Passing limits through integrals

[rsq_a]

This seems to be an elementary question, but I rarely have to deal with rigor these days, so excuse me if there's a simple answer.

I'm looking for a theorem that will allow me to state,

$$F(z) = \int_0^b \frac{f(\tau)}{\tau - z} d\tau \sim \int_0^b \frac{f(\tau)}{\tau} d\tau$$

as $z \to 0$ and $f(\tau): \mathbb{C} \to \mathbb{C}$, but it's real along the path of integration.

I believe [itex]f(\tau)[/tex] satisfies these properties:

• $f(0) = 0$ and $f(b) = 0$
• $f(\tau)$ is Holder continuous along the path of integration

Note that F(z) is well-defined and Holder continuous along the line because of the values of f at the endpoints.

An example would be,

$$\int_0^1 \frac{\sqrt{\tau}}{\tau - z} d\tau$$

 

[CompuChip]

What is it that you are looking for. Do you want a theorem that will tell you that the integral you wrote down is a primitive of f(z), or do you want a theorem that will tell you that you can replace t - z by t if z goes to zero.
Because the latter simply follows from continuity of F(z) at z = 0.

 

[rsq_a]

Huh. I guess it's that simple. Sorry, tripped over myself there.

 

[CompuChip]

Would it make you feel less stupid to know that I usually look such things up, like when I really need to know the conditions of passing limits through integrals? Otherwise I usually write something like: "I don't know the exact theorem, but f is continuously differentiable which is - although overkill - definitely a sufficient condition."

 

[gammamcc]

Your example doesn't have f(b)=0...whuuuu?
How is f(x) Holder continuous along the real line segment [0,1]?
The derivative of the square root function is unbounded as x-> 0.

A basic convergence thm I would consider is the Lebesgue Dominated Convergence Thm.

Tough going to just deem something continuous, quote a thm and call it a day.
You have to have gone to a high-tuition college to get away with that and still you
may be wrong.