I am looking simply for the formula of what's in the title, namely $$ \Bbb E\int_0^t1_{[0,u]}(|B_s|)\mathrm ds $$ where $B$ is the standard Brownian motion. I have found this article but it does not provide anything readily useful for the case of the first moment. I also have found this unanswered question so whoever can help me, perhaps could also answer that question. I'd be happy to know an answer to it too.
So I've done some thinking, and changing the order of integration we would get $$ A(u, t) = \Bbb E\int_0^t1_{[0,u]}(|B_s|)\mathrm ds = \int_0^t \Bbb P(|B_s|\leq u)\mathrm ds $$ which further simplifies to $A(u, t) = t J(u/\sqrt t)$ where $$ J(x) = 2\int_0^1\Phi\left(\frac{x}{\sqrt{s}}\right)\mathrm ds - 1 $$ and $\Phi$ is the CDF of the standard normal distribution. Can this be simplified further using perhaps some special function? WolframAlpha says nothing about this integral.