Explicit formula for zeta function with special type of weight

Question

Consider the following line of thinking:

$$\pi(x) = \operatorname{R}(x) - \sum_{\rho}\operatorname{R}(x^{\rho}) - \frac1{\ln x} + \frac1\pi \arctan \frac\pi{\ln x} $$

Here, $\operatorname{R}(x) = \sum_{n=1}^{\infty} \frac{ \mu (n)}{n} \operatorname{li}(x^{1/n})$

Now by Stieltjes Integral:

$\int_a^bf(x)d(π(x))=\sum_{a\leq{p}\leq{b}}f(p)$

i.e. the integral counts $f$-weighted primes

For our simplicity let's consider $f$ to be analytic for our domain

Now consider the following function :

$$f(x) = \sin^2\left(\frac{π\Gamma(x)}{2x}\right)\\\\$$

For every integer other than prime $f(x)$ is zero

Now as we can see for each prime $p\geq5$

$$\sin^2\left(\frac{π\Gamma(p)}{2p}\right)\ = \cos^2\left(\frac{π}{2p}\right)\\\\$$

Define: $\cos^2\left(\frac{π}{2x}\right)= F(x)$

Now consider the following two equations:

Eq1: $\int_5^bF(x)d(π(x))=\sum_{5\leq{p}\leq{b}}F(p)==\sum_{5\leq{p}\leq{b}}f(p)$

(π(x) is as mentioned above)

Eq2: Finite version of Abel- Plana Summation on f(x):

$$\begin{align}f(x) = {} & \sin^2\left(\frac{π\Gamma(x)}{2x}\right)\\ \sum_{k=5}^b f(k)= {} & \frac{f(5) +f(b)}2 + \int_5^b f(x) \, dx \\ & {}+ i\int_0^∞\frac{f(5+iy) − f(5−iy)}{e^{2πy }− 1} \, dy +i \int_0^∞\frac{f(b-iy) − f(b+iy)}{e^{2πy }− 1} \, dy \end{align}$$

As we can see for every integer $b$ ( only integer)

Eq(1)=Eq(2)

Here , what we have done is creating a duality with two (seemingly) different pieces of information ( prime sieving by zeta zeros and prime sieving by gamma function) using two different methods or tools ( Stieltjes Integral and Abel Plana ) So here I'm curious if something comes out of it .

Also we can further modify the weight at our service but I'm not gonna go into that because it's irrelevant for my instant aim .

Questions:

Is this line of thinking any "good" ?

Can we derive anything about non trivial zeros from above duality?

Is there any way to manipulate/transform Eq(2) into equation Eq(1)

(Also if anything wrong with the analysis please correct me I'm just a student )

I'm highly doubtful about all this but some people in my local maths community think this is original , but i know original doesn't mean fruitful so asking here.

Can we treat the above duality with operator theory such that the zeros of Zeta comes as eigenvalues of some operator (after suitable scaling and transformation of the equation)?