# how to plot an approximate Riemann Zeta function beyond $t=180$?

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 ($$Im(z)=0, Re(z)=t>=12$$): \[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 $$\Xi_2(t)$$.  You can clearly see that $$\Xi_2(t)$$ does not behave properly beyond $$t=180$$.

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

• Try with Build function: Plot[-RiemannSiegelZ[x], {x, 150, 200}]. May 31, 2020 at 9:18
• 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.
– mike
May 31, 2020 at 16:15
• 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)
– mike
Jun 2, 2020 at 6:09
• f(t)=-(1/8+t^2/2)\pi^{-1/4}\exp[\Re(\log\Gamma(1/4+it/2))]<0
– mike
Jun 2, 2020 at 10:53

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] 