# Is it possible to isolate and thereby by integration compute a zeta zero gap when accentuating the zeros and counting them?

The starting point is this integrable formula for the von Mangoldt function:

$$\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}$$

We can plot the Dirichlet series:

$$f(N,t)=\Re\left(\zeta(1/2+it)\sum\limits_{\substack{n=1 \\ d|n}}^{N} \frac{\mu(d)}{n \cdot t \cdot d^{1/2+it-1}}\right)$$

on the critical line:

where the red curve is:

$$g(N,t)=\frac{H_{\text{N}}+\frac{\partial \vartheta (t)}{\partial t}}{t}$$

where $\vartheta (t)$ is the Riemann-Siegel theta function, and $H_{\text{N}}$ is a Harmonic number.

Dividing the Dirichlet series with the asymptotic we get:

$$\frac{f(N,t)}{g(N,t)}$$

Part of the question is if this division with $g(N,t)$ can be avoided and the then possibly integrable function later still can be achieved.

Now consider the integrable (zeta zero) counting function:

which was generated by integrating the Euler-Maclaurin formula over the Möbius $\mu(d)$ function of the divisors of $n$ and adding Riemann-Siegel theta $\vartheta(t)$ as R. Manzoni does:

$$N(t)=\frac{1}{\pi}\left(\vartheta (t)-\Re\left(\sum _{n=1}^{\text{nn}} \frac{1}{n^c} \left(\underset{d \mid n}{\sum\limits_{d=1}^{n}} \left(f(\frac{1}{2}+it, d)-f(\frac{1}{2}+i0, d) \right)\right)\right)\right)$$

where:

$$f(s,d)=-i \mu (d) \left(\sum _{n=0}^{q-1} \left(\sum _{i=0}^{2 n+1} -\frac{d B_{2 (n+1)} k^{-2 n-1} \left|S_{2 n+1}^{(i)}\right| \log ^{-i-1}(d k) \Gamma (i+1,s \log (k d))}{(2 (n+1))!}\right)+\sum _{n=2}^k -\frac{d^{1-s} n^{-s}}{\log (d)+\log (n)}+\text{If}\left[d=1 \text{ then } 0\text{ else }-\frac{1^{-s} d^{1-s}}{\log (d)+\log (1)}\right]+\frac{d^{1-s} k^{-s}}{2 (\log (d)+\log (k))}+\text{Ei}(-(s-1) \log (d)-(s-1) \log (k))\right)$$

If something was lost in translation then the correct formula is found in the Mathematica program here at the end of this answer. And $c=1$.

Now as already pointed out by H.M. Edwards in his book, the analytic continuation of the above is:

$$N(t)=\frac{\vartheta (t)}{\pi }+\frac{\Im\left(\log \left(\zeta \left(i t+\frac{1}{2}\right)\right)+i \pi \right)}{\pi }$$

But as far as I know $\log (\zeta(s))$ is not integrable, which is what this whole question is about. Anyways for faster computation I will use the analytic continuation to produce the following plot:

The (Mathematica) program for this plot is:

(*Mathematica start*)
scale = 600;(*scale=5000 gives the plot below*)
Print["Counting to 60"]
f[t_] = D[RiemannSiegelTheta[t], t];
Monitor[g1 =
ListLinePlot[
Table[Re[
1/(f[t] + HarmonicNumber[scale])*Zeta[1/2 + I*t]*
Total[Table[
Total[MoebiusMu[Divisors[n]]/
Divisors[n]^(1/2 + I*t - 1)]/(n), {n, 1, scale}]]*(-1)^
Round[RiemannSiegelTheta[t]/Pi +
Im[Log[Zeta[1/2 + I*t]] + I*Pi]/Pi]], {t, 0 + 1/1000, 60,
N[1/30]}], DataRange -> {0, 60}, PlotRange -> {-3.5, 3.5},
PlotStyle -> {Thickness[0.004]}, ImageSize -> Large], Floor[t]]
(*end*)


where

(-1)^Round[
RiemannSiegelTheta[t]/Pi + Im[Log[Zeta[1/2 + I*t]] + I*Pi]/Pi]]


is in latex:

$$(-1)^{\text{Round}\left[\frac{\vartheta (t)}{\pi }+\frac{\Im\left(\log \left(\zeta \left(i t+\frac{1}{2}\right)\right)+i \pi \right)}{\pi }\right]}$$

Now it is not neccessary, and for the function to be integrable one must not add the Absolute Value function, but I do it here for beautification and clarity (which this question lacks - somewhat), like this:

$$(-1)^{\left|\text{Round}\left[\frac{\Im\left(\log \left(\zeta \left(i t+\frac{1}{2}\right)\right)+i \pi \right)}{\pi }+\frac{\vartheta (t)}{\pi }\right]\right|}$$

By then changing the multiplying factor to:

$$(-1/100000000)^{\left|\text{Round}\left[\frac{\Im\left(\log \left(\zeta \left(i t+\frac{1}{2}\right)\right)+i \pi \right)}{\pi }+\frac{\vartheta (t)}{\pi }\right]\right|}$$

I, with this program:

scale = 150;(*scale=5000 gives the plot below*)
g = 1;
Print["Counting to 60"]
f[t_] = D[RiemannSiegelTheta[t], t];
Monitor[g1 =
ListLinePlot[
Table[Re[
1/(f[t] + HarmonicNumber[scale])*Zeta[1/2 + I*t]*
Total[Table[
Total[MoebiusMu[Divisors[n]]/
Divisors[n]^(1/2 + I*t - 1)]/(n), {n, 1, scale}]]*(-1/
100000)^Abs[
Round[-g + 1 + RiemannSiegelTheta[t]/Pi +
Im[Log[Zeta[1/2 + I*t]] + I*Pi]/Pi]]], {t, 0 + 1/1000, 60,
N[1/30]}], DataRange -> {0, 60}, PlotRange -> {-3.5, 3.5},
PlotStyle -> {Thickness[0.004]}, ImageSize -> Large], Floor[t]];
g21 = Graphics[
Rotate[{Text[
N[Im[ZetaZero[If[g == 1, 1, g - 1]]]], {Im[
ZetaZero[If[g == 1, 1, g - 1]]], -3/2}]}, Pi/2]];
g22 = Graphics[
Arrow[{{Im[ZetaZero[If[g == 1, 1, g - 1]]], -1}, {Im[
ZetaZero[If[g == 1, 1, g - 1]]], -1/7}}]];
g31 = Graphics[
Rotate[{Text[N[Im[ZetaZero[g]]], {Im[ZetaZero[g]], -3/2}]}, Pi/2]];
g32 = Graphics[
Arrow[{{Im[ZetaZero[g]], -1}, {Im[ZetaZero[g]], -1/7}}]];
Show[g1, g21, g22, g31, g32]


plotted this:

(the first zeta zero gap)

and this:

(the second zeta zero gap)

and this:

(the third zeta zero gap)

Question What I am trying to say before I now go to bed is:

Is it possible to construct an integrable function (maybe by not dividing with the red curve - the Riemann-Siegel theta and the Harmonic numbers) that gives the value of the $n$-th Riemann zeta zero gap?

This since the zeta zero counting function is integrable and:

Series[(-1)^x, {x, 0, 6}]


gives: $$1+i \pi x-\frac{\pi ^2 x^2}{2}-\frac{1}{6} i \pi ^3 x^3+\frac{\pi ^4 x^4}{24}+\frac{1}{120} i \pi ^5 x^5-\frac{\pi ^6 x^6}{720}+O\left(x^7\right)$$

$1/100000000$ is just a small number meant to go to zero (but never reaching it).