If you just want to see how fast it blows up, it shouldn't be too hard. First integrate by parts:

∫_{0}^{1} f(z)/(z-ε) dz = f(1)log(1-ε) - f(0)log(-ε) - ∫_{0}^{1} f'(z)log(z-ε) dz.

For the integral on the right-hand side, note that when you set ε to 0, you get ∫_{0}^{1} f'(z)log(z) dz, which should converge (to a finite value) as long as f'(z) is bounded, so let's rewrite the integral as

∫_{0}^{1} f'(z)log(z-ε) dz = ∫_{0}^{1} f'(z)log(z) dz +
∫_{0}^{1} f'(z)(log(z-ε) - log(z)) dz.

The second integral looks like it should converge to 0 as ε goes to 0. To confirm this, it seems advantageous to deal with the singularity at z=0 first (there may be a cleaner way). Make the change of variables z = u^{2}:

∫_{0}^{1} f'(z)(log(z-ε) - log(z)) dz = 2∫_{0}^{1}f'(u^{2}) u(log(u^{2}- ε) - log(u^{2})) du,

and now it shouldn't be too hard to show that the integrand converges uniformly to 0 as ε goes to 0 if f'(z) is bounded. This gives the estimate

∫_{0}^{1} f(z)/(z-ε) dz = -f(0)log(-ε) - ∫_{0}^{1} f'(z)log(z) dz + o(1).