I wanted to solve a parametric double integral. Matlab requires me to set these parameters (for my case x0, x1, x2, x3, x4, x5, x6, x7, x8 and x9). The variables are r and t. Is there a solution with Matlab to solve a double integral set without sett

I wanted to solve a parametric double integral. Matlab requires me to set these parameters (for my case x0, x1, x2, x3, x4, x5, x6, x7, x8 and x9).

The variables are r and t.

Is there a solution with Matlab to solve a double integral set without setting any values of the parameters.

here is the code:

------------------------------------

 syms t r x0 x1 x2 x3 x4 x5 x6 x7 x8 x9;
 tmin = 0; 
 tmax = 1; 
 rmin = 0; 
 rmax = 2*pi; 

 g = +(x0*cos(-2*t)-x5*sin(-2*t))*(1-r^(-2))+(x1*cos(-1*t) - x6*sin(-1*t))*(1-r^(-1))+(x2*cos(0*t)-x7*sin(0*t))*(1-r^(0))+(x3*cos(1*t)-x8*sin(1*t))*(1-r^(1))+(x4*cos(2*t)-x9*sin(2*t))*(1-r^(2));

 h = +(x0*sin(-2*t)+x5*cos(-2*t))*(1-r^(-2))+(x1*sin(-1*t)+x6*cos(-1*t))*(1-r^(-1))+(x2*sin(0*t)+x7*cos(0*t))*(1-r^(0))+(x3*sin(1*t)+x8*cos(1*t))*(1-r^(1))+(x4*sin(2*t)+x9*cos(2*t))*(1-r^(2));

 f = r*(g^2+h^2);

 fun = @(t,r) f; 
 format long 
 J = integral2(fun,tmin,tmax,rmin,rmax,'Method','iterated','AbsTol',1e-12 ,'RelTol',1e-10)

---------------end-------------------

thank you in advance

Answer 1

[Walter Roberson]

You are already using syms so you have the symbolic toolbox. Use int() to do the integration.

int(int(f, t, tmin, tmax), r, rmin, rmax)

The result cannot be a numeric value because of the free variables.

Note: the solution is going to be +/- infinity depending on the value of some of the x* coefficients, because you have r^(-2) and 0 is in the range of r: you have singularities in your integral.

Answer 2

[Babacar Ndiaye]

thank you for the answer. That works now.... I see that the value will be +/- infinity depending on the value of x0,...,x9. I have to check the expression. I have a sum which varies from -n to n, with n = 2. Due to the singularities, I have to change the expression from 0 to 2n + 1. thank you for everything

[Walter Roberson]

Maple says that the result of what you originally posted will be

signum((1/2)*x0^2+(1/2)*x5^2)*infinity

which turns out to depend only on x0 and x5

[Babacar Ndiaye]

I change the first value in the sum (from 0 to N=5), but I can't obtained a value. I don't known what is the problem.

Here is the code:

-------------------------

syms t r x0 x1 x2 x3 x4 x5 x6 x7 x8 x9;
 tmin = 0; 
 tmax = 2*pi; 
 rmin = 0; 
 rmax = 1; 
 g = +(x0*cos(0*t)-x5*sin(0*t))*(1-r^(0))+(x1*cos(1*t)-x6*sin(1*t))*(1-r^(1))+(x2*cos(2*t)-x7*sin(2*t))*(1-r^(2))+(x3*cos(3*t)-x8*sin(3*t))*(1-r^(3))+(x4*cos(4*t)-x9*sin(4*t))*(1-r^(4));
 h = +(x0*sin(0*t)+x5*cos(0*t))*(1-r^(0))+(x1*sin(1*t)+x6*cos(1*t))*(1-r^(1))+(x2*sin(2*t)+x7*cos(2*t))*(1-r^(2))+(x3*sin(3*t)+x8*cos(3*t))*(1-r^(3))+(x4*sin(4*t)+x9*cos(4*t))*(1-r^(4));
 f = r*(g^2+h^2);
 fun = @(t,r) f; 
 J = int(int(fun,t,tmin,tmax),r,rmin,rmax)

-----------------------

can you have any idea for that.

thank on advance.

[Walter Roberson]

Although the result, J, shows up in terms of int(), if you simplify(J) then the closed form solution is returned.

[Babacar Ndiaye]

it works well now. thank you for the indication.