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Riemann zeta function $\zeta(s)$ is related to Riemann Xi function $\Xi(z)$ via: $$s=\frac12+ iz,\qquad \Xi(z):=\frac12s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s),\tag{1}$$

We found the following function $\Xi_2(t)$ to approximate $\Xi(z)$ on the critical line ($Im(z)=0, Re(z)=t>=12$):

enter image description here

    \[CapitalXi]2[t_] := Sum[(k^2*Pi)^(-(1/4) + (I*t)/2)*
        ((-(1/4) - (I*t)/2)*Gamma[5/4 - (I*t)/2, (k*Pi)^2/t] + 
        ((k*Pi)^2/t)^(5/4 - (I*t)/2)*Exp[-((k*Pi)^2/t)]) + 
   (k^2*Pi)^(-(1/4) - (I*t)/2)*((-(1/4) + (I*t)/2)*
   Gamma[5/4 + (I*t)/2, (k*Pi)^2/t] + ((k*Pi)^2/t)^(5/4 + (I*t)/2)*
   Exp[-((k*Pi)^2/t)]), {k, 1, t/Pi}]

Let $M(t)=t^2\exp(-\pi t/4)$ be a normalization function:

    M[t_] := t^2*Exp[-(t*(Pi/4))]

Here are the two plots of $\Xi(t),\Xi_2(t)$ vs. $t$ at different $t$. Green color is for $\Xi(t)$ and Red color is for $\Xi_2(t)$. enter image description here

enter image description here

You can clearly see that $\Xi_2(t)$ does not behave properly beyond $t=180$.

Question: how can we plot $\Xi_2(t)$ beyond $t=180$?

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    $\begingroup$ Try with Build function: Plot[-RiemannSiegelZ[x], {x, 150, 200}]. $\endgroup$
    – Mariusz Iwaniuk
    May 31, 2020 at 9:18
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    $\begingroup$ RiemannSiegelZ(x)*$(-1/2)(1/2+ix)(1/2-ix)=\Xi(x)$ defined in (1). Both built-in function Zeta(1/2+ix) and RiemannSiegelZ(x) worked fine for x in (150,200). However our purpose is to study a different approximation to Riemann $\zeta(s)$ function. Not using Hardy-Littlewood's approximate functional equation and Riemann-Siegel equation. $\endgroup$
    – mike
    May 31, 2020 at 16:15
  • $\begingroup$ correction: $\Xi(t)=f(t)RiemannSiegelZ(t)$ where $f(t)=-(1/8+t^2/2)\pi^{-1/4}\exp[\Re(\log\Gamma(-3/4+it/2))]<0$. (See p.119 of Edwards' book) $\endgroup$
    – mike
    Jun 2, 2020 at 6:09
  • $\begingroup$ f(t)=-(1/8+t^2/2)\pi^{-1/4}\exp[\Re(\log\Gamma(1/4+it/2))]<0 $\endgroup$
    – mike
    Jun 2, 2020 at 10:53

1 Answer 1

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Use higher precision than 60

Ξ2[t_?NumericQ] := 
 Sum[(k^2*Pi)^(-(1/4) + (I*t)/2)*((-(1/4) - (I*t)/2)*
       Gamma[5/4 - (I*t)/2, (k*Pi)^2/t] + ((k*Pi)^2/t)^(5/4 - (I*t)/2)*
       Exp[-((k*Pi)^2/t)]) + (k^2*Pi)^(-(1/4) - (I*t)/2)*((-(1/4) + (I*t)/2)*
       Gamma[5/4 + (I*t)/2, (k*Pi)^2/t] + ((k*Pi)^2/t)^(5/4 + (I*t)/2)*
       Exp[-((k*Pi)^2/t)]), {k, 1, t/Pi}]

M[t_] := t^2*Exp[-(t*(Pi/4))]

Plot[Ξ2[t]/M[t], {t, 150, 200},
 WorkingPrecision -> 100]

enter image description here

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