Integral squared sinus of the Brownian motion is infinite I'm looking to prove that, almost surely:
$$\int_0^\infty \sin^2(B_s) \, ds = \infty $$
where $B_s$ is a Brownian motion, without using the Ito Formula. 
I had two initial ideas:


*

*We know that the Brownian motion visits zero infinitely many times, but countable, so the Lebesgue measure of these visits is zero. Therefore we can hope that it stays nonzero and positive (cause $\sin^2$ is positive) long enough infinitely many times.

*Following up on the first idea, if I can prove that
$$ P \left( B_s \in \bigcup_{n \in \mathbb{Z}} \left[\frac{\pi}{4} + n\pi,\frac{3\pi}{4} + n\pi \right] \right)$$
is big enough (maybe something like using Borel-Cantelli's theorem?), I could lower bound the integral by infinity.  
I'm not quite sure of these ideas and their formality, every help is much appreciated. Thank you!
 A: Your idea is fine, and it is possible to make it rigorous using the strong Markov property of Brownian motion. However, it's a bit involved (at least as far as I see) because we need to define iteratively suitable stopping times. Here is a slightly different approach which seems simpler to me:
Define
$$\tau_k := \inf\{t>0; B_{t+\tau_0+\ldots+\tau_{k-1}}=0\}, \qquad \tau_0=0,$$
and 
$$B_t^{(k)} := B_{t+\tau_0+\ldots+\tau_{k-1}}, \qquad t \geq 0, \, k \geq 1.$$
By the strong Markov property of Brownian motion, the process $(B_t^{(k)})_{t \geq 0}$ is a Brownian motion for every $k \geq 1$. If we set $\sigma_k := \sum_{i=0}^k \tau_i$, then
\begin{align*} \int_0^{\infty} \sin^2(B_s) \, ds &= \sum_{k=0}^{\infty} \int_{\sigma_k}^{\sigma_{k+1}} \sin^2(B_s) \, ds \\ &= \sum_{k=0}^{\infty} \underbrace{\int_0^{\tau_{k+1}} \sin^2(B_s^{(k)}) \, ds}_{=:X_k}. \end{align*}
Since each $B^{(k)}$ is a Brownian motion, the random variables $X_k$ are identically distributed. Moreover, the strong Markov property yields that $X_k$, $k \geq 1$, are independent. The stopping time $\tau_1$ is strictly positive with probability $1$, and this implies that there exists a constant $c>0$ such that
$$\mathbb{P}(X_1 \geq c)> \frac{1}{2},$$
and hence
$$\mathbb{P}(X_k \geq c)> \frac{1}{2}, \qquad k \geq 1.$$
Consequently,
$$\sum_{k \geq 1} \mathbb{P}(X_k \geq c) =\infty,$$
which implies by the Borel Cantelli lemma that $X_k \geq c$ infinitely often with probability $1$. Thus, $\sum_{k=1}^{\infty} X_k=\infty$ almost surely.
Remark: The above reasoning shows, more generally, that 
$$\int_{(0,\infty)} f(B_s) \, ds = \infty \quad \text{a.s.}$$
for every non-negative, measurable function $f$ which is strictly positive on some interval $(0,\epsilon)$. Integrals of the form $\int_{(0,\infty)} f(B_s) \, ds$ are called perpetual integrals.
