Question

Let $n$ be a positive integer. For a real number $x$, let $\left[x\right]$ denote the largest integer not exceeding $x$ and $\left\{x\right\}=x-\left[x\right]$. Then ${\int }_1^{n+1}\frac{{\left(\left\{x\right\}\right)}^{\left[x\right]}}{\left[x\right]}dx$ is equal to

• A. ${\log }_e\left(n\right)$
• B. $\frac{1}{n+1}$
• C. $\frac{n}{n+1}$
• D. $1+\frac{1}{2}+\cdots +\frac{1}{n}$

Answer 1

The correct option is C $\frac{n}{n+1}$

Let us first evaluate the integral $I\left(k\right)=\underset{k}{\overset{k+1}{\int }}\frac{\left(\right\{x\}{)}^{\left[x\right]}}{\left[x\right]}dx$, where $k$ is an integer.

By the properties of $\left[x\right]$, we have $\left[x\right]=k$ for $x\in [k,k+1)$

Therefore, $I\left(k\right)=\underset{k}{\overset{k+1}{\int }}\frac{\{x{\}}^{\left[x\right]}}{\left[x\right]}dx=\underset{k}{\overset{k+1}{\int }}\frac{(x-k{)}^k}{k}dx$

Substituting $t=x-k$, we have $I=\underset{0}{\overset{1}{\int }}\frac{t^k}{k}dt=\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$

We know that $\underset{1}{\overset{n+1}{\int }}\frac{\{x{\}}^{\left[x\right]}}{\left[x\right]}dx=\sum _{k=1}^n\underset{k}{\overset{k+1}{\int }}\frac{\{x{\}}^{\left[x\right]}}{\left[x\right]}dx=\sum _{k=1}^{n}I\left(k\right)=\sum _{k=1}^n\frac{1}{k}-\frac{1}{k+1}=1-\frac{1}{n+1}=\frac{n}{n+1}$