Asymptotic expressions for incomplete Gamma

Question

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"}]
 ]

Answer 1

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!