Approximation to the prime counting function

The best analytic built-in approximation is the Riemann Prime Counting Function; it is implemented in Mathematica as RiemannR. So far we know exact values of $\pi$ prime counting function for n < 10^25, however in Mathematica its counterpart PrimePi[n] can be computed exactly to much lower values i.e. up to 25 10^13 -1, see e.g. What is so special about Prime? for more details.

RiemannR[10^1000] // N

4.344832576401197453*10^996

See e.g. Prime-counting function, it claims that the best estimator is $ \operatorname{R}(x) - \frac1{\ln x} + \frac1\pi \arctan \frac\pi{\ln x}$

let's define it as APi:

APi[x_] := RiemannR[x] - 1/Log[x] + 1/Pi ArcTan[ Pi/Log[x]]

it provides the same approximation as RiemannR, at least IntegerPart does, in fact

Grid[ Table[ IntegerPart @ { 5 10^k, RiemannR[5 10^k] - PrimePi[5 10^k], 
                             APi[5 10^k] - PrimePi[5 10^k]}, {k, 3, 13}],
      Frame -> All, Alignment -> Left]

We can realize how good RiemanR is plotting related errors of interesting approximations where we know exact values of the prime counting function (see e.g. "Mathematica in Action" by Stan Wagon):

Plot[{ LogIntegral[x] - PrimePi[x], RiemannR[x] - PrimePi[x], 
        x/(Log[x] - 1) - PrimePi[x]},  {x, 2, 3 10^5}, MaxRecursion -> 3, 
      Frame -> True, PlotStyle -> {{Thick, Red}, {Thick, Darker @ Green}, 
                                  {Thick, Darker @ Cyan}}, 
      PlotLegends -> Placed["Expressions", {Left, Bottom}], 
      ImageSize -> 800, AxesStyle -> Thick]

Plot[{ LogIntegral[x] - PrimePi[x], RiemannR[x] - PrimePi[x], 
       x/(Log[x] - 1) - PrimePi[x]}, {x, 10^6, 10^7}, MaxRecursion -> 3, 
       Frame -> True, PlotStyle -> {{Thick, Red}, {Thick, Darker @ Green}, 
                                    {Thick, Darker @ Cyan}}, 
       PlotLegends -> Placed["Expressions", {Left, Bottom}], 
       Axes -> {True, False}, AxesStyle -> Thick, ImageSize -> 800]

RiemannR seems to be a better choice than LogIntegral based on this plot:

Plot[{PrimePi[n], LogIntegral[n], RiemannR[n]}, {n, 1, 5000}, PlotStyle -> {Black, Blue, Red}]

RiemannR[1.*10^1000]

4.344832576401197453*10^996