How to evaluate integral from 0 to inf of besselj(0,kr) * besselj(0,kR) * 1/k * (2 - e^-kz - ek(z-L)) dk
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Hello,
I am trying to evaluate the integral given in equation 6 in this paper (see also eq 10 for g()):
"Coulomb potential and energy of a uniformly charged cylindrical shell"
I've tried it symbolically with Python Sympy library, and the code never finished running. I tried numerically with scipy.integrate.quad , but the margin of error in the answer was huge, 1/4 of the answer. Finally I tried in MATLAB, and got the same errors that Python was giving.
syms r R z L k
f = @(k) besselj(0, r * k) * besselj(0, R * k) * (1/k) * (2 - exp(-k*z) - exp(k*(z-L)))
f =
function_handle with value:
@(k)besselj(0,r*k)*besselj(0,R*k)*(1/k)*(2-exp(-k*z)-exp(k*(z-L)))
int(f,k,0,inf)
ans =
int(-(besselj(0, R*k)*besselj(0, k*r)*(exp(-k*z) + exp(-k*(L - z)) - 2))/k, k, 0, Inf)
Where:
k = Variable of Integration
R = Radius of Charged Cylinder
L = Length of Cylinder
r = r-coordinate of Point at which the Potential is being measured, (In Cylindrical Coordinates)
z = z-coordinate of Point
Is it possible to get a symbolic solution? If not, how could I get a numerical solution, if I specied values for R, L, r, and z?
Thank you!
2 commentaires
cr
le 19 Déc 2022
You cannot per se compute a definite integral from 0 to Inf but for a convergent summation you can get a very good approximation. Did you try symbolics and is it not working?
Réponse acceptée
Fabio Freschi
le 19 Déc 2022
Modifié(e) : Fabio Freschi
le 19 Déc 2022
For the numerical integration, you can use integral, that also accepts Inf as integration upper bound
clear variables, close all
% some randoms values for the params
r = 1;
R = 2;
z = 1.5;
L = 2;
% function handle
f = @(k)besselj(0,r*k).*besselj(0,R*k).*(1./k).*(2-exp(-k*z)-exp(k*(z-L)));
% indefinite integral
Vinf = integral(f,0,Inf);
One could worry about the warning message. However, without playing with the tolerances and other optional inputs of integral, it seems that the result is reasonable:
% definite integral with increasing upper bound
N = 100;
V = zeros(N,1);
for i = 1:N
V(i) = integral(f,0,max([r,R,z,L])*i);
end
figure, hold on
plot(1:N,V)
plot([1 N],[Vinf,Vinf],'--')
legend('V','Vinf')
2 commentaires
Fabio Freschi
le 22 Déc 2022
@Sean: my pleasure! if this answer solves your original problem, please accept it
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