Hi Jacky Wang,
I understand that you want value of a complex function that need to be integrated twice.
To obtain a numerical solution for the double integral, the following might be a possible workaround:
Use the `vpa` function to evaluate the symbolic expression at a given precision. And then use the `double` function to convert the result to a double-precision floating-point number.
Here's an example of how the code can be modified to obtain a numerical solution:
syms a b x y
fun1 = exp(-( (a-x0).^2+(b-y0).^2 ).^2).*...
exp(-1i.*(k.* ( (a-x0).^2+(b-y0).^2 )./2) ) .*...
exp(-1i.*2.*pi.*(a.*(x-x0)+b.*(y-y0)));
% Define limits and precision
limit = 10;
precision = 10;
% Evaluate double integral using symbolic math
f0_sym = int(fun1,a,-limit,limit);
f0_sym = int(f0_sym,b,-limit,limit);
% Evaluate symbolic expression at given precision
f0_precise = vpa(f0_sym, precision);
% Convert result to double-precision floating-point number
f0_numeric = double(f0_precise);
In this example, the `vpa` function is used to evaluate the symbolic expression at a precision of 10 decimal digits. Adjust the `precision` argument as needed to obtain the desired level of accuracy. The `double` function is then used to convert the resulting vpa object to a double-precision floating-point number.
For more understanding, kindly go through following links:
