Question

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?

• A. ${\log }_{e}2$
• B. $e^2$
• C. $0$
• D. $\frac{2}{e}$

Answer 1

Answer: (A)

${\log }_{e}2$

$I={\int }_0^{\infty }\left[\frac{2}{e^x}\right]dx$
Substitute $e^x=t\Rightarrow x=\log t\Rightarrow dx=\frac{1}{t}dt$
Then
$I={\int }_1^{\infty }\left[\frac{2}{t}\right]\frac{dt}{t}={\int }_1^2\left[\frac{2}{t}\right]\frac{dt}{t}+{\int }_2^{\infty }\left[\frac{2}{t}\right]\frac{dt}{t}={\int }_1^2\frac{dt}{t}+{\int }_2^{\infty }0\frac{dt}{t}={\left[\log t\right]}_1^2=\log 2$