evaluate $\int_0^1(\int_x^1 \frac{e^{-y}}{y} dy) dx$
3. if we reverse the order of $dx$ and $dy$ , the integration becomes $\int_0^1(\int_0^y \frac{e^{-y}}{y} dx) dy$
so the final answer comes to be $1 - \frac{1}{e}$. thanks