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) <br /> = 1-\alpha \int_0^{\infty} <br /> e^{\alpha t} <br /> \frac{\frac{1}{1+s^2}}<br /> {e^t(1- \frac{1}{1+s^2}) + \frac{1}{1+s^2}}dt

<br /> = 1-\alpha \int_0^{\infty} <br /> e^{(\alpha -1)t} <br /> \frac{1}<br /> {(1+s^2- 1) + 1}dt

<br /> = 1-\alpha \int_0^{\infty} <br /> \frac{e^{(\alpha -1)t}}<br /> {1+s^2}dt<br />

 

[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) <br /> = 1 + \alpha \Sum_{n=1}^{infty} \frac{(-s^{-2})^n}{n+\alpha}<br /> = 1 + \alpha \Sum_{n=1}^{infty} \frac{(-1)^n s^{-2n}}{n+\alpha}<br />

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