Denote by $\tau(x) := \inf \{t \ge 0, W_t=x\},$ where $W_t$ is a Wiener process started at $W_0 = w_0 > 0$ and
I would like to show that for any $a>1$ it almost surely holds that
$$\int_{0}^{\tau(0)} \frac{1}{W^a_t}dt = \infty.$$
This more or less means to show that $W_t,$ when it approaches zero, it does that (almost surely) at least linearly fast (since for all $a \ge 1,$ $\int_{0+}\frac{1}{y^a} dy=\infty$). I have little idea how to show such a thing for something that is nowhere differentiable.
Maybe one thing one might do is to take a look at time reversal of $W_t$ starting from $\tau(0)$ and try to apply the law of iterated logarithm, which gives bounds for almost all paths of the process near $0$. However, here we'd have to condition on it hitting $w_0$ and also staying positive until it does so. However, this would probably not help anyway, because the $\sqrt{2t\log\log(1/t)}$ does not go to $0$ fast enough (for $t \rightarrow 0+$) for its inverse to diverge.
Are there any Brownian motion path properties that tell us something useful about its behaviour near $\tau(0)$?
EDIT:
The following statement comes from Brownian Motion and Stochastic Calculus by Karatzas and Shreve (Proposition 3.6.27., Engelbert's 0-1 law):
Let $f:\mathbb R \mapsto [0,\infty]$ be Borel measurable. Then the following is equivalent.
1) $f$ is a locally integrable function.
2) $\mathbb{P}^0[\int_0^t f(W_t) < \infty, 0 \le t < \infty]>0$
3) $\mathbb{P}^0[\int_0^t f(W_t) < \infty, 0 \le t < \infty]=1$
This is not exactly what I want. First of all, I am not starting from $0$, but from a positive point. That is not really a big deal though. What is more troublesome is that I do not integrate to a constant time $t$, or at least an independent stopping time, but to something that heavily influences the path. So even though I have a nice uniform result for the constant times, it is not as nice with the stopping time I have. I am now trying to fit the result to what I have, but I'm a bit stuck. My $f$ can be seen as $f(x)=\frac{1}{|x|^a},$ where $a$ is positive and therefore it is not locally integrable. Still, there is the problem with the stopping time.