I imagine integration works a bit like a Russian doll where you have to open every doll/function in order to get to the core doll/function to do the actual integration on a power/Dirichlet series of $x$-s. I might be wrong on this but the integral I am trying to do is:


$$14.13472514173469...=\int _0^{1000}\frac{1}{2} \left(1-\text{sgn}\left(\frac{\vartheta (t)+\Im\left(\log \left(\zeta \left(i t+\frac{1}{2}\right)\right)\right)}{\pi }-n+\frac{3}{2}\right)\right)dt$$

for $n=1$ to $n=126$.

That I borrowed from Raymond Manzonis answer here.

As a Mathematica program the integral above can be computed:

Clear[n, t, gamma];
gamma = 15; 
 Do[gamma = 
   N[NIntegrate[(1/2)*(1 - 
        Sign[(RiemannSiegelTheta[t] + Im[Log[Zeta[I*t + 1/2]]])/Pi - 
          n + 3/2]), {t, 0, gamma + 15}, PrecisionGoal -> 45, 
     MaxRecursion -> 220, WorkingPrecision -> 50], 40]; 
  Print[gamma], {n, 1, 1}]]

But because there is the Riemann Siegel theta function $\vartheta(t)$ that I don't know if it will work inside the sign function I have for $n=1$ to $n=126$ instead tried the similar integral:


$$\small 14.13472514173469...=\int_0^{1000}\frac{1}{2} \left(1-\text{sgn}\left(\left\lfloor \frac{t \log \left(\frac{t}{2 e \pi }\right)}{2 \pi }+\frac{7}{8}\right\rfloor +\frac{1}{2} \left(-1+\text{sgn}\left(\Im\left(\zeta \left(\frac{1}{2} + i t\right)\right)\right)\right)-n+\frac{3}{2}\right)\right)dt$$

which has different singularities but otherwise is the same and which I arrived at by using the Franca-LeClair asymptotic of the zeta zeros here.

Looks more complicated but fond as I am of my own(?) novelty I tried to express the sign function as in this answer by Chappers here:

$$\text{sgn(x)}=\tanh{(x \cdot \text{large number})} = 1 + 2\sum_{k=1}^{\infty} (-1)^k(e^{-2x \cdot \text{large number})})^k$$

It is a limit actually where the large number goes to infinity but for writing a Mathematica program I would like to keep it as some large number.

The exponential function I intend to write as:

(*Mathematica start*)
Clear[x, y]
exp[x_] = Normal[Series[Exp[x], {x, 0, 12}]]
x = N[exp[1/20000]^200000, 30]
y = N[Exp[10], 30]

as given by user Myself in the comment to the question here.

The logarithm I intend to write as:

$$\log(\frac{t}{2 e \pi }) = \lim\limits_{s \rightarrow 1} \zeta(s)\left(1 - \frac{1}{(\frac{t}{2 e \pi })^{(s - 1)}}\right)$$

The division with $2 e \pi $ might turn out to be troublesome though since there is no rational valued Dirichlet series that corresponds to it.

Then finally before the other sign function there is the floor function, this I thought could be borrowed from ybeltukov's answer here from which I constructed this floor function by adding $x$ the the sawtooth function:

$$\text{floor(x+1)}=\frac{1}{2} \left(\frac{2 \tan ^{-1}\left(\frac{\sin \left(\frac{2 \pi x}{2}\right)}{\delta }\right) \left(1-\frac{2 \cos ^{-1}\left((1-\delta ) \sin \left(\frac{2}{4} \pi (2 x+1)\right)\right)}{\pi }\right)}{\pi }+1\right)+x$$

$\delta$ is some small number. In Mathematica this is:

\[Delta] = 0.001;
Plot[x + (1 + (1 - 
        2 ArcCos[(1 - \[Delta]) Sin[
             2 \[Pi] (2 x + 1)/4]]/\[Pi])*2 ArcTan[
        Sin[2 \[Pi] x/2]/\[Delta]]/\[Pi])/2, {x, -4, 4}]

So what do you think, can either of these numerical integrals (1) or (2) be solved symbolically?

ybeltukov's sawtooth as a step function:

\[Delta] = 1/1000;
Plot[-1/2 + 
  x - (ArcSin[(\[Delta] - 1)*Cos[Pi*x]]*Tanh[Sin[Pi*x]/\[Delta]])/
   Pi, {x, -2, 2}]

simplified a bit.

  • $\begingroup$ The first zeta zero 14.13472514173469379045725... is the case where $n=1$. The formulas are a bit misleading because I have written $n$ and not simply $1$. $\endgroup$ – Mats Granvik Apr 14 '17 at 12:49
  • $\begingroup$ Related: mathoverflow.net/a/261734/25104 $\endgroup$ – Mats Granvik Apr 14 '17 at 12:51

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