If $\left[x\right]$ denotes the greatest integer less than or equal to x then $\underset{0}{\overset{\infty }{\int }}\left[\frac{{\displaystyle 2}}{{\displaystyle e^x}}\right]dx$ is equal to 

$$\begin{gathered} {\int }_0^{\infty }\left[\frac{2}{{\mathrm{e}}^{\mathrm{x}}}\right] \\ 0<\mathrm{x}<\infty \\ \Rightarrow {\mathrm{e}}^0<{\mathrm{e}}^{\mathrm{x}}<{\mathrm{e}}^{\infty } \\ \Rightarrow 1<{\mathrm{e}}^{\mathrm{x}}<\infty \\ \Rightarrow 1>\frac{1}{{\mathrm{e}}^{\mathrm{x}}}>0 \\ \Rightarrow 2>\frac{2}{{\mathrm{e}}^{\mathrm{x}}}>0 \\ \Rightarrow 0<\frac{2}{{\mathrm{e}}^{\mathrm{x}}}<2 \end{gathered}$$

$$\begin{gathered} if \frac{2}{{\mathrm{e}}^{\mathrm{x}}}=1 \Rightarrow 2=e^{\mathrm{x}} \\ \Rightarrow \log 2_{\mathrm{e}}=\mathrm{x} \\ \mathrm{now} \mathrm{I}={\int }_0^{\log 2} \left[\frac{2}{{\mathrm{e}}^{\mathrm{x}}}\right]\mathrm{dx}+{\int }_{\log 2}^{\infty } \left[\frac{2}{{\mathrm{e}}^{\mathrm{x}}}\mathrm{dx}\right] \\ ={\int }_0^{\log 2} 1 \mathrm{dx}+{\int }_{\log 2}^{ \infty } 0 \mathrm{dx} \\ =\log 2+0 \\ ={\log }_e 2_{ } \end{gathered}$$