Consider the following function:
$$F(z) = \omega(z)\sin^2\left(\frac{c\Gamma(z)}{z}\right)$$
Here, $\omega(z)$ is a weight we have to construct and $c$ is a constant.
The following three conditions should meet for $\omega(z)$:
- $$\omega(z)>\frac{1}{z},\ \forall z\in\mathbf{R}$$
( 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 $x\to\infty$.
Question : Can we Explicitly construct $\omega(z)$.
If anyone could omit the first condition and could find the weight satisfying condition 2 and 3 please mention.
(Is it even possible?)
See this MSE post ( see the part after UPDATE) and this MSE post for more details.
The question is inspired by following analysis:
\begin{align}f(x) = {} & \sin^2\left(\frac{π\Gamma(x)}{2x}\right)\\ \sum_{k=2}^p f(k)= {} & \frac{f(2) +f(p)}2 + \int_2^p f(x) \, dx \\ & {}+ i\int_0^∞\frac{f(2+iy) − f(2−iy)}{e^{2πy }− 1} \, dy +i \int_0^∞\frac{f(p-iy) − f(p+iy)}{e^{2πy }− 1} \, dy \end{align}
As we can see the above summation is 'sort of' a prime counting function.
I'm trying to prove the infinitude of primes by trying to show that the above sum diverges as p tends to infinity.
In doing so I faced the following: The complex integral in the RHS show oscillation of increasing amplitude. So to convert the summation into real integral I tried to attach weight such that the complex integral tends to zero as p tends to infinity and so we only remained with real integral and some constant terms.
