sLnR(ns)/R(4-q)s

That,s an integral for PI(x) exact as you can see know there is no problem with the term Pi(x)/x^4 you have only neeed to know the integral with an accuracy of 0.1 (rror term=0.1) to have PI(x)...

Before critizying my method i would like to say that the "real" value of Pi(x) set by Pi(x)=x/ln(x) has also fails you need for example to know ln(x) with an accuracy of 10^-100 to calculate pi(10^100) and so on...

P.D:if someone can help me to put it into latex i would be very grateful :)

what do you think of this new integral?...