# 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

## 1 Answer

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).

• Thank you, I see. I wonder, are there absolutely-convergent series for zeta? – crow Jul 8 '18 at 23:40
• Yes of course, see arxiv.org/pdf/1208.1440.pdf – crow Jul 14 '18 at 18:12