# ${\bf E}[Y]$ of a joint distribution

So, I have that a joint probability density function is given by the formula: $$5e^{-5x} / x, \quad 0 < y < x < \infty$$ and I have to find the $\operatorname{Cov}(X,Y)$.

I know that $\operatorname{Cov}(X,Y) = {\bf E}[XY] - {\bf E}[X]{\bf E}[Y]$. I've been able to find ${\bf E}[XY]$ and ${\bf E}[X]$ ($1/25$ and $1/5$ respectively, hopefully I'm correct there), but I've been unable to find the ${\bf E}[Y]$... Can someone help?

$E[Y]=\int_0^\infty \int_0^x y\cdot \frac{5}{x}e^{-5x} dy dx$. This formula is obtained after inserting into general form $E[Y]=\int_{\mathbb{R}^2} y \cdot pdf(x,y) dydx$.