solving a triple sigma equation (or triple integral)

above equation is what i want to express by mathlab code

for definite the A(x,y,z)

A(x,y,z) : random size 3-D vector 00x00x00 (but example is 6x61x17)

below is my try, but i think that is not correct

it is hard to make code... please somebody help me

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

clear

% for A = 6 x 61 x 17  (3-d vector)
A = ones(6,61,17);

k = linspace(10.4, 31.4, 6);
pii = linspace(0, pi, 61);           % pii = 𝝓
E = linspace(-2, 2, 17);

for x = 1:6
    for y = 1:61
        for z = 17
            for f = 1:6
                for p = 1:61
                    for e = 1:17
                        A(x,y,z) = A(x,y,z) + (k(f)).^2 * (1/(norm([x y z]-[cos(pii(p)) sin(pii(p)) E(e)]))) * (1-cos(pii(p))-sin(pii(p))) * exp(1i*k(f)*(((norm([x y z]-[cos(pii(p)) sin(pii(p)) E(e)])))-1));
                    end
                end
            end
        end
    end
end

[Torsten]

And how does the triple integral look like ?

[형준 이]

you mean the original function expressed as an integral?

[Torsten]

I don't understand your triple integral. E.g. integration with respect to k from 0 to Inf - why should it exist with a k^2-term ?

[형준 이]

thankyou for your reply

I understand your question

In the original triple integral equation, the integral range is from 0 to inf, 0 to 2*pi, -inf to inf,

but the actual values inserted on equation are finite discrete samples.

Therefore, I transformed the expression to the sigma symbol instead of the integral.

There are 6 smaples of k, 61 of 𝝓, 17 of E which are actually used.

[Torsten]

When you approximate an integral by a sum, there must appear terms of the form

(k(i+1)-k(i))*(phi(j+1)-phi(j))*(E(n+1)-E(n))

for the volume element in which you evaluate the function.

I don't see this volume element in your finite sum.

Answer 1

[Chunru]

% for A = 6 x 61 x 17  (3-d array)
A = ones(6,61,17);

k = linspace(10.4, 31.4, 6);
pii = linspace(0, pi, 61);           % pii = 𝝓
E = linspace(-2, 2, 17);

% reshape to 3d array
k3d = reshape(k, 6, 1, 1);          % along 1st dimension
p3d = reshape(pii, 1, 61, 1);       % along 2nd dimension
e3d = reshape(E, 1, 1, 17);         % along 3rd dimension

for x = 1:6
    for y = 1:61
        for z = 1:17
            R = sqrt((x-cos(p3d)).^2 +(y-sin(p3d)).^2 + (z-e3d).^2);
            tmp = k3d.^2 ./ R .*  (1-x*cos(p3d)-y*sin(p3d)) .* exp(1i * k3d .*(R-1));
                    
            A(x,y,z) = sum(tmp, 'all');
        end
    end
end
imagesc(20*log10(abs(A(:, :, 8))));   % slice for z=8

[형준 이]

i appreciate your response