TITLE: Help with a mellin-type integral 
QUESTION [5 upvotes]: greetings . i've been trying to do this integral for many days now, with no clue on how to attack it . the integral is a mellin inverse of some kind, and appears in analytic number theory . 
$$I(x)=\lim_{T\rightarrow \infty}\frac{1}{2\pi i}\int_{2-iT}^{2+iT}\frac{x^{s}}{s}\left(\sum_{k=1}^{\infty}\frac{\zeta(ks)-1}{k} \right )^{n}ds$$
$\zeta(s)$ : is the Riemann zeta function. 
$n \geqslant 2 $
any insights are more than welcome .

REPLY [2 votes]: Continued from my earlier post: I would therefore suggest approaching it in the ``counting domain'' as follows: from the explicit definition of the coefficients given by GH above, you can easily see that your Dirichlet series in the case $n=1$ is
   $$\sum_{n=2}^{\infty}\left(\sum_{n=n_i^{k_i}(n)}\log n_i(n)\right)\frac{1}{n^s\log n}=\sum_{n=2}^{\infty}\frac{\log\prod_{i}n_i(n)}{n^s\log n}.$$
From here you get 
    $$I(x)=\sum_{n\leq x}\frac{\log\prod_{i}n_i(n)}{\log n}.$$