Hi, I am trying to solve a complicated integral for my project and until now I couldn’t solve it by Mathematica. Can you help me to find a way to solve it? I send the code that I have written.
Thank you for helping me solve this integral
k = 0.235;
q = 1;
Ki = 910980;
\[Theta]e = 90 Degree; (*\[Theta]e is a variable that takes different values*)
(*Step 2:Define \[Theta]1 and Kf*)
\[Theta]1 = ArcCos[(Ki - k Cos[\[Theta]e])/Kf];
Kf = Sqrt[Ki^2 + k^2 + k *Ki *Cos[\[Theta]e]];
(*Step 3:Define vectors Kf,Ki,and q*)
KfVec = {Kf Sin[\[Theta]1], 0, Kf Cos[\[Theta]1]};
KiVec = {0, 0, Ki};
qVec = KiVec - KfVec;(*Vector q*)
kVec = {k Sin[\[Theta]e], 0, k Cos[\[Theta]e]};
(*Step 4:Calculate dot products*)
KfDotKi = Dot[KfVec, KiVec];
KfDotq = Dot[KfVec, qVec];
qDotKi = Dot[qVec, KiVec];
kDotq = Dot[qVec, kVec];
\[Theta]2 = ArcCos[kDotq/k*q];
q' = k^2 + q^2 - 2*k*q*Cos[\[Theta]2];
q2 = Dot[KfVec, (kVec - qVec)];
(*Step 5:Define A,B,C,D*)
A = (q^2 + \[Lambda]^2) (k^2 + \[Beta]^2/
4) x^2 + (\[Beta] (q^2 + \[Lambda]^2) +
2 \[Lambda] (k^2 + \[Beta]^2/4)) x + (q' + (\[Lambda] + \[Beta]/
2)^2)
B = -2 (I \[Lambda] Kf - KfDotq) (k^2 + \[Beta]^2/4) x^2 -
2 (( \[Beta] (I \[Lambda] Kf - KfDotq) +
I (k^2 + \[Beta]^2/4) Kf)) x -
2 (q2 + I (\[Lambda] + \[Beta]/2) Kf)
cExpr = -2 (k^2 + I \[Beta]/2 k) (q^2 + \[Lambda]^2) x^2 -
2 ((2 \[Lambda] (k^2 + I \[Beta]/2 k) +
I (q^2 + \[Lambda]^2) k) ) x -
2 (k^2 + I (\[Lambda] + \[Beta]/2) k - kDotq)
dExpr = 4 I \[Lambda] (KfDotq) (k^2 + I \[Beta]/2 k) x^2 +
4 (I (k + \[Beta]/2) k Kf + I \[Lambda] KfDotq) x + (2 *
Dot[KfVec, kVec] - k Kf)
simplifiedA = Simplify[A];
simplifiedB = Simplify[B];
simplifiedcExpr = Simplify[cExpr];
simplifieddExpr = Simplify[dExpr];
(*Define the integrand*)
mT = 7*1873;(*amu*)
mp = 16*1873;(*amu*)
\[Mu]pT = (mT*mp)/(mT + mp)
zp = 6;
zT = 2.8683;
\[Alpha]pT = \!\(TraditionalForm\`\[Mu]pT*zp*zT/Kf\)
\[Alpha]T1 = 1*2.8683/0.235simplifiedA = Simplify[A];
simplifiedB = Simplify[B];
simplifiedcExpr = Simplify[cExpr];
simplifieddExpr = Simplify[dExpr];
integrand = (1/ A)*((A/(A + B))^(-I \[Alpha]pT))*((A/(A + cExpr))^(-I \[Alpha]T1))*
Hypergeometric2F1[-I \[Alpha]pT, -I \[Alpha]T1,
1, (B cExpr - A dExpr)/((A + B) (A + cExpr))];
integrandsimple = FullSimplify[integrand];
F[_\[Beta], \[Lambda]] =
Integrate[integrandsimple, {x, 0, \[Infinity]},
Assumptions -> {\[Lambda] > 0, \[Beta] > 0}]
The line
\[Alpha]T1 = 1*2.8683/0.235simplifiedA = Simplify[A];
does not make sense. Perhaps you meant
\[Alpha]T1 = 1*2.8683/0.235simplifiedA;
You cannot start a definition with
F[_\[Beta], \[Lambda]] =
Perhaps you meant
F[\[Beta]_, \[Lambda]_] =
I have tried plotting your function with numeric values for the parameters:
ReImPlot[Evaluate[integrandsimple /. {\[Lambda] -> 1, \[Beta] -> 1}],
{x, 5.5, 5.55}]
and it appears to be wildly oscillating. Have you good reason to think that the function is integrable from 0 to infinity?
