Checking regular variance around 0, hypergeometric fucntion

[jjhyun90]

Homework Statement

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.

Homework Equations

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

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.

 

[lanedance]

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

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

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

 

[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...

 

[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.

 

[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