If $$\left[ x \right]$$ denotes the greatest integer less than or equal to $$x$$ then $$\int_0^\infty {\left[ {\frac{2}{{{e^x}}}} \right]dx} $$    is equal to :

Solution :
$$\eqalign{ & I = \int_0^\infty {\left[ {\frac{2}{{{e^x}}}} \right]dx} \cr & {\text{Substitute }}{e^x} = t \Rightarrow x = \log \,t \cr & dx = \frac{1}{t}dt \cr & {\text{Then}} \cr & I = \int_1^\infty {\left[ {\frac{2}{t}} \right]\frac{{dt}}{t}} \cr & = \int_1^2 {\left[ { - \frac{2}{t}} \right]\frac{{dt}}{t}} + \int_2^\infty {\left[ {\frac{2}{t}} \right]\frac{{dt}}{t}} \cr & = \int_1^2 {\frac{{dt}}{t}} + \int_2^\infty {0.} \frac{{dt}}{t} \cr & = \left[ {\log \,t} \right]_1^2 \cr & = \log \,2 - \log \,1 \cr & = \log \,2 - 0 \cr & = \log \,2 \cr & = {\log _e}2 \cr & \therefore \,I = {\log _e}2 \cr} $$