I have a random variable:
\begin{equation} \mathcal{A} = \frac{cos(2\pi\theta)}{\alpha} \tag{1} \end{equation}
where $\theta \sim U(0,1)$, $\alpha \sim (-1)^{X} \cdot U(0.1,10)$ and $P(X=0)=0.5 \land P(X=1)=0.5$
From this it follows that $\mathbb{E}[cos(2\pi\theta)]=\mathbb{E}[\alpha]=0$ but I am not sure how to calculate:
\begin{equation} \mathbb{E}[\mathcal{A}] \tag{2} \end{equation}
If I do the calculation I obtain an indeterminate form i.e. $\mathbb{E}[\mathcal{A}]= \frac{0}{0}$ but my intuition tells me that the actual value might be 1. Might there be a rigorous method for settling this matter?
Note: I now realise that my misconception was that $\mathbb{E}[\frac{1}{\alpha}]=\frac{1}{\mathbb{E}[\alpha]}=\frac{1}{\infty}$.
You need to calculate $E [1/\alpha]$, not $E[\alpha]$. Assuming, of course, that $\theta$ and $\alpha$ are independent. Under this assumption, the random variables $X=\cos(2\pi\theta)$ and $Y=1/\alpha$ are also independent.
In general the expectation of the product, $E[XY]$, of independent random variables is given by the product of their expectstions: $$E[XY] =E[X]E[Y].$$ In your case, you have $$E\mathcal A =E[\cos(2\pi\theta)(1/\alpha)]= E[\cos (2\pi \theta)] E[1/\alpha].$$