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eljose

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**an integral expression for Pi(x)...**

hello i have discovered a new method to calculate [tex]\pi(e^x) [/tex] it runs in O(x^d) operations d>0 it is very simple:

Forst of all we have the integral for Pi(x) as knows:

[tex]Ln\zeta(s)=s\int_0^{\infty}\frac{\pi(x)}{x^{s}-1}dx [/tex]

we make the change of variable x=exp(exp(t)) and apply the integral transform to both sides:

[tex] \int_{-\infty}^{\infty}ds(2+is)^{-iw} [/tex]

now we have a double integral we express 2+is as exp(ln(2+is) we make the change of variable t+ln(2+is)=u t=v so finallly we would have:

[tex] \int_{-\infty}^{\infty}ds(2+is)^{-iw-1}Ln\zeta(2+is)=\int_{-i\infty}^{i\infty}\frac{r^{-iw}}{exp(r)-1}*\int_{-\infty}^{\infty}g(v)e^{iwv} [/tex]

that have the solution:(with g(t)=Pi(exp(exp(t))

[tex]g(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}dwe^{-iwt}F(w)/G(w) [/tex]

where we call F(w) and G(w) to:

[tex] G(w)=\int_{-i\infty}^{i\infty}\frac{r^{-iw}}{exp(r)-1} [/tex]

[tex] F(w)=\int_{-\infty}^{\infty}ds(2+is)^{-iw-1}Ln\zeta(2+is) [/tex]

so we have an expression for [tex]\pi(e^{e^t}) [/tex] so to calculate pi(x) we have to take Ln(Ln(t)) in our integral...

Edit:there are some integrals which lacks on the dx simbol well the symbol is the same as the variable (if the integral appears r is dr ,if is s then is ds..and so on)...

why my method is better than other?..well i would say several things...

a)it is an analityc method...you solve it by solving three integrals...

b)time employed:if the time to calculate the three integrals goes like O(x^d) d>0 it seems big but to calculate for example Pi(exp(exp(100) you only need to calculate the integral upto t=100 so is faster than other methods that give the equation for Pi(x).

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