Checking regular variance around 0, hypergeometric fucntion

1. Jun 20, 2011

jjhyun90

1. The problem statement, all variables and given/known data
A function g is $\alpha$-regularly varying around zero if for all $\lambda > 0$, $\lim_{x\to 0} \frac{g(\lambda x)}{g(x)}=\lambda^{\alpha}$
For real s and $\alpha \in (0,1)$, define f:
$f(s)=1-\alpha \int_{0}^{\infty} e^{\alpha t} \frac{\frac{1}{1+s^2}}{e^t(1-\frac{1}{1+s^2})+\frac{1}{1+s^2}} dt = 1 + \alpha \sum_{n=1}^{\infty} \frac{(-s^{-2})^{n}}{n+\alpha}$.
Show f is $2\alpha$-regularly varying around zero.

2. Relevant equations
Note $\frac{1}{1+s^2}$ is a characteristic function of Laplace distribution.

3. The attempt at a solution
I am not familiar with hypergeometric series. For example, I do not know how to show that the series converges to $-\frac{1}{\alpha}$ when s goes to 0. Are there any properties of hypergeometric series that might be useful for proving this? Attempt to directly compute the limit failed.

Last edited: Jun 20, 2011
2. Jun 20, 2011

lanedance

so, i may be reading it wrong but can you make the following simplifications
$$f(s) = 1-\alpha \int_0^{\infty} e^{\alpha t} \frac{\frac{1}{1+s^2}} {e^t(1- \frac{1}{1+s^2}) + \frac{1}{1+s^2}}dt$$

$$= 1-\alpha \int_0^{\infty} e^{(\alpha -1)t} \frac{1} {(1+s^2- 1) + 1}dt$$

$$= 1-\alpha \int_0^{\infty} \frac{e^{(\alpha -1)t}} {1+s^2}dt$$

Last edited: Jun 21, 2011
3. Jun 20, 2011

lanedance

then i would be thinking consider expanding in the function in either integral or summation for around s=0 in terms of s

substitute $s = s\lambda$ and look at taking the limit or something along those lines...

4. Jun 20, 2011

jjhyun90

Thank you for the reply, but the $e^t$ on the denominator is multiplied to the first term only, so the simplification doesn't quite work.

5. Jun 21, 2011

lanedance

ok so that points you in the direction of using the sum, which is looking easier anyway
$$f(s) = 1 + \alpha \Sum_{n=1}^{infty} \frac{(-s^{-2})^n}{n+\alpha} = 1 + \alpha \Sum_{n=1}^{infty} \frac{(-1)^n s^{-2n}}{n+\alpha}$$

now when s is small you can approximate f(s) by
$$f(s) \approx f(0) + \frac{df(0)}{ds}s$$

try and come up with a similar approximation for $f(s\lambda)$ then consider the limit of the ratio

Last edited: Jun 21, 2011