Suppose $X$ and $Y$ are non-negative independent random variables. Then how can I evaluate $\mathbb E[\ X\ |\ XY > \alpha\ ]$ for some $\alpha > 0$?
If I know the precise distributions of both $X$ and $Y$, does this evaluate to a number, or a random variable?
The notation may give rise to confusion. I presume that you want to consider for given $\alpha>0$ the measurable set $H=H_\alpha=\{XY>\alpha\}$ and then want to evaluate the expectation of $X$ conditioned to this set, i.e. $$ E(X \ | \ H_\alpha) = \int_{XY>\alpha} X \; d\mu \ \ / \int_{XY>\alpha} \; d\mu$$ where $\mu$ is the probability measure. This is just a function of $\alpha$ and not a random variable as such.
To add to the confusion(?), the expression $Z=E(X\ |\ XY)$ is on the other hand a random variable. Here, the properties of $Z$ is that it is constant on level curves of $XY$ and for any $0\leq a<b\leq +\infty$ we have
$$ \int_{a<XY<b} Z \; d\mu = \int_{a<XY<b} XY \; d\mu. $$ Hope it makes sense?
In order to calculate the quantity in question write $F_Y(y)=P(Y>y)$ and note that independence implies that the distribution of $(X,Y)$ is a direct product measure $dP_X \; dP_Y$. We then have $$ \int_{XY>\alpha} X \; d\mu = \int_0^\infty x \left( \int_{\alpha/x}^\infty dP_Y(y) \right) dP_X(x) = \int_0^\infty x F_Y(\alpha/x) dP_X(x). $$ But I don't know if this is useful to you?