# Approximation of integral for small boundary

1. Jun 4, 2012

### Fraggler

This problem arises in a paper on population genetics (Kimura 1962).

1. The problem statement
Let $f(p) = \int_0^p ((1 - x)/x)^k dx$.

For a small value of p, we have approximately
f(p) = (p ^ (1-k)) / (1-k)

How is this obtained?

2. My attempt at a solution
I tried to expand the f(p) around p = 0. However, f'(p) = ((1 - p)/p)^k is undefined at p=0. Furthermore, it does not seem that this approach can yield the form p^(1-k) / (1-k). I must be missing something.

I would appreciate any insights. Thanks.

2. Jun 4, 2012

### Villyer

$\frac{d}{dp} \frac{p^{1-k}}{1-k} = \frac{1}{p^{k}}$

And the derivative of your original function you said was $(\frac{1-p}{p})^{k}$

Does that help at all?

3. Jun 4, 2012

### Fraggler

Thanks for your reply Villyer. I've just solved it, with some help from a friend.

Here's the solution.

For $x \approx 0$, $(\frac{1 - x}{x})^k = x^{-k}$. Therefore, $f(p) = \int_0^p x^{-k} dx = \frac{1}{1-k} p^{1 - k}$, as required.

Cheers!

4. Jun 4, 2012

Exactly.