Are the more or less closed formula for Ito integral
$$ \int_0^t B_{s/2} d B_s $$
where $B_s$ is standard 1-dimensional Brownian motion?
For example, $\int_0^t B_s d B_s = (B_t^2 - t) / 2$, are there similar result for integral above?
2026-03-30 09:50:33.1774864233
Ito integral of $\int_0^t B_{s/2} d B_s$
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