4D integral using integral and integral3

I tried 4D integral using integral and integral3 like the following link: https://jp.mathworks.com/help/matlab/ref/integral3.html

however, it does not work when a function used become complecated.

Here I show a simplified code I used.

f = @(a, b, c, d) a+b+c+d;
I = @(a, b, c) integral(@(d) f(a, b, c, d), 0, 1);
f2=@(a, b, c) I(a,b,c)+1;
I2=integral3(f2, 0, 1, 0, 1, 0, 1);

It showed an error like "The array size does not fit this operation." (translated from Japanese).

But I did not figure out which array it said.

My question is what is the acutual issue and how to solve it to achieve the 4D integral.

Answer 1

[Matt J]

f = @(a, b, c, d) a+b+c+d;
I = @(a, b, c) integral(@(d) f(a, b, c, d), 0, 1,'ArrayValued',1);
f2=@(a, b, c) I(a,b,c)+1;
I2=integral3(f2, 0, 1, 0, 1, 0, 1)

I2 = 3.0000

[亮介 桶谷]

Thank you!!

'ArrayValued' also solved problem in my actual code!!