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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Sean on 19 Dec 2022
Commented: Fabio Freschi on 22 Dec 2022
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!
cr on 19 Dec 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?
Sean on 19 Dec 2022
Yes

Fabio Freschi on 19 Dec 2022
Edited: Fabio Freschi on 19 Dec 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);
Warning: Reached the limit on the maximum number of intervals in use. Approximate bound on error is 3.4e-06. The integral may not exist, or it may be difficult to approximate numerically to the requested accuracy.
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')
Torsten on 19 Dec 2022
@Sean comment moved here:
Thank you Fabio! :)
Fabio Freschi on 22 Dec 2022

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