How to find integral $\int e^{-2x} /x\,dx\,$?
You need not evaluate that integral. $EY=\int_0^{\infty} y\int_y^{\infty} 2e^{-2x}/x dx$. Interchange the integrals to get $EY=\int_0^{\infty}(\int_0^{x} ydy)2e^{-2x}/x dx$. This becomes $\int_0^{\infty} xe^{-2x}dx$ and you can evaluate this by a simple integration by parts.