# Asymptotic expressions for incomplete Gamma

I have an expression below and want to approximate it with a simpler form by taking $$d$$ to infinity

$$f(s)=\frac{s^{1/p} \Gamma \left(-\frac{1}{p},(d+1)^{-p} s\right)}{p}$$

Taking Asymptotic returns unevaluated.

expr = (t^(1/p) Gamma[-(1/p), (1 + d)^-p t])/p ;
Asymptotic[expr, {d, Infinity, 2}]


Are there tricks in Mathematica to get a simpler expression for this in the infinite $$d$$ limit?

With[{p = 101/100, d = 10^9},
expr = (t^(1/p) Gamma[-(1/p), (1 + d)^-p t])/p ;
Plot[expr, {t, 1, d}, AxesLabel -> {"time", "error"}]
]


• Adding Assumptions -> d > 0 && p > 0 && t > 0 (assuming those are desired restrictions) results in a response from Asymptotic (Windows 10, Mathematica 13.2.0.0).
– JimB
Apr 19, 2023 at 3:41
• The 2nd order approximation appears to be 1 + d + ((1 + d)^(1 - p) t)/(-1 + p) + ((1 + d)^(1 - 2 p) t^2)/(2 - 4 p) + ((1 + d)^(1 - 3 p) t^3)/(-6 + 18 p) - t^(1/p) Gamma[(-1 + p)/p]. And the form for the $k$-th order approximation is $\sum _{i=0}^{k+1} \frac{(-1)^{i+1} t^i (d+1)^{1-i p}}{i! (i p-1)}-t^{1/p} \Gamma \left(1-\frac{1}{p}\right)$.
– JimB
Apr 19, 2023 at 4:50
• Asymptotic does produce some result, but it doesn't seem to fit the graph in $0<t<d$ regime which is what I'm trying to find Apr 19, 2023 at 6:34
• Please include your requirements for a good fit. The following produces a ratio of the true to approximate value from 0.999997 to 1: a = With[{k = 6}, -t^(1/p) Gamma[1 - 1/p] + Sum[(-1)^(i + 1) t^i (1 + d)^(1 - i p)/(i! (i p - 1)), {i, 0, k + 1}]]; Plot[expr/a /. {d -> 10^9, p -> 101/100}, {t, 1, 10^9}, PlotRange -> All].
– JimB
Apr 19, 2023 at 13:29

First substituting u -> (1 + d)^-p t and assuming u->0 for d->Infinity Mathematica (v12.2 Windows11) evaluates

expr = (t^(1/p) Gamma[-(1/p), (1 + d)^-p t])/p
zw = Asymptotic[Gamma[-(1/p), u], {u, 0, 3}]
asy = t^(1/p)/p zw /. u -> (1 + d)^-p t // Simplify(* third order asymptote*)


$$\frac{t^{\frac{1}{p}} \left(\frac{1}{6} p \left(\frac{t^3 (d+1)^{-3 p}}{3 p-1}+\frac{3 t^2 (d+1)^{-2 p}}{1-2 p}+\frac{6 t (d+1)^{-p}}{p-1}+6\right) \left(t (d+1)^{-p}\right)^{-1/p}+\Gamma \left(-\frac{1}{p}\right)\right)}{p}$$

Block[{p = 101/100, d = 10^9},
Plot[{expr, asy}, {t, 1, d}, AxesLabel -> {"time", "error"},PlotRange -> {{0, Automatic}, All}[[-1]],PlotStyle -> {Automatic, Dashed}]]


But initial assumption u->0 doesn't hold for z=O[d]! That's why approximation applies only to t=o[d]!

Hope it's what you're looking for!

• ah, thanks, indeed, I was looking for something tight in the range of $t=o(d)$ Apr 20, 2023 at 2:56
• background for this problem is given here Apr 21, 2023 at 5:52