  It is well known that $\int_{0}^{1} t^{-\theta} d t<\infty$ for $\theta \in
(0,1)$ and $\int_{0}^{1} t^{-\theta} d t=\infty$ for $\theta \in [1,\infty)$.
Since $t$ can be taken as an $\alpha$-stable subordinator with $\alpha=1$, it
is natural to ask whether $\int_{0}^{1} t^{-\theta} d S_{t}$ has a similar
property when $S_{t}$ is an $\alpha$-stable subordinator with $\alpha \in
(0,1)$. We show that $\theta=\frac 1\alpha$ is the border line such that
$\int_{0}^{1} t^{-\theta} d S_{t}$ is finite a.s. for $\theta \in (0, \frac
1\alpha)$ and blows up a.s. for $\theta \in [\frac1\alpha,\infty)$. When
$\alpha=1$, our result recovers that of $\int_{0}^{1} t^{-\theta} d t$.
Moreover, we give a $p$-th moment estimate for the integral when $\theta \in
(0,\frac 1\alpha)$.
