Find the mean of $Y=\frac{1}{A}(e^{A}-1)$ where $A\sim N(0,\sigma^2)$

Question

A sub-question of an exercise from my probability class is as follows:

Let $A\sim N(0,\sigma^2)$ and $X(t)=e^{At}$. Find the mean of $Y=\int_0^1X(t)dt$.

I figured that we have $Y=\frac{1}{A}(e^{A}-1)\cdot\unicode{x1D7D9}_{\{A\neq 0\}}$ so that the mean is equal to

\begin{align} \mathbb{E}[Y] & = \frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^\infty\frac{e^x-1}{x}\exp{(-\frac{x^2}{2\sigma^2})}dx. \end{align}

I do, however, have not the slightest idea as to how I should proceed from here. I am doubting whether I am even approaching the question in the right way.

Answer 1

Use the MGF viz.$$\Bbb E\left[\int_0^1X(t)dt\right]=\int_0^1\Bbb E[X(t)]dt=\int_0^1e^{\sigma^2t^2/2}dt.$$Write that in terms of the error function.