# 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?

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
?

• Note that $$\zeta(s)=\sum^\infty_{n=1}\frac1{n^s}$$ is true only for $\Re s\ge1$. Here, the real part is $\frac12$, so you cannot blindly use this summation form of zeta function. Consider the analytical continuation of zeta by eta. – Szeto Jul 9 '18 at 0:15
• It is not that the integral diverges. It’s just you made a mistake in the first line due to not caring about convergence: the integral is not equal to the sum. – Szeto Jul 9 '18 at 0:18
• Yes, I see my mistake now. I'm looking over arxiv.org/abs/1208.1440 for some other possibilities – crow Jul 9 '18 at 0:19

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.