Problem with tolerances in integral

While evaluating an integral with expensive computation of the integrand, I tried the adaptive integration provided by the integral function. Unfortunately the function seems not to accept the desired (lower) tolerance argument, if the integral is not improper as in the example from the help file. Instead of taking the desired tolerance, integral seems to work with default tolerance in any case. I give a minimal example for illustration:

if true
  % code
f = @(t) log(t); 
I_true1 = -1;
I_true2 = 3*log(3)-2*log(2)-1;
for k = 2:2:12
  epsi = 10^(-k);
  I1 = integral(f,0,1,'RelTol',0,'AbsTol',epsi);
  I2 = integral(f,2,3,'RelTol',0,'AbsTol',epsi);
  fprintf('%1.1e    %1.4e   %1.4e\n', epsi, abs(I1-I_true1), abs(I2-I_true2))
end
end

producing the output

1.0e-02 1.7536e-07 1.1102e-16

1.0e-04 1.7536e-07 1.1102e-16

1.0e-06 1.0960e-08 1.1102e-16

1.0e-08 1.7124e-10 1.1102e-16

1.0e-10 6.6880e-13 1.1102e-16

1.0e-12 1.0214e-14 1.1102e-16

Has anyone an idea how to change this behaviour? Thanks in advance for your help!

Answer 1

[Mario]

In the meanwhile I found the answer: integral works initially with 150 evaluations of the integrand. In the example I found problematic, machine precision is reached already in this initial configuration.