Checking regular variance around 0, hypergeometric fucntion

jjhyun90
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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.
 
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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 />
 
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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...
 
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.
 
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
 
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