How to integrate a vector-valued function?

I am trying to calculate the attached integral. I have data as vectors r and f(r) tht I load as r and fr, respectively.. The vector x should have the same range as r. Is the following correct? I'm not sure how I should incorporate the fact that f(r) is a function of r, or how the integral depends on x as well.

I try

x = linspace(0.05,17.95,180)';
fun = @(x,r) (r.^2).*(fr-1).*sin(x.*r)./(x.*r);
eq1 = integral(@(r) fun(x,r),0,r(end),'ArrayValued',1);
eq2 = cumtrapz(r,(r.^2).*(fr-1).*sin(x.*r)./(x.*r));

But I get two different answers. Again, r and fr (= f(r)) are both vectors of numbers. Is either eq1 or eq2 one correct? Is neither? I have limited understanding of anonymous functions and numerical integration in MATLAB.

Answer 1

[Ameer Hamza]

You need to integrate this function w.r.t. 'r' so you shouldn't use x as a vector in your code. Both of your approaches seem to be doing what you might not intend to do. Check this code, it shows how to do this with integral and trapz. Both will give similar results.

R = linspace(1, 100, 1000); % example r
FR = exp(-R); % example fr
x = linspace(0.05,17.95,180)';

fr = @(r) interp1(R, FR, r, 'pchip');
integrand = @(r,x) r.^2.*(fr(r)-1).*sin(x.*r)./(x.*r);

fx_int = @(x) integral(@(r) integrand(r,x), 0, max(R));
fx_int_v = zeros(size(x));
for i=1:numel(x)
    fx_int_v(i) = fx_int(x(i));
end

fx_trapz = @(x) trapz(R, integrand(R, x));
fx_trapz_v = zeros(size(x));
for i=1:numel(x)
    fx_trapz_v(i) = fx_trapz(x(i));
end

[N/A]

Thanks, Ameer. One follow-up question: My input 'r' vector is the same as my 'x' vector (which is why I arbitrarily chose the 'x' vector that I did, not sure if that was the correct thing to do or not). Anyway, as a result the following line: 'fr = @(r) interp1(R, FR, r, 'pchip');' doesn't work. What do you suggest I do instead? Thanks again.

[Ameer Hamza]

Is there an error?

From the image you shared, the 'x' being same length as 'r' should be a coincidence. You can see from the equation that 'r' is the variable of integration, and it is independent of the value of 'x'.

[N/A]

Never mind, it works now. Thank you!

[Ameer Hamza]

I am glad to be of help.