If this could be achieved then we are able to get the estimates on primes using purely analytic information ( no number theoretic information like Euler product ) . And this seems (although extremely hard but,) possible as due to relatively elementary nature of summand and integrals .
Question : Can we get an 'Sharp' estimates on the sum w.r.t parameters $p$ and $s$?
TheI try to construct the $\omega(z)$ s.t.
$$\sum_2^n F(n)= \int_2^n F(x)dx + A$$
Here A is constant.
For this to be true the following twothree conditions should meet for $\omega(z)$ in context of APSF:
$$\lim_{ y→∞}|F(x ± iy)|e^{−2πy }= 0$$
$$\int_0^\infty |F(x + iy) − F(x − iy)|e^{−2πy} dy<+\infty$$ for every $x≥1$ and tends to zero as $x\to\infty$.
$$\omega(z)>\frac{1}{z},\ \forall z\in\mathbf{R}$$
( More generally this condition is added for divergence of $\int_c^\infty F(x)dx$ So , $\omega(z)$ can even be complex valued for real domain as long as the given integral is divergent )
$$\lim_{ y→∞}|F(x ± iy)|e^{−2πy }= 0$$
$$\int_0^\infty |F(x + iy) − F(x − iy)|e^{−2πy} dy<+\infty$$ for every $x≥1$ and tends to zero as $x\to\infty$.
Can we Explicitly construct $\omega(z)$?
Even if one can omit 1st condition and able to construct the weight s.t. it follows condition 2,3 please mention. ( i.e in this case the integral $\int_2^\infty F(x)dx$ is convergent.
Also one can generalize further:
$F(z) = {\phi(\sin^2[π\Gamma(z)/(2z)])}$
S.t.
(1) ϕ(x)=0 if x is zero ; and 'suitably' finite otherwise
(Here , 'suitably' means a value which guarantees the expected divergence of sum (very close to 1 or greater than or equal to 1) )
(2) condition (3) holds for such function.
Could we make above analysis workable?
Also, I think the condition (3) is very hard to achieve because of the complex roots of equations $\Gamma(z)/(2z)$=even integers.
HowIf this could be achieved then we are able to constructget the estimates on primes using purely analytic information $\omega(z)$?( no number theoretic information like Euler product ) . And this seems (although extremely hard but,) possible as due to relatively elementary nature of summand and integrals .
Note: I know this question received negative reviews (due to both my behavior and insufficient information ) but please consider the importance of question.