Expected value of the stochastic integral $\int_0^t e^{as} dW_s$ I am trying to calculate a stochastic integral 
$\mathbb{E}[\int_0^t e^{as} dW_s]$. I tried breaking it up into a Riemann sum 
$\mathbb{E}[\sum e^{as_{t_i}}(W_{t_i}-W_{t_{i-1}})]$, but I get expected value of $0$, since $\mathbb{E}(W_{t_i}-W_{t_{i-1}}) =0$. But I think it's wrong.  Thanks!
And I want to calculate $\mathbb{E}[W_t \int_0^t e^{as} dW_s]$ as well, I write $W_t=\int_0^t dW_s$ and get  $\mathbb{E}[W_t \int_0^t e^{as} dW_s]=\mathbb{E}[\int_0^t e^{as} dW_s]$.
Is that ok?
($W_t$ is brownian motion.)

The expectation of the Ito integral $\mathbb{E}( \int_0^t \mathrm{e}^{a s} \mathrm{d} W_s )$ is zero as George already said.
To compute $\mathbb{E}( W_t \int_0^t \mathrm{e}^{a s} \mathrm{d} W_s )$, write $W_t = \int_0^t \mathrm{d} W_s$. Then use Ito isometry:
$$
 \mathbb{E}( W_t \int_0^t \mathrm{e}^{a s} \mathrm{d} W_s ) = \mathbb{E}\left( \int_0^t \mathrm{d} W_s \cdot \int_0^t \mathrm{e}^{a s} \mathrm{d} W_s \right) = \int_0^t (1 \cdot \mathrm{e}^{a s}) \mathrm{d} s = \frac{\mathrm{e}^{a t} - 1}{a} \phantom{hhhh}
$$
