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How can I evaluate a certain integral numerically?

Mathematica Asked on February 12, 2021

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The problem arises as value of $gamma_0$ is very high (on the order of 10^13), whereupon $exp (-gamma_0,h_a^2)$ is becoming very small and ultimately zero. Can anyone help me to evaluate this using NIntegrate?

$k$, $z$, $gamma_0$ are all constants.

Edit 1:

As some people asked for the code that I used, it is following:

Pt = 0.2511
R = 1
sigma = 10^(-7)
gammaa = (2*Pt^2*R^2)/(sigma^2)
beta = 12.06
l = 200
l + 5
z = 4.343/(beta*l)
k = 5.49
res = NIntegrate[
  Gamma[k, z*Log[Exp[1], (l/h)]]*Exp[-gammaa*h^2], {h, 0, 1}, 
  MinRecursion -> 30,  MaxRecursion -> 70, AccuracyGoal -> 350,   
  Method -> "AdaptiveQuasiMonteCarlo"]
val = Sqrt[gammaa/Pi]/Gamma[k]

( Here I have used "AdaptiveQuasiMonteCarlo" method just because only this way I was getting any value except Zero)

2 Answers

Make use of the AsymptoticIntegrate command:

AsymptoticIntegrate[Gamma[2, 3*Log[1/t]]*Exp[-γ[0]*t^2],{t, 0, 1}, {γ[0],Infinity,1}]
(*-(E^-γ[0]/(2 γ[0]))*)

In more complicated cases Laplace's method may be applied.

Answered by user64494 on February 12, 2021

Changing substitution to integral and expanding with infinte sum we have: $$int_0^1 Gamma left(k,z ln left(frac{1}{t}right)right) exp left(-gamma t^2right) , dt=int_0^{infty } Gamma (k,z t) exp (-gamma exp (-2 t)-t) , dt=frac{sqrt{pi } text{erf}left(sqrt{gamma }right) Gamma (k)}{2 sqrt{gamma }}+sum _{j=0}^{infty } frac{(-1)^{1+j} z^k (1+2 j+z)^{-k} gamma ^j Gamma (k)}{j!+2 j j!}$$

Closed-form for series probably not exist.

$Version
(*"12.1.1 for Microsoft Windows (64-bit) (June 9, 2020)"*)

f[k_] := (Sqrt[[Pi]] Erf[Sqrt[[Gamma]]] Gamma[k])/(2 Sqrt[[Gamma]]) + 
Sum[((-1)^(1 + j) z^k (1 + 2 j + z)^-k [Gamma]^j Gamma[k])/(j! + 2 j j!), {j, 0, Infinity}];
Table[f[k], {k, 1, 5}] // MatrixForm // Simplify

(*1/2 [Gamma]^(-(1/2) - z/2) Gamma[(1 + z)/2, 0, [Gamma]] ... *)

Formula only works for: $k > 0$ and $kin mathbb{Z}$

EDITED:

f[k_, [Gamma]_, z_] := (Sqrt[[Pi]] Erf[Sqrt[[Gamma]]] Gamma[k])/(2 Sqrt[[Gamma]]) + 
Sum[((-1)^(1 + j) z^k (1 + 2 j + z)^-k [Gamma]^j Gamma[k])/(j! + 2 j j!), {j, 0, Infinity}];

f[2, 10^16, 1] // Simplify(* for: [Gamma] = 10^16; k = 2; z = 1 *)

(*(2 - 2/E^10000000000000000 + EulerGamma - 
ExpIntegralEi[-10000000000000000] + 
Log[10000000000000000])/40000000000000000*)

N[f[2, 10^16, 1] // Simplify, 20](* for: [Gamma] = 10^16; k = 2; z = 1 *)

(*9.85464428820156595122359384126*10^-16*)

Answered by Mariusz Iwaniuk on February 12, 2021

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