Let $W$ a standart brownian motion.
For $t\in (0,1]$ I would like to prove that $$\Bbb{E}\bigl(\int_{0}^t \vert W(s)\frac{ds}{s}\vert\bigr)<\infty$$
I am tempted to do $\Bbb{E}\bigl(\int_{0}^t \vert W(s)\frac{ds}{s}\vert\bigr)=\int_{0}^t\Bbb{E}(\frac{\vert W(s)\vert}{s})ds$ but not sure how can I justify this and how can I continue.
By Tonelli's theorem, $\mathbb{E} \int = \int \mathbb{E}$ whenever the integrand is nonnegative, which is the case here, so your "temptation" is justified.
You can now pull $\frac{1}{s}$ out of the expected value, since it is deterministic. Now you just have to handle $\mathbb{E}|W(s)|$, i.e. the expectation of the absolute value of a Gaussian. Note that $W(s)$ is normal with mean zero and variance $s$; in other words, it is $s^{1/2}$ times a standard normal. Let $c = \mathbb{E}|Z|$ be the expected absolute value of a standard normal. Now the desired quantity is $c \int_0^t s^{-1/2}\,ds$ which is clearly finite.