# How to do asymptotics for integrals?

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 ∫0a 1/(z-ε)dz is growing when ε->0, ε∈C, I can do this:

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

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

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My first guess is that you might need something slightly stronger than being differentiable (perhaps bounded variation in some neighbourhood of 0). –  Yemon Choi Nov 2 '09 at 7:45
I rephrased the question, so hopefully its more clear. –  Sevak Mkrtchyan Nov 2 '09 at 8:20

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).

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Thanks. That's exactly what I was looking for. I had been looking in the wrong direction. –  Sevak Mkrtchyan Nov 2 '09 at 8:40