Let $X_t$ be the solution of the SDE $$ dX_t=f(t,X_t)dt+\sigma(X_t)dW_t.$$ I need to check if $$ \mathbb{E}\biggl[(\sigma(X_t)+\sigma(X_a))\int_a^t\sigma'(X_s)\sigma(X_s)dW_s\biggr]=0.$$ If this is not the case, it is enough for me to prove that it is equal to $\mathcal{O}(t-a)$ if possible.
Thank you in advance!
Using $$\Bbb E\left[\int_a^t u(X_s)dW_s\right]=0 \tag1$$ and $$\Bbb E\left[\int_a^t u(X_s)dW_s\int_a^t v(X_s)dW_s\right]=\int_a^bu(X_s)v(X_s)ds \tag2$$ together with $$ σ(X_t)-σ(X_a)=\int_a^t\left(σ'(X_s)f(s,X_s)+\tfrac12σ''(X_s)σ(X_s)^2\right)\,ds + \int_a^tσ'(X_s)σ(X_s)\,dW_s \tag3 $$ it follows that the non-zero terms are for the given expectation are, $$ \int_t^a\Bbb E\left[σ'(X_s)^2σ(X_s)^2\right]\,ds,\tag4 $$ so no, its value is not zero. And yes, provided the integrand is bounded you get that it is $O(t-a)$.