Plot of Meijer-G function $g(t)$ disagrees with $\lim_{t\to \infty} g(t)$

Question

My function is

$$g(t)=\frac{2}{\sqrt\pi t}-\frac{\pi t}{2}\,G_{0,3}^{3,0}\left(\frac{\pi^2 t^2}{4}\middle|-1,-\frac12,0\right)$$ where $G^{a,b}_{c,d}$ is the Meijer $G$ function. I want to know if $g(t) \to \frac{1}{t}$ as $t \to \infty$, i.e. I want to find

$$\lim_{t\to \infty} \frac{g(t)}{\frac1t}.$$

On Mathematica:

g[t_] := 2/(Sqrt[Pi] t) - 1/2 Pi t MeijerG[{{}, {}}, {{-1, -(1/2), 0}, {}}, (Pi^2 t^2)/
4]
Limit[g[t]/(1/t), t -> Infinity]

gives $\dfrac{2}{\sqrt{\pi}}$ so I know that indeed $g(t)$ decays like $\frac1t$ as $t \to \infty$.

However, plotting $g(t)$ tells a different story:

and so does just evaluating $g(t)$ for larger and larger values of $t$.

How can I tell whether $g(t)$ decays as $\frac1t$ or diverges as $t \to \infty$? And why is Mathematica giving contradicting results?

Answer 1

MeijerG often cannot be evaluated accurately except with high WorkingPrecision. Here is the plot with WorkingPrecision -> 60.

Plot[g[t], {t, 1, 100}, WorkingPrecision -> 60, PlotRange -> All, 
    ImageSize -> Large, AxesLabel -> {t, g}, LabelStyle -> Directive[12, Bold, Black]]

The curve is nearly indistinguishable from 2/(t Sqrt[Pi]).

Answer 2

As a partial answer:

Note that you can simplify your Meijer-$G$ expression a fair bit to make it amenable to asymptotic analysis. Using the absorption and flip identities, as well as switching to the generalized $G$-function, we have the equivalent

$$g(t)=\frac{2}{\sqrt\pi t}-G_{3,0}^{0,3}\left(\frac2{\pi t},\frac12\middle|{{\frac12,1,\frac32}\atop{\text{—}}}\right)$$

or in Mathematica form:

g[t_] := 2/(Sqrt[π] t) - MeijerG[{{1/2, 1, 3/2}, {}}, {{}, {}}, 2/(π t), 1/2]

This should now be more amenable to asymptotic analysis; your limit of interest is now equivalent to

$$\frac{2}{\sqrt\pi}-\frac2{\pi}\lim_{t\to\infty} G_{3,0}^{0,3}\left(\frac2{\pi t},\frac12\middle|{{0,\frac12,1}\atop{\text{—}}}\right)$$

Plotting still needs the assistance of a high WorkingPrecision setting, however:

Plot[g[t], {t, 1, 100}, WorkingPrecision -> 45, PlotRange -> All]

Unfortunately, I quickly run into trouble when I directly use Series[] or Limit[]; this seems to be an inherent problem of the underlying Meijer-$G$ implementation when the arguments are too large or too small for a given set of parameters. Altho the terms in Series[MeijerG[{{0, 1/2, 1}, {}}, {{}, {}}, 2/(π t), 1/2], {t, ∞, 2}] give rise to Indeterminate terms upon direct evaluation, some experimentation with ridiculously high precision evaluations seems to indicate that the coefficients do drop to zero remarkably quickly, e.g.

Block[{$MaxExtraPrecision = 100}, 
      N[MeijerG[{{0, 1/2, 1}, {}}, {{}, {}}, 1*^-4, 1/2], 1*^4]] // N[#, 20] &
   1.4065564604009371892*10^-603

(Note the exponent.)

Some more analytical work will be needed for a rigorous proof, which I might append to this answer if I find the time.