# The effect of roots of Dirichlet's $\beta$ function condenses to $\frac12\left(1+ie^{i2\pi\frac{p}4}\right)$

With the help of Raymond Manzoni and Greg Martin I was able to derive an explicit formula for the number of primes of the form $4n+3$ in terms of (sums of) sums of Riemann's $R$ functions over roots of Riemann's $\zeta$ resp. Dirichlet $\beta$ function: \begin{align*} \pi^*(x;4,3)&=\sum_{k=0}^\infty 2^{-k-1}\left( \operatorname{R}(x^{1/2^{k}})-\sum_{\rho_\zeta} \operatorname{R}(x^{\rho_\zeta/2^k}) +\sum_{\rho_\beta} \operatorname{R}(x^{\rho_\beta/2^k}) \right) \end{align*}

Ilya helped a lot to derive a formula for summing over General Functions of Primes $$\sum_{p\le x}f(p)=\int_2^x f(t) d(\pi(t))\tag{1}$$ and applying this to Prime $\zeta$ function this simplifies (haha) to $$P_{x;4,1}(r)+P_{x;4,3}(r)= \sum_{p<\color{red}x} \frac{1}{p^{ir}} =\sum_{n=1}^{\infty}\frac{ \mu (n)}{n}\sum_{z\in\{1,\rho\}}(-1)^{1-\delta_{1z}} \left[ {\rm li}(t^{\frac zn-ir}) \right]^{\color{red}x}_2$$

So I think it's possible to combine these two partial results and come up with something

$$P_{x;4,3}(r)=\sum_{k=0}^\infty 2^{-k-1} \sum_{n=1}^{\infty}\frac{ \mu (n)}{n}\sum_{z\in\{1,\rho(\zeta),\rho(\beta) \}/2^k}\alpha(z) \left[ {\rm li}(t^{\frac zn-ir}) \right]^{\color{red}x}_2,\tag{2}$$ where $\alpha(z)=\cases{\phantom{-} 1 ,\text{if$z=1/2^k$or$z=\rho_\beta/2^k$},\\ -1 , \text{if$z=\rho_\zeta/2^k$}}$.

But I also found another, pretty simple, way to represent $P_{x;4,3}(r)$: $$P_{x;4,3}(r)= \sum_{p<x} \frac12\left(1+ie^{i2\pi\frac{p}4}\right)p^{ir},$$ where $\left(1+ie^{i2\pi\frac{p}4}\right)$ just cancels for primes of the form $4n+1$. We have to apply $(1)$ and insert $$\pi(x) = \operatorname{R}(x^1) - \sum_{\rho(\zeta)}\operatorname{R}(x^{\rho(\zeta)}) \tag{3}$$ to get something comparable to $(2)$, but the roots of Dirichlet's $\beta$ function wouldn't obviously show up here.

Is it possible to show that the effect of $\rho(\beta)$ in $(2 )$ can be condensed to $\frac12\left(1+ie^{i2\pi\frac{p}4}\right)$?

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