Error while executing integral function

function A = bessel_coefficients(F,alpha)
A = zeros(size(alpha));
s = size(alpha);
N = s(2);
for ii = 1:N
    a = 2/(besselj(1,alpha(ii)))^2;
    b =  @(r) besselj(0,alpha(ii)*r);
    fun = @(r) F(r)*b*r;
    A(ii) = a * integral(@(r) fun,0,1,"ArrayValued",true);
end

The error message I have recieved :

Operator '.*' is not supported for operands of type 'function_handle'
Error in integralCalc/iterateArrayValued (line 156)
                fxj = FUN(t(1)).*w(1);

Error in integralCalc/vadapt (line 130)
            [q,errbnd] = iterateArrayValued(u,tinterval,pathlen);

Error in integralCalc (line 75)
        [q,errbnd] = vadapt(@AtoBInvTransform,interval);

Error in integral (line 88)
Q = integralCalc(fun,a,b,opstruct);

Error in bessel_coefficients (line 10)
        A(i) = a * integral(@(r) fun,0,1,"ArrayValued",true);

I cant seem to figure out the solution to this issue. Any hints on what i can do?

[madhan ravi]

Would you mind providing the datas of F and alpha?

[Mahmoud Abdalhamid]

F is a (vectorized) function handle defining the initial position F(r)

alpha is a 1 × N row vector

the values of F and alpha change but here is an example of what F might be F = @(r) 0.2*cos(7*pi*r/2); N = 30;

Answer 1

[Ameer Hamza]

Change the line

fun = @(r) F(r)*b*r;

to

fun = @(r) F(r)*b(r)*r;

and replace

A(ii) = a * integral(@(r) fun,0,1,"ArrayValued",true);

with

A(ii) = a * integral(fun,0,1,"ArrayValued",true);

[Mahmoud Abdalhamid]

I still get the same error message

Thanks though

[Ameer Hamza]

yes, I suspected that the input F is not function handle. Can you specify the type of F? How do you call the function bessel_coefficients?

[Mahmoud Abdalhamid]

F is a (vectorized) function handle defining the initial position F(r)

the values of F and alpha change but here is an example of what F might be F = @(r) 0.2*cos(7*pi*r/2);

[Ameer Hamza]

Please check the updated answer.

[Steven Lord]

In addition to changing the definition of fun as Ameer Hamza stated, you also need to change your integral call.

% Evaluate both F and b and multiply the results together
fun = @(r) F(r)*b(r)*r;

% Your integrand function should return the value obtained by evaluating fun
% not fun itself
A(ii) = a * integral(@(r) fun(r),0,1,"ArrayValued",true);

% Actually, since fun is already a function handle
% you can write this more simply without an extra level of anonymous function
A(ii) = a * integral(fun,0,1,"ArrayValued",true);
% This is effectively equivalent to
% A(ii) = a * integral(@(r) F(r)*b(r)*r, 0, 1, "ArrayValued", true);

[Mahmoud Abdalhamid]

Thanks Ameer that was very helpful

[Ameer Hamza]

Steven, thanks for your input.

Mahmoud, I am glad to be of help.