I have been having trouble solving/simplifying this stochastic integral: $$\mathop{\mathbb{E}}(\int_0^T 1_{\{S_t>K\}}dS_t)$$ where $dS=rSdt+\sigma SdW$. What is the best way to approach this?
$\textbf{Attempted Solution:}$
$$\mathop{\mathbb{E}}(\int_0^T 1_{\{S_t>K\}}dS_t)$$ $$=\mathop{\mathbb{E}}(\int_0^T 1_{\{S_t>K\}}(rS_t dt+\sigma S_t dW_t))$$ $$=\mathop{\mathbb{E}}(\int_0^T 1_{\{S_t>K\}}rS_t dt)+\mathop{\mathbb{E}}(\int_0^T 1_{\{S_t>K\}}\sigma S_t dW_t)$$ Then exchanging the expectation and the first integral assuming basic regularity conditions $$=\int_0^T r \mathop{\mathbb{E}}(1_{\{S_t>K\}}S_t) dt+\mathop{\mathbb{E}}(\int_0^T 1_{\{S_t>K\}}\sigma S_t dW_t)$$ Then using the known moments of a truncated log-normal random variable, the expectation is equal to: $$=\int_0^T r S_0 e^{(r+\frac{\sigma^2}{2})t} \Phi(\frac{(r+\sigma^2)t-ln(K/S_0)}{\sqrt{\sigma^2 t}}) dt+\mathop{\mathbb{E}}(\int_0^T 1_{\{S_t>K\}}\sigma S_t dW_t)$$ Finally the second integral is equal to zero as $\int_0^T 1_{\{S_t>K\}}\sigma S_t dW_t$ is a martingale on $[0,T]$, and therefore the expecation of it is zero. Therefore: $$\mathop{\mathbb{E}}(\int_0^T 1_{\{S_t>K\}}dS_t)=\int_0^T r S_0 e^{(r+\frac{\sigma^2}{2})t} \Phi(\frac{(r+\sigma^2)t-ln(K/S_0)}{\sqrt{\sigma^2 t}}) dt$$
Is all of that correct?