Reimann sum and numerical integration

I am trying to solve the functions below numerically using MATLAB

This is my code where I have used "integral" command to solve for the second integral, while I used Reimann sum to solve for the first. Take for example c=1/128.

    x= -1/128:0.0001:1/128;

    for j=1:length(x)

        den= @(y)  pi^2.* sqrt(1- (x(j)+y).^2) .* sqrt(1- y.^2);
        fun= @(y) 1./den(y);
        ymin= max(-0.999,-0.999-x(j));
        ymax= min(0.999,0.999-x(j));
        pdfX(j)=integral(fun,ymin,ymax);

    end

    sum1=0;
    for j=1:length(omegat)
       func1(j)=pdfX(j)*0.0001;
       sum1=sum1+func1(j);
    end

    F=sum1;

Do you think it is correct? Are there any other ways to make it more accurate?

Thanks

Answer 1

[Star Strider]

I would likely use either trapz or cumtrapz to calculate ‘F’, but otherwise I see no problems.

[DM]

can you please explain to me why? and what would I write instead... Thanks

[DM]

Also did you notice I have 0.999 instead of 1, so as to avoid the singularity point, is that OK ?

[Star Strider]

Personal preference, really. The trapz function is designed for numerical integration:

F = trapz(x, pdfX)

Avoiding the singularity is likely a good move. I doubt that substituting -0.999 would produce significant error. You might want to experiment with (1-1E-8) instead if you are curious.