Suppose that $B$ is $d$-dimensional Brownian motion, where each of the $d$ coordinates $B^i$ is an independent Brownian motion over $\mathbb R$. Then, can the integral below be understood as an Itô integral?
$$\int_s^t B^i dB^j, 0\le s < t$$
I know that it can in the case that $i=j$, and it yields $\tfrac{1}{2} (B^i)^2 - \tfrac{t-s}{2}$. What about the other case?
Partial answer:
By the integration-by-parts formula (which is a special case of the two-dimensional Ito-formula) $$ B^i_tB^j_t-B^i_sB^j_s=\int_s^t B^i_u\,dB^j_u+\int_s^t B^j_u\,dB^i_u+\langle B^i,B^j \rangle_t -\langle B^i,B^j \rangle_s\,. $$ In your case the covariation terms vanish.