Trouble with MATLAB Triple Integral

I'm attempting to evaluate this integral (and equation) where 'a' and 'x' are just inputs, and x' is (x,y,z). also: x will have the form [a,b,c]. In my 'int_func' below, I have taken the dot product of the transpose and the original, to yield the ^2 power that is observable below, for example in: (X(:,1)-x).^2

The code I have so far is:

int_func = @(x,y,z) func([x,y,z]).*exp((-3/(2*a^2)).*((X(:,1)-x).^2 + (X(:,2)-y).^2 + (X(:,3)-z).^2));

val = integral3(int_func,-inf,inf,-inf,inf,-inf,inf);

disp(val * (3/(2*pi*a^2))^(3/2));

where:

X = [0, 0, 0];

a = 1;

With this code, I should be getting an answer of 0.0370, but instead I am getting 0.5556

Can someone suggest another form of evaluating this integral or spot what is wrong with what I have listed?

[David Goodmanson]

Hi,

what does func({x,y,z]) look like? WIth that set to 1, the integral of the gaussian equals 1 as it should.

[Walter Roberson]

The original poster comments:

This my comment. I messaged MATLAB asking them to remove it and they said they would for me but it is still up. Please remove my question.

[Walter Roberson]

The policy is not to remove questions with a valid attempt an an answer from a volunteer, unless the question is abusive or not permitted by law; on rare occassions, some intellectual property claims are also accepted.

The process here is that public Questions get free public responses from volunteers, and that both are left in public view to be of potential assistance to whomever bothers to look at the material later.

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The most active volunteers encounter a lot of instances in which people expect them to give free major labour for the gain of one individual, and most of them have become pretty blunt with people that there are limits.

[Walter Roberson]

You indicated that the answer did not work for you. What difficulty did you have with the proposed solution?

Answer 1

[David Goodmanson]

Hi,

The problem with the func([x,y,z]) approach is that there is too much expectation that [x,y,x] is 1x3. Once the integration takes over, that's not so clear. So instead just address x,y,z directly:

X = [0, 0, 0];
a = 1;
int_func = @(x,y,z) func(x,y,z).*exp((-3/(2*a^2)).*((X(1)-x).^2 + (X(2)-y).^2 + (X(3)-z).^2));
val = integral3(int_func,-inf,inf,-inf,inf,-inf,inf);
z = val * (3/(2*pi*a^2))^(3/2)
function result = func(x,y,z)
result = x.^2.*y.^2.*z.^2;
end

z = 0.0370

I also replaced X(:,1) by X(1) etc, for related reasons.