Consider the function $f(x)$ defined in formula (1) below and the related function $F(s)$ defined in formula (2) below (which corresponds to the integral of the Riemann zeta function $\zeta(s)$).
(1) $\quad f(x)=\sum\limits_{n=2}^x\frac{1}{\log(n)}-\delta (x-1)$
(2) $\quad F(s)=\int \zeta(s)\,ds=-s\int\limits_0^\infty f(x)\ x^{-s-1}\,dx=s-\underset{N\to \infty }{\text{lim}}\ \sum\limits_{n=2}^N\frac{1}{\log(n)\,n^s}\,,\quad\Re(s)>1$
The following plot illustrates formula (2) for the function $F(s)=\int\zeta(s)\,ds$ evaluated at $N=1000$, $N=10000$, and $N=100000$ in blue, orange, and green respectively.
Figure (1): Illustration of formula (2) for the function $F(s)=\int\zeta(s)\,ds$
Note $F(s)=\int\zeta(s)\,ds$ has a zero somewhere in the range $1<s<2$.
Question 1: Is there a closed expression for the location of the zero of $F(s)=\int\zeta(s)\,ds$ in the range $1<s<2$?
Note formula (2) for $F(s)=\int\zeta(s)\,ds$ converges for $\Re(s)>1$.
Question (2): Does the function $F(s)=\int\zeta(s)\,ds$ have an analytic continuation for $\Re(s)\le 1$ and if so, what is known about the locations of the zeros of $F(s)=\int\zeta(s)\,ds$?
Best Answer
An analytic continuation of $$F(s) = s-\sum_{n\ge 2} \frac{n^{-s}}{\log n}$$ is $$F(s) = F(2+i\Im(s))+\int_{2+i\Im(s)}^s \zeta(z)dz$$ Note that $F(s)-\log(s-1)$ is a primitive of the entire function $\zeta(s)-\frac1{s-1}$ so it is entire.
Then $F(2+i\Im(s))= i\Im(s)+O(1)$ and for $\Re(s)\ge 0, \zeta(z)=O(|\Im(z)|^{1/2+\epsilon})+\frac1{z-1}$, whence there is some $T$ such that $F(s)$ has no zeros for $\Re(s)\in [0,2], \Im(s)\ge T$.
$F(s)$ also has no zeros for $\Re(s)\ge 10$.
On a neighborhood of $s=1$ we have $F(s)=\log(s-1)+O(1)$ so it has no zeros as well.
The finitely many zeros on $\Re(s)\ge 0$ (finitely for the principal branch of $\log$) are found with the argument principle.
With the functional equation of $\zeta(s)$ we can probably say something about $\Re(s)\le 0$ as well.