Assistance with integration limits and implementing it in Matlab

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Jesse
Jesse 2015-3-5
评论： Torsten 2015-3-6
Greetings all,
Back at it again with a little more complicated issue that I tried, but stuck in neutral here.
I'm going to introduce the functions first, and then how I coded them, and then ask my question because, while it makes sense to me equation wise, coding it is another thing (per usual)!
So I have the following integral function (sorry I can't do symbols here):
Integral with limits Psi_n-1 (read "Psi sub n-1) to infinity Phi(Psi)dPsi - Integral with limits Psi_n to infinity Phi(Psi)dPsi
where Phi(Psi) = 4*Psi/(4(Psi)^2)^3/2
and Psi = z/r.
Now for how I coded this:
r=3.66;
z=[1.3336 1.3370 1.3438 1.3594 1.3700 1.3577 1.3507 1.3559 1.2900 1.3525];
Psi=z/r;
n=10;
Phi=@(Psi) (4*Psi_/(4*(Psi).^2+1).^3/2;
q_Phi_a=integral(Phi,n-1,inf, 'ArrayValued', true);
q_Phi_b=integral(Phi,n,inf, 'ArrayValued', true);
Now here's the thing. q_Phi_a and q_Phi_b appear to be the same thing, so when you subtract them, there going to be zero, which is not what I thought. The only thing I can think of is how I am doing the limits.
In the equation, the bottom limit is either Psi sub n-1 or Psi sub n. I don't think I am doing this right is Matlab. Any advice on this? Do I need another function handle in the limits of the integral function?
Let me know if I need to attach anything from MathType for a clearer picture.
Thanks!
-J

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回答（1 个）

Torsten
Torsten 2015-3-5
As written, your integral is simply
1/(4*Psi(n-1)^2+1)^0.5 - 1/(4*Psi(n)^2+1)^0.5
(an antiderivative of f(x)=4x/(4x^2+1)^1.5 is F(x)=-1/(4x^2+1)^0.5)
Best wishes
Torsten.
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Jesse
Jesse 2015-3-5
编辑：Jesse 2015-3-5
Ah, ok I see what's going on, but what about the fact that Psi changes with respect to z? I mean Psi is a vector, so if I set this equation equal to q, q should also be a vector.
Torsten
Torsten 2015-3-6
The above value of the integral is
Integral with limits Psi_n-1 (read "Psi sub n-1) to infinity Phi(Psi)dPsi - Integral with limits Psi_n to infinity Phi(Psi)dPsi.
Make a loop over n to get the vector you are searching for:
for n=2:numel(Psi)
value_integral(n-1)=1/(4*Psi(n-1)^2+1)^0.5 - 1/(4*Psi(n)^2+1)^0.5;
end
Best wishes
Torsten.

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