# Matlab: function handle integration with several variables

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Pro B on 9 Jun 2017
Commented: Walter Roberson on 12 Jun 2017
My goal here is to build an array (cell array since I'm working with function handles) via a for loop and take the integral of each element, plug in a value and get an array. But I get the following error:
Input function must return 'double' or 'single' values. Found 'function_handle'.
The error occurs on the line when I'm trying to plug in the value 1 (or any scalar value) for x_2. Any tips on how to "handle" this error? Note a(1,1) and c(1,1) are both scalar values (eg. 0 and 1).
Here is the code:
FUN_1 = @(y_1,y_2,x_1,x_2)sum(heaviside(y_1-a_k(1:m,1)).*dirac(1,y_2-a_k(1:m,2))).*(-1/2.*log((x_1-y_1).^2+(x_2-y_2).^2))+(x_1-y_1).^2./((x_1-y_1).^2)+sum(dirac(y_1-a_k(1:m,1)).*dirac(y_2-a_k(1:m,2))).*(-1/2.*log((x_1-y_1).^2+(x_2-y_2).^2))+(x_1-y_1).*(x_2-y_2)./((x_1-y_1).^2+(x_2-y_2).^2);
Q_1 = @(x_1,x_2)integral2(@(y_1,y_2)FUN_1(y_1,y_2,x_1,x_2),a(1,1),c(1,1),a(1,2),c(1,2));
FUN_2 = @(y_1,y_2,x_1,x_2)sum(heaviside(y_1-a_k(1:m,1)).*dirac(1,y_2-a_k(1:m,2))).*(-1/2.*log((x_1-y_1).^2+(x_2-y_2).^2))+(x_1-y_1).*(x_2-y_2)./((x_1-y_1).^2)+sum(dirac(y_1-a_k(1:m,1)).*dirac(y_2-a_k(1:m,2))).*(-1/2.*log((x_1-y_1).^2+(x_2-y_2).^2))+(x_2-y_2).^2./((x_1-y_1).^2+(x_2-y_2).^2);
Q_2 = @(x_1,x_2)integral2(@(y_1,y_2)FUN_1(y_1,y_2,x_1,x_2),a(1,1),c(1,1),a(1,2),c(1,2));
k = zeros(1,2*M);
n=0;
for n = 0:2*M-1
S = @(x_1,x_2)Q_1(x_1,x_2)*2*n*(x_1+1i*x_2)^(n-1) + Q_2(x_1,x_2)*2*n*1i*(x_1+1i*x_2)^(n-1);
R = @(x_2)integral(@(x_1)S,a(1,1),c(1,1));
k(1,n+1) = R(1);
end
disp(k);
end

Guillaume on 9 Jun 2017
A guess as I've not really tried to understand what your functions are doing:
R = @(x_2) integal(@(x_1) S(x_1, x_2), a, c)
The error you get makes sense. (@(x_1) S is a function that returns the function handle S regardless of the input. My modification returns the result of S(x_1, x_2) for input x_1 received from integral.
Note that if a and c are scalar, there's absolutely no point in writing them as a(1, 1) and c(1, 1) other than puzzling the reader and making them wonder if you meant to pass instead some other variable that was a 2D array.
Walter Roberson on 12 Jun 2017
You posted a Question about this; I replied there. You missed that integral() passes in a vector of values.