We know the following from Euler product:
$$\sum_{n,p}\frac{1}{n(p)^{nz}}= \log(\zeta(z))$$
$$\sum_{n,p} \frac{\log p}{p^{nz}} = -\frac{\zeta'(z)}{\zeta(z)} $$
I want to find the alternate representation for the following sum:
$$\sum_{n,p}\frac{\log(p)}{n(ap)^{nz}}$$
Here '$a$' is a constant greater than 1
What is the representation of above sum in terms of Zeta and other allied function?
The first formula is incorrect: it's halfway in between $$ \sum_{p^n} \frac1{np^{nz}} = \log\zeta(z) \quad\text{and}\quad \sum_{p^n} \frac{\log p}{p^{nz}} = -\frac{\zeta'(z)}{\zeta(z)} $$ for $\Re z>1$. Analogous to the first formula, we would have \begin{align*} \sum_{p^n} \frac1{n(ap)^{nz}} &= \sum_p \log\big( 1-(ap)^{-z} \big)^{-1} \\ &= \log \prod_p \big( 1-(ap)^{-z} \big)^{-1} = \log \sum_{n=1}^\infty \big( a^{-\Omega(n)}n \big)^{-z} \end{align*} in the same half-plane, where $\Omega(n)$ is the number of prime factors of $n$ counted with multiplicity.