I was solving a linear SDE and there I got the solution in this form $X_t = X_0 e^{-(\theta + \frac{1}{2}\sigma^2 ) t + \sigma W_t} + \theta\,\mu\,e^{-(\theta + \frac{1}{2}\sigma^2 ) t + \sigma W_t}\int_0^t e^{(\theta + \frac{1}{2}\sigma^2 ) s - \sigma W_s} ds$. I was trying to find its mean. Since $\mathbb{E}(e^{\sigma W_t})=e^{\frac{\sigma^2~ t}{2}}$, So I got $ \mathbb{E}(X_t) = X_0 e^{-(\theta + \frac{1}{2}\sigma^2 ) t} e^{\frac{1}{2}\sigma^2 t} + \theta\,\mu\,e^{-(\theta + \frac{1}{2}\sigma^2 ) t} e^{\frac{1}{2}\sigma^2 t}\int_0^t \mathbb{E}\left\{e^{(\theta + \frac{1}{2}\sigma^2 ) s - \sigma W_s} ds\right\}=X_0 e^{-\theta t} + \theta\mu\,e^{-\theta t} \int_0^t \mathbb{E}\left\{e^{(\theta + \frac{1}{2}\sigma^2 ) s - \sigma W_s} ds\right\}=X_0 e^{-\theta t} + \theta\mu\,e^{-\theta t} \int_0^t e^{(\theta + \frac{1}{2}\sigma^2 ) s} \mathbb{E}\left(e^{- \sigma W_s}\right) ds$.
I stuck at $\mathbb{E}\left(e^{- \sigma W_s}\right)$. Kindly tell me what will be its expectation. Also, I am not sure whether my steps are right or wrong? Please help me.