The error of using command "integral"

I am trying to integrate a function , but the result is wrong. The strange thing is, if I integrate from -5 to 5, it is correct (0.0117). If I integrate it from -15 to 5, it will wrong (get 0). And then, if I integrate it from -15 to -5, the result is still 0.... I don't know where does matlab steal my 0.0117.............

The test code:

theta=4;
SP=@(X,D) univariate2dirboxspline_fast(cos(theta/180*pi),sin(theta/180*pi),X,D);
P2=0.394147314513724;
P3=1.391711364773548;
integral(@(X)SP(X,P2).*SP(X,P3),-5,5) % from -5 to 5 is 0.0117
integral(@(X)SP(X,P2).*SP(X,P3),-15,5) % from -15 to 5 is 0
integral(@(X)SP(X,P2).*SP(X,P3),-15,-5) % from -15 to -5 is 0

Related function, it can create a 2 dimension box spline, where D is the shifting value

function v = univariate2dirboxspline_fast(xi1, xi2, x,d)
%% Expects xi1, xi2, and x to be row vectors
xi = [xi1; xi2];
xi(abs(xi)<sin(1/180*pi))=0;
% x = abs(x - sum(xi)/2); % Symmetrize around 0
x=x-d;
x = abs(x);

l1 = vecnorm(xi, 1);
v = (l1/2-x) ./ abs(prod(xi));

h = 1 ./ vecnorm(xi, Inf);

v = max(v, 0);
v = min(v, h);

%v = piecewise(v > 0, v, 0);
%v = piecewise(v > h, h, v);
end

[Walter Roberson]

You need to add waypoints .

[Ziyu Shu]

excuse me, what is waypoints?

Answer 1

[Cris LaPierre]

I can't answer why it's doing that, but when I plot your function, I can understand why it may be missing the peak. The peak you are trying to integrate is basically an impulse.

Integral likely automatically sets the integration step size (at least there is no way to set it), and it appears to be stepping right over the peak, missing it entirely in the integration. That may not be exactly true, but it makes sense to me.

To have it return the correct point, use the 'waypoints' name, value pair to force the integration to include specific points in the integration. I used the peak of the impulse, and now integral returns the correct value.

integral(@(X)SP(X,P2).*SP(X,P3),-5,5,'Waypoints',0.89)

ans = 0.0117

integral(@(X)SP(X,P2).*SP(X,P3),-15,5,'Waypoints',0.89)

ans = 0.0117

integral(@(X)SP(X,P2).*SP(X,P3),-15,-5,'Waypoints',0.89)

ans = 0

I understand that you will not likely always know what the peak point is. Perhaps you could differentiate the function and solve for the zero point? Anyway, hopefully this helps you get started.

[Ziyu Shu]

Thank you so much!

It turns out if the integrate distance is smaller than 10, such as -5,5 -6,4, the result is correct...