Dirichlet Series of a given sequence

Question

I am trying to calculate the dirichlet generating function of $(p(n)q( \log(n)))_{n \geq 1}$ where $p,q$ are arbitary polynoms. First I calculated the dirichlet genarating function of $(p(n))_{n \geq 1}$ which is \begin{align*} \sum_{n \geq 1}\frac{p(n)}{n^s} &= \sum_{n \geq 1}\frac{1}{n^s}\sum^{k}_{l = 0}p_l n^l \\ &= \sum_{n \geq 1}\sum^{k}_{l = 0}\frac{p_l}{n^{s - l}} \\ &= \sum^{k}_{l = 0} \sum_{n \geq 1}\frac{p_l}{n^{s-l}} \\ &= \sum^{k}_{l=0}p_l \zeta(s - l) \end{align*} where I used the absolute convergence for $\Re(s - k) > 1$ where $\mbox{deg}(p) = k$. Also \begin{align*} \sum_{n \geq 1}\frac{\log(n)}{n^s} &= \sum_{n \geq 2}\frac{\log(n)}{n^s} \\ &= \sum_{n \geq 2}-\frac{d}{ds}\frac{1}{n^s} \\ &= -\frac{d}{ds} \left( \sum_{n \geq }\frac{1}{n^s} \right) \\ &= -\frac{d}{ds} \zeta(s) \\ &= -\zeta'(s). \end{align*} I also calculated \begin{align*} \sum_{n \geq 1}\frac{q(\log n)}{n^s} &= \sum_{n \geq 1}\frac{1}{n^s} \sum^{r}_{l=0}q_l \log(n)^l \\ &= \sum_{n \geq 1}\sum^{r}_{l=0}(-1)^lq_l\frac{d^l}{ds^l} \frac{1}{n^s} \\ &= \sum^{r}_{l=0}(-1)^l q_l \zeta^{(l)}(s). \end{align*} but I am not sure if I need this. Now I am not sure if there is some result to calculate the dirichlet generating function of $(p(n)q(\log(n))_{n \geq 1}$ based on the previously calculated dirichlet generating functions.
Edit: From the comment I got \begin{align*} \sum_{n \geq 1}\frac{p(n) q(\log n)}{n^s} &= \sum_{n \geq 1} \left( \sum^{k}_{l = 0} q(\log n) \frac{p_l}{n^{s-l}} \right) \\ &= \sum^{k}_{l = 0}p_l \sum_{n \geq 1}n^l \frac{q(\log n)}{n^{s}} \\ &= \sum^{k}_{l = 0}p_l \sum_{n \geq 1}\frac{n^l}{n^s} \sum^{r}_{m=0}q_m \log(n)^m \\ &= \sum^{k}_{l = 0}p_l \sum^{r}_{m = 0}q_m \sum_{n \geq 1}\frac{1}{n^{s-l}} \log(n)^m \\ &= \sum^{k}_{l = 0}p_l \sum^{r}_{m = 0}q_m \sum_{n \geq 1}(-1)^m \frac{d^m}{ds^m} \frac{1}{n^{s-l}} \\ &= \sum^{k}_{l=0}p_l \sum^{r}_{m=0}(-1)^m q_m\zeta^{(m)}(s - l) \end{align*} I used the absolute convergence for $\Re(s - k - r) > 1$ where $\mbox{deg}(q) = r$ to interchange the order of summation. I am not sure if this is correct.