# MMA does not provide the correct asymptote for an integral function

Given is the function $$f(x)=\int_0^\infty \mathbb{exp}\left(-\frac{x^2}{2t^2}-t\right)\mathbb{d}t$$

Mathematica returns for the asymptotic behavior $$x\to\infty$$ using Series or Asymptotic

$$g(x)=2\sqrt{\frac{2\pi}{3}}x^{1/3}\mathbb{exp}\left(-\frac{3}{2}x^{2/3}\right)$$

However the asymptotic behavior is better modeled by $$g(x)/2$$ as seen in the graphs below. Why does MMA give a worse approximation and how MMA calculates it? Maybe there are different approaches that deliver different results.

Absolute error:

Relative error:

As the relative error for $$g(x)$$ approaches $$\approx 1$$ it follows from $$\dfrac{g(x)-f(x)}{f(x)}\approx 1$$ that for $$x\to \infty$$ $$f(x)\approx g(x)/2$$ but not $$f(x)=g(x)$$ as stated by MMA.

f = Integrate[Exp[-x^2/(2 t^2) - t], {t, 0, \[Infinity]}];
(* f can also be expressed as: MeijerG[{{}, {}}, {{0, 1/2, 1}, {}}, x^2/8]/Sqrt[\[Pi]]*)
g1 = Assuming[x > 0, Normal[Series[f, {x, \[Infinity], 0}]]]
g2 = Simplify[Asymptotic[f, x -> \[Infinity]], x > 0]

Edit

The factor $$2$$ affects the whole series expansion, not only the first term.

MMA 12.3

• This question has been asked before Aug 18 '21 at 4:24
• @CarlWoll: As neither the other question nor the cross linked wolfram community post have "official" answers: what is the "official" reaction from WRI to this? Is it considered a bug, will it be corrected? Aug 18 '21 at 8:04
• @AlbertRetey It is a known bug. As for prognosis...I spent a chunk of yesterday diving into the relevant code, comparing with the formulas at functions.wolfram.com. It's not something I can fix. I sent details to people with more expertise, but if the stuff at f.w.c is amiss, I don't hold out great hope here. For what it's worth, the trouble case seems to be when the total length of the b lists, q, exceeds that of the a lists (p) by 3. Aug 19 '21 at 14:06
• @DanielLichtblau: OK, thanks for the update. At least we now know whats going on... Aug 20 '21 at 15:15