Let $\xi$ be the Riemann Xi function (in the sense of Landau), then $\xi(z):=\frac{1}{2}z(1-z)\pi^{-s/2}\Gamma(\frac{z}{2})\zeta(z)$ for $z\in\mathbb{C}$ satisfying the symmetry $\xi(z)=\xi(1-z).$
For $\xi(z)=\xi(0)\prod_n\left(1-\frac{z}{\rho_n}\right)$ (where the product extends over $\rho_n$, the non-trivial zeros of the zeta function in order of $|Im(\rho_n)|$) and $\frac{\xi^{\prime}(z)}{\xi(z)}=\sum_n\frac{1}{z-\rho_n}$, is it possible to express $\frac{\xi^{\prime}(z)}{\xi(z)}$, the logarithmic derivative of $\xi(z),$ as an exponential $\frac{\xi^{'}(z)}{\xi(z)}=e^{\Phi(z)}$, where $\Phi(z)$ does not make explicit reference to the non-trivial zeros of the zeta function, for $z\in\mathbb{C}$?
Thanks in advance!
No, $\Phi(z) = \log \frac{\zeta'(z)}{\zeta(z)}+\log \frac{\chi'(z)}{\chi(z)}$ is the best you can do.
For example $\phi(z) = \log \frac{\zeta'(z)}{\zeta(z)}$ is not a Dirichlet series anymore, so you can't even express in terms of the primes :
$\frac{\zeta'(z)}{\zeta(z)} = -\sum_{n=2}^\infty \Lambda(n) n^{-z} = - 2^{-z}\Lambda(2) \ \sum_{n=2}^\infty \frac{\Lambda(n)}{\Lambda(2)} (n/2)^{-z} = 2^{-z}F(z)$ . For $z$ large enough : $\log F(z) = \sum_{k=1}^\infty \frac{(1-F(z))^k}{k}$ where $F(z)$ and hence $(1-F(z))^k$ are generalized Dirichlet series.
But $(1-F(z))^k = \sum_{n=2^k}^\infty a_{n,k}(n/2^k)^{-z}$ so that $\log F(z) = \sum_{k,n} c_{n,k}(n/2^k)^{-z}$ .
But $\{n/2^k\}$ is dense in $\mathbb{Q}_{\ge 1}$ so $\log F(z)$ is not a generalized Dirichlet series : you can't compute its first few coefficients.