Can This 4-Dimensional Integral Provide a New Way to Calculate Pi(x)?

[eljose]

let,s suppose we know a method to calculate $$pi(e^t)$$ by means of a double complex integral..then we will need a time O(t^a) a>0 so to know the value of $$pi(u)$$ we will need time O(ln^a(u)) so this method will be better than other that goes like O(t^a).

 

[matt grime]

and...?

 

[eljose]

the final equation is:

$$\pi(e^u)=\frac{1}{2i\pi}\int_{c-i\infty}^{c+i\infty}t^p\frac{\int_0^{\infty}ln\zeta(x)x^{p-1}/x}{\int_0^{\infty}x^{p-1}/(exp(x)-1)}$$

now the time to calculate goes like O(ln(t)^a) with a>0

 

[eljose]

sorry it was the integral

$$\pi(2^u)=\frac{1}{2i\pi}\int_{c-i\infty}^{c+i\infty}t^p\frac{\int_0^{\infty}ln\zet a(x)x^{p-1}/x}{\int_0^{\infty}x^{p-1}/(exp(x)-1)}$$

 

[eljose]

$$\pi(e^t)=\frac{1}{2i\pi}\int_{c-i\infty}^{c+i\infty}t^p\frac{\int_0^{\infty}ln\zet a(x)x^{p-1}/x}{\int_0^{\infty}x^{p-1}/(exp(x)-1)}dp$$

there is and error where it puts lna(x) is $$ln\zeta(x)$$

i have obtained this integral from solving the integral eqauation for $$\pi(e^t)$$ that,s it:

$$ln\zeta(s)/s=\int_0^\infty}dt\pi(e^t)/(exp(st)-1)$$ now i think i,m the first in solving this integral equation