Yes, you are right, Camille. If alpha < 1, your function is integrable, but the infinity in its integrand at x = 0 is apparently proving to be too difficult for matlab's 'integral' function to evaluate it properly.
However, if you make the following simple change of variable, you should encounter no such difficulty. To simplify notation use 'a' instead of 'alpha'. Define variable z as:
z = x^(1-a)
Then we have
dz = (1-a)/x^a*dx
and
(1/x-x)^a*dx = (1-x^2)^a/x^a*dx = (1-z^(2/(1-a)))^a/(1-a)*dz
Hence you should evaluate the integral
F = @(z) (1-z^(2/(1-a)))^a/(1-a);
y = integral(F,0,r^(1-a));
Matlab should have no trouble with this integral provided a < 1, since it has no singularities in its integrand.
