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 $$\alpha > 1$$?

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!

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 $$f(cx)/f(x)$$.