I am stuck to show that for some $\eta>0$ we have:
$$\int_{0}^{\eta} f(\mathcal{D}(\epsilon,d)) \, d\epsilon<\infty $$
where for some constant $K>0$, $f$ is defined by $f(x) = \sqrt{\ln(1+x)} + K\ln(1+x)$ ($x\geq 0$) and $\mathcal{D}(\epsilon, d)$ is the packing number of $[a,b] \subset \mathcal{R}$ (i.e. the maximum number of $\epsilon$-separated points in $[a,b]$) with respect to the metric $d(s,t) = |s-t|^\alpha$, where $\alpha \in ]0,0.5]$.
My approach is to make use of the covering number:
Let $\mathcal{N}(\epsilon, d)$ be the covering number of $[a,b]$ (i.e. the minimal number of $\epsilon$-Balls to cover $[a,b]$). It is then well known that $$\mathcal{N}(\epsilon, d) \leq \mathcal{D}(\epsilon, d) \leq \mathcal{N}(\frac{1}{2}\epsilon, d).$$
Since the nonnegative function $f(x)$ is strictly increasing it would be sufficient to show that $$\int_{0}^{\eta} f(\mathcal{N}(\frac{1}{2}\epsilon,d))\, d\epsilon<\infty $$ for some $\eta>0$.
A $d$-Ball around $t$ is the interval $[t-\epsilon^{1/\alpha}, t+ \epsilon^{1/\alpha}]$. Hence [a,b] can be covered by $\mathcal{N}(\frac{1}{2}\epsilon,d) =2^{\frac{1}{\alpha}-1}(b-a) \epsilon^{-\frac{1}{\alpha}}$ balls of radius $\frac{1}{2}\epsilon$.
Setting $c=2^{\frac{1}{\alpha}-1}(b-a)$ one then has to show that there exists a $\eta>0$ such that
$$\int_{0}^\eta \sqrt{\ln{(1+c\epsilon^{-\frac{1}{\alpha}}})} + K \ln{(1+c\epsilon^{-\frac{1}{\alpha}}})\, d\epsilon<\infty.$$
Unfortunately, I don't know how to proceed from here.
My guess is that for each $\alpha \in ]0,0.5[$ there exists a constant $L$ such that $\ln(1+cx^{-\frac{1}{\alpha}})$ can be bounded above by $L x^{-\alpha}$ for $x>0$ small enough, from which the assertion would immediately follow.
Edit: Found one solution: set $\eta = \min(d^\alpha, 1)$ then for $x\leq \eta$ we have $1 \leq d x^{-\frac{1}{\alpha}}$. Since $\ln(x)$ is a increasing function, we then have for $x\leq \eta$ $$\ln(1+dx^{-\frac{1}{\alpha}}) \leq \ln(2dx^{-\frac{1}{\alpha}}) = \ln(2d) -\frac{1}{\alpha} \ln(x).$$
Hence $$\int_{0}^\eta \ln(1+dx^{-\frac{1}{\alpha}}) \leq \int_{0}^\eta \ln(2d) -\frac{1}{\alpha} \ln(x).$$ But $$\int_0^\eta \ln(2d) -\frac{1}{\alpha} \ln(x) \, dx = Const - \frac{1}{\alpha}(\eta(\ln(\eta)-1)) < \infty,$$ which validates one part of the assertion.
The assertion itselfs then follows for example by setting $\eta= \min(1, (\frac{d}{\exp(1)-1})^\alpha) \leq \min(1,d^\alpha)$.