Hello ,I thanks for your answer.
By plotting the function, it seems that the function diverges in this interval. Right?
We used several rules to finally arrive at the simplest form of the integral like that and After taking the integral, we must take the derivative with respect to Beta and Landa, and betha=1.8683 then the limit of the term for Landa tends to zero from positive values.
in 2F1(a,b,c,z i find the condition for Convergence :abs(x)<1
this information can help to solve problem?
and for our project in one essay i had this
The obtained Gaussian hypergeometric function is much more useful for numerical solution due to its symmetry compared to α_T"1" and α_PT
Hello
Thanks for correcting the code. This is the integral that we obtained by simplifying, and in solving the integral in the articles, the range is from zero to infinity.
We actually want to obtain the transition amplitude in the single ionization of the lithium atom, and the transition amplitude is obtained from the integral of which this integral is a part.
Hello again. I ran the code on my system but it gives this error:
The integrand integrand[1,1,x] has evaluated to non-numerical values for all sampling points in the region with boundaries {{[Infinity],0.}}.
what is the reason?!
for me works fine on Mathematica 14.1.
Try:
ClearAll["`*"]; Remove["`*"];
(*code above*)
integrand[1, 1, x]
(*1/(136627 + 104418 x +
24418 x^2)40000 ((
136627 + 104418 x +
24418 x^2)/((127791 -
28200 I) + (95582 - 56400 I) x + (15582 -
18800 I) x^2))^(-28683 I/2350) ((
136627/40000 + (52209 x)/20000 + (12209 x^2)/20000)/(
127791/40000 - (3 I Sqrt[33195382416002209])/200 + (52209 x)/
20000 - (12209 I Sqrt[33195382416002209] x)/
4000000 + ((-2209 - 200 I Sqrt[33195382416002209]) x)/20000 + (
12209 x^2)/20000 - (
12209 (2209 + 200 I Sqrt[33195382416002209]) x^2)/800000000))^(-((
18051015024 I)/(575 Sqrt[33195382416002209])))
Hypergeometric2F1[-((28683 I)/2350), -((18051015024 I)/(
575 Sqrt[33195382416002209])),
1, (94 ((10000 (-94 + Sqrt[33195382416002209]) -
200 I (100 Sqrt[33195382416002209] +
47 (-200 + Sqrt[33195382416002209])) x - (220900 -
103823 I) x^2) (136627 + 104418 x +
24418 x^2) - ((300 -
94 I) + (600 - 94 I) x + (200 -
94 I) x^2) (80000 (-2209 I +
150 Sqrt[33195382416002209]) +
200 (12209 Sqrt[33195382416002209] +
200 (-2209 I + 200 Sqrt[33195382416002209])) x +
12209 (-2209 I +
200 Sqrt[33195382416002209]) x^2)))/(((127791 -
28200 I) + (95582 - 56400 I) x + (15582 -
18800 I) x^2) (20000 (87791 -
400 I Sqrt[33195382416002209] +
200 (200 - I Sqrt[33195382416002209])) +
200 (12209 (200 - I Sqrt[33195382416002209]) +
200 (37791 - 200 I Sqrt[33195382416002209])) x +
12209 (37791 - 200 I Sqrt[33195382416002209]) x^2))]*)
Hello ,
Thank you very much for your help. Finally I managed to run the code. In order to solve the problem that there is a partial derivative with respect to beta and then with respect to Landau, and then the limit of the obtained derivative when Landau tends to zero, maybe you can help me. Because the numerical integral is taken, I have a problem in getting the derivative.