How to do asymptotics for integrals?

Question

What's a good way to find how fast the integral of a function is growing near a pole of the function? Here is what I mean on an example.

Look at 1/z. If I want to find out how fast ∫₀^a 1/(z-ε)dz is growing when ε->0, ε∈C, I can do this:

∫₀^a 1/(z-ε)dz = ln((a-ε)/ε)=-ln(-ε)+ln(a)+ε/a+O(ε).

What if I have ∫₀^a f(z)/(z-ε) dz , where f(z) is finite?

Answer 1

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

∫₀¹ f(z)/(z-ε) dz = f(1)log(1-ε) - f(0)log(-ε) - ∫₀¹ f'(z)log(z-ε) dz.

For the integral on the right-hand side, note that when you set ε to 0, you get ∫₀¹ 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

∫₀¹ f'(z)log(z-ε) dz = ∫₀¹ f'(z)log(z) dz + ∫₀¹ 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 = u²:

∫₀¹ f'(z)(log(z-ε) - log(z)) dz = 2∫₀¹f'(u²) u(log(u²- ε) - log(u²)) 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

∫₀¹ f(z)/(z-ε) dz = -f(0)log(-ε) - ∫₀¹ f'(z)log(z) dz + o(1).