Integral on interval $[-\infty,W_t]$, $W_t$ is Brownian motion Basicaly I have an expectation of an integral on the interval which contains Brownian motion and it look like this.
$$
E\left[e^{W_t}\cdot\int_{-\infty}^{W_t} e^{-z^2}dz\right]
$$
$W_t$ is Brownian motion, $t$ is fixed. My question is: Can I write this is any other form or, can calculate this?
Any help would be appreciated. Thanks!
 A: Hint::
For the function
$$
f(x):=\int\limits^{x}_{-\infty}e^{-r^2}\mathrm d r\,,
$$
using Leibniz's rule and Ito's lemma tells us
$$
\mathrm df(W_t) = e^{-W_t^2}\mathrm dW_t-W_t~e^{-W_t^2}\mathrm dt\,.
$$
Therefore,

$$
\int\limits^{W_t}_{-\infty}e^{-r^2}\mathrm d r = \dfrac{\sqrt{\pi}}{2}{}{}+{}\int\limits_{0}^{t}e^{-W_u^2}\mathrm dW_u-\int\limits_{0}^{t}W_u~e^{-W_u^2}\mathrm du\,.
$$

This puts the argument of the expectation in a form that is "more appealing". Can you proceed from here?
Edit:
This approach leads to a successive evaluation of expectations by successive use of Ito's lemma. The result is a series representation of the desired expectation. So, for instance,
expansions like 
$$
\mathbb{E}\left[e^{W_t}\int\limits_{0}^{t}e^{-W_u^2}\mathrm dW_u\right] = \mathbb{E}\left[\int\limits_{0}^{t}e^{-W_u^2}\mathrm dW_u\right] + \mathbb{E}\left[\int\limits_{0}^{t}e^{W_u-W_u^2}\mathrm du\right] + \frac{1}{2}\mathbb{E}\left[\int\limits_{0}^{t}e^{W_u}\mathrm du\int\limits_{0}^{t}e^{-W^2_u}\mathrm dW_u\right]
$$ 
follow from observing
$$
e^{W_t}=1 + \int\limits_{0}^{t}e^{W_u}\mathrm dW_u + \frac{1}{2}\int\limits_{0}^{t}e^{W_u}\mathrm du\,.
$$
This is akin to situations in ordinary calculus where integrals are solved by obtaining a Taylor expansion for the solution. However, this is more involved than it needs to be. Certainly, a more direct approach would be to use the properties of the standard normal CDF, $\Phi$, and $W_t\sim\mathcal{N}\left(0,t\right)$ to justify the following:
$$
\begin{eqnarray*}
\mathbb{E}\left[e^{W_t}\int\limits^{W_t}_{-\infty}e^{-u^2}\mathrm du\right] = \mathbb{E}\left[\sqrt{\pi}e^{W_t}\Phi(\sqrt{2}W_t)\right] = \sqrt{\pi}\frac{1}{\sqrt{2\pi t}}\int\limits_{-\infty}^{\infty}\Phi(\sqrt{2}x)e^{-\frac{1}{2}\frac{x^2}{t}+x}\mathrm dx = \sqrt{\pi}\frac{e^{\frac{1}{2}t}}{\sqrt{2\pi(2t)}}\int\limits^{\infty}_{-\infty}\Phi(y)e^{-\frac{1}{4t}\left(y-\sqrt{2}t\right)^2}\mathrm dy = \sqrt{\pi}e^{\frac{1}{2}t}\Phi\left(\frac{\sqrt{2}t}{\sqrt{1+2t}}\right),
\end{eqnarray*}
$$
resulting in a final form for the solution originally suggested by @Did.
