Please help, I have no idea how to do this integral in matlab
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Walter Roberson
am 2 Nov. 2023
No closed form solution. Runs the risk of having a singularity if a_0 > theta_0
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Torsten
am 2 Nov. 2023
Bearbeitet: Torsten
am 3 Nov. 2023
The integral seems to exist for those values of theta0 with cos(theta0) ~= 0, thus theta0 ~= (2*k+1)*pi/2 for k in Z.
In this case, the Taylor expansion of sin(theta0)-sin(theta) around theta0 shows a behaviour like 1/sqrt(x) around x=0.
If cos(theta0) = 0, the Taylor expansion of sin(theta0)-sin(theta) around theta0 shows a behaviour like 1/x around x = 0 which means that the integral diverges.
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Walter Roberson
am 2 Nov. 2023
Bearbeitet: Walter Roberson
am 3 Nov. 2023
syms a_0 positive
syms theta real; assumeAlso(theta >= 0);
theta_0 = sym(30.5);
F = 1/(sind(theta_0) - sind(theta))
a_0 = sym([5:5:25 26 27 28 29]).';
values = arrayfun(@(A) int(F, theta, 0, A), a_0);
values(1) %for example
valued = double(values);
plot(a_0, valued)
4 Kommentare
Walter Roberson
am 3 Nov. 2023
@Torsten you are right, I did forget the sqrt() !
It looks like MATLAB is able to integrate even so, but with a more complicated formula.
The imaginary components of the results are so small that they are surely due to round-off error in the numeric calculations.
format long g
syms a_0 positive
syms theta real; assumeAlso(theta >= 0);
theta_0 = sym(30.5);
F = 1/sqrt(sind(theta_0) - sind(theta))
a_0 = sym([5:5:25 26 27 28 29 30:.1:30.4]).';
values = arrayfun(@(A) int(F, theta, 0, A), a_0);
values(1) %for example
valued = double(values);
plot(a_0, real(valued), a_0, imag(valued))
legend({'real', 'imaginary'})
isimag = imag(valued) ~= 0;
a_0(isimag)
valued(isimag)
Torsten
am 3 Nov. 2023
Your values seem to converge towards
syms theta theta0
theta0 = 30.5*pi/180;
f = 1/sqrt(sin(theta0)-sin(theta));
sol = vpaintegral(f,theta,0,theta0)
sol = sol * 180/pi
vpa(sol)
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