I did the specialization $x=\frac{1}{p^3}$, where $p$ is a prime number at a standard identity, that is the integration of the partial sum of a geometric of ratio $x$. After the change of variable $x=\frac{1}{p^3}u$ and taking the sum over all prime numbers you can recognize : $$\sum_{p\text{ prime}}\int_0^1\sum_{k=0}^n u^k\frac{1}{p^{3k+3}}du=\log\zeta(3)-\sum_{p\text{ prime}}\int_0^1\frac{u^{n+1}}{p^{3n+3}(p^3-u)}du,$$
Thus taking the exponential in both sides I get when I've combined with the definition of the Prime Zeta function from this Wikipedia entry, that since $$\sum_{k=0}^n\frac{\mathcal{P}(3k+3)}{k+1}=\log\zeta(3)-\sum_{p\text{ prime}}\int_0^1\frac{u^{n+1}}{p^{3n+3}(p^3-u)}du$$ then
$$\zeta(3)= \left( \prod_{k=0}^n e^{\frac{\mathcal{P}(3k+3)}{k+1}} \right) \left( \prod_{p\text{ prime}}e^{\int_0^1\frac{u^{n+1}}{p^{3n+3}(p^3-u)}du } \right) \tag{1}.$$
Question. Criticize the usefulness of factorization $(1)$. Thanks in advance.
What I am asking, if it is acceptable as a Question, is to discuss if such factorization (in geometric factors) is interesting or potentially useful for the unsolved problems related to $\zeta(3)$ ; or argue that is artificious, wrong or isn't interesting. If you can improve the calculations from RHS to do more interesting also is welcome.
Thus if the question is acceptable and intesting, I am waiting an answer.