Mathematica Asked on February 12, 2021
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)
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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