real analysis - For which values of $alpha geq 0$ is the function $exp(log^alpha(x))$ varying regularly?

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 $\alpha > 1$?

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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!

Answer

Observe that for a fixed $c$,

$$
\frac{f(cx)}{f(x)}=\exp\left(\left(\log x\right)^\alpha\left( \left(1+\frac{\log c}{\log x}\right)^\alpha-1\right) \right).
$$
Since $\lim_{t\to 0}\left(\left(1+t\right)^\alpha-1\right)/t=\alpha$, we can write
$$
\left(1+\frac{\log c}{\log x}\right)^\alpha-1 =\frac{\log c}{\log x}\left(\alpha +\varepsilon(x)\right),
$$
where $\varepsilon(x)\to 0$ as $x$ goes to infinity. Therefore,
$$
\frac{f(cx)}{f(x)}=\exp\left(\left(\log x\right)^{\alpha-1}\log c\left(\alpha +\varepsilon(x)\right) \right).

$$
If $0\leqslant \alpha<1$, the term in the exponential goes to $0$ as $x$ goes to infinity hence $f$ is slowly varying; for $\alpha=1$ we get $c$ and for $\alpha>1$ the term in the exponential goes to infinity hence so does $f(cx)/f(x)$.