Question

The value of integral $\underset{0}{\overset{\infty }{\int }}[n\cdot e^{-x}]dx$ is equal to $($where $[\cdot ]$ denotes the greatest integer function and $n\in N,n>1)$

• A. $\ln \left(\frac{n^{n-1}}{n!}\right)$
• B. $\ln \left(\frac{n^n}{n!}\right)$
• C. $\ln \left(\frac{n^{n+1}}{n!}\right)$
• D. $0$

Answer 1

The correct option is B $\ln \left(\frac{n^n}{n!}\right)$

$I=\underset{0}{\overset{\infty }{\int }}[n\cdot e^{-x}]dx$
Let $e^{-x}=t$
$\Rightarrow I=\underset{0}{\overset{1}{\int }}\frac{\left[nt\right]}{t}dtI=\underset{0}{\overset{\frac{1}{n}}{\int }}\frac{\left[nt\right]}{t}dt+\underset{\frac{1}{n}}{\overset{\frac{2}{n}}{\int }}\frac{\left[nt\right]}{t}dt+\cdots +\underset{\frac{n-1}{n}}{\overset{1}{\int }}\frac{\left[nt\right]}{t}dt=0+\underset{\frac{1}{n}}{\overset{\frac{2}{n}}{\int }}\frac{1}{t}dt+\underset{\frac{2}{n}}{\overset{\frac{3}{n}}{\int }}\frac{2}{t}dt+\cdots +\underset{\frac{n-1}{n}}{\overset{1}{\int }}\frac{(n-1)}{t}dt={\left[\ln |t|\right]}_{1/n}^{2/n}+2{\left[\ln |t|\right]}_{2/n}^{3/n}+\cdots +(n-1){\left[\ln |t|\right]}_{\frac{n-1}{n}}^1=\ln \frac{2}{n}-\ln \frac{1}{n}+2(\ln \frac{3}{n}-\ln \frac{2}{n})+\cdots +(n-1)\ln \frac{n}{n-1}=\ln \frac{2}{1}+2\ln \frac{3}{2}+\cdots +(n-1)\ln \frac{n}{n-1}=\ln (2\times \frac{3^2}{2^2}\times \frac{4^3}{3^3}\times \cdots \times \frac{n^{n-1}}{(n-1{)}^{n-1}})=\ln \left(\frac{n^{n-1}}{1\times 2\times 3\times \cdots \times (n-1)}\right)=\ln \left(\frac{n^n}{n!}\right)$