I am trying to prove for which values of $\alpha \geq 0$ the following function $f(x) := \exp(\log^\alpha(x))$ is varying regularly (or slowly) and if it is varying, I need to determine the index of variation.
- How do I prove that $\exp(\log^\alpha(x))$ is varying regularly for $\alpha \in (0,1)$ and what is the index? (The problem in particular is the handling of $\log^\alpha(\cdot)$ for non-integer $\alpha$.)
- How do I prove that $\exp(\log^\alpha(x))$ is $\underline{\text{not}}$ varying regularly for $\alpha > 1$?
With some research I have already found out, that this function is varying slowly (thus regularly with index = 0) for values $0 < \alpha < 1$, but without proof. Further, I have already found out that $f(x)$ is varying slowly for $\alpha = 0$ and regularly for $\alpha$ = 1, because:
For Case 1 ($\alpha = 0$):
$$\lim_{x\to\infty} \frac{f(\lambda x)}{f(x)} = \lim_{x\to\infty}\frac{\exp(\log^0(\lambda x))}{\exp(\log^0(x))} = \lim_{x\to\infty}\frac{\exp(1)}{\exp(1)} = 1 = \lambda^0,$$ thus varying slowly (regularly with index 0).
For Case 2 ($\alpha = 1$):
$$\lim_{x\to\infty} \frac{f(\lambda x)}{f(x)} = \lim_{x\to\infty} \frac{\exp(\log^1(\lambda x))}{\exp(\log^1(x))} = \lim_{x\to\infty}\frac{\lambda x}{x} = \lambda^1,$$ thus varying regularly with index $1$.
Thank you very much!