Does $z (s) = \int_0^s \zeta \left( \frac{1}{2} + i t \right) d t = s + \sum_{n = 2}^{\infty} \frac{i (n^{- i s} - 1)}{\ln (n) \sqrt{n}}$ converge ?
If we take the termwise integral for $n^{-s}$ with $s={\frac{1}{2}+it}$ we get
$\int_{0}^{s}\!{n}^{-\frac{1}{2}-it}\,{\rm d}t={\frac {i \left( {n}^{-is}-1 \right) }{\ln \left( n \right) \sqrt {n}}}$
is it valid to take the termwise integral of each summand like this? The summand for $n=1$ is singular but the limit at this point is $s$ hence the summation starts at 2.
A graph of the numerically computed integral vs the sums with truncated at N=2000 are shown.. it looks close but not exact.. the oscillations never seem to cancel.. ?
is there some transform that could be used to develop a series for the integral? It seems there might be some way to derive an error term for the truncation or something
?
It seems that what you are asking is if $$ \sum_{n \geq 1} \frac{1 - n^{it}}{\log n\sqrt n}$$ converges. It is possible to see heuristically why this sum diverges (in a way which can be made rigorous if one really wants to).
Note that $\lvert n^{it} \rvert = 1$ and thus $\mathrm{Re}(1 - n^{it}) \geq 0$. Further, the values of $\mathrm{Re}(n^{it})$ are regularly distributed in $[-1, 1]$, and so for a (large) positive proportion of $n$ we should expect $\Re(1 - n^{it}) > \frac{1}{2}$ (this is an arbitrarily chosen value).
The sum simply over these $n$ will diverge, and thus the overall sum diverges.
One shouldn't expect for termwise integration to work unless the original series converges absolutely. (Where you are looking, it doesn't even converge conditionally).