I have an equation: $\mathbb{E}[\exp({\epsilon}) (x \epsilon + x)]$, where $\epsilon \sim p(0 | \sigma^2)$ is a normal distribution. I know what $\mathbb{E}[\epsilon] = 0$ and $\mathbb{E}[\epsilon^2] = \sigma^2$.
I would like to get rid of the expectation here. My initial thought was $\mathbb{E}[\exp({\epsilon}) (x \epsilon + x)] = x$, but I feel like this is not true. I would be grateful if you provide some insight.
Use the normal distribution function
$$f(t)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{t^2}{2\sigma^2}} $$
to evaluate,
$$E[e^t]=\int_{-\infty}^{\infty}e^tf(t)dt=\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty}e^{t-\frac{t^2}{2\sigma^2}}dt=e^{\frac12\sigma^2}$$
$$E[te^t]=\int_{-\infty}^{\infty}te^tf(t)dt=\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty}te^{t-\frac{t^2}{2\sigma^2}}dt=\sigma^2e^{\frac12\sigma^2}$$
Therefore,
$$E[e^\epsilon (x \epsilon + x)]=x(E[\epsilon e^\epsilon]+E[e^\epsilon])=xe^{\frac12\sigma^2}(\sigma^2+1)$$