# Definite integral of $\zeta(s)$ over the critical strip

Take the following definite integral:

$$f(s):=\int_s^{1-s} \zeta(x) \mathrm{d} x$$

with $s \in \mathbb{C}$, $s=\sigma \pm ti$, $0<\sigma<1$ and $t,\sigma \in \mathbb{R}$.

The graph of $|f(s)|$ shows a monotonically increasing function for $\sigma=\frac12$ (as expected, it 'plateaus' exactly at the $\rho$s) and an apparently strictly increasing function when $\sigma\ne\frac12$.

There is however a small 'dip' in the area $1 < t < 3$, that unexpectedly induces a zero at $\frac12 \pm 2.50056818181399528638615277529..i$. For $\sigma\ne\frac12$ there are no zeros.

Is there anything known about this zero? Could it be proven that it only exists for $\sigma=\frac12$?

Thanks!

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There is a zero $s=1/2$ and at least few outside the critical strip. – joro Apr 15 '12 at 8:05
Hi Joro. Thanks for the search (again!). I did also find the $s=\frac12$, but did not search beyond the critical strip. Also guess that a definitive integral over a certain range yields unambiguous outcomes, however I found for the primitive integral: "the fact that zeta has a pole with non-vanishing residue at $1$ means that it does not have a primitive integral in various sets; the integral is multivalued". That zero in the critical strip is intriguing though, esp. since the real and imaginary curves of its 'derivative' (i.e. $\zeta(s)$) do not reveal anything peculiar at that point. – Agno Apr 15 '12 at 9:45
$|f(s)|$ is not monotonically increasing after the second zero. Check at $14.1$ and $14.6$ or examine the graph in this interval. – joro Apr 17 '12 at 8:47
You are right, Joro. There are dips around the 'plateau' at $\rho_1$. It appears though that this only happens at the first $\rho$. Did you find similar ones? In the mean time I also realized, that the derivative of this particular integral is $-\zeta(1-s)-\zeta(s)$ and not $\zeta(s)$. The same integral for $\eta(s)$ has no zero. This integral can be written as as a sum: $1-2s+\sum _{n=2}^{\infty }{\frac { \left( -1 \right) ^{n-1} \left( {n}^{-s}-{n}^{-1+s} \right) }{\ln \left( n \right) }}$. Guess the multiplication $\frac{1}{1-2^{1-s}}$ term gives the 0 at $2.50..i$. – Agno Apr 17 '12 at 11:37
According the computations it is decreasing not only at the first zeta zero. Also at zeros: 3,5,8,10,11 – joro Apr 17 '12 at 12:39