Integral with fractional part

Let $n \in \mathbb{N}$ and let $\{ \cdot \}$ denote the fractional part. Prove that:

$$\int_0^1 x^n \left\{ \frac{1}{x} \right\}^n \, {\rm d}x = 1- \frac{1}{n+1} \sum_{k=2}^{n+1} \zeta(k)$$

where $\zeta$ is the Riemann zeta function.

Solution

 \begin{align*}
\int_{0}^{1}x^n \left \{ \frac{1}{x} \right \}^n \, dx &=\sum_{k=1}^{\infty}\left [ \int_{1/(k+1)}^{1/k} x^n \left \{ \frac{1}{x} \right \}^n \right ]\, dx \\
 &= \sum_{k=1}^{\infty}\left [ \int_{1/(k+1)}^{1/k}x^n \left ( \frac{1}{x}-k \right )^n \right ]\,dx\\
 &=\sum_{k=1}^{\infty}\left [ \int_{1/(k+1)}^{1/k}\left ( 1-kx \right )^n \right ]\, dx \\
 &= \sum_{k=1}^{\infty}\frac{1}{k}\left ( \int_{1/(k+1)}^{1/k} \left ( 1-y \right )^n \, dy  \right )\\
 &= \frac{1}{n+1}\sum_{k=1}^{\infty}\frac{1}{k(k+1)^{n+1}}
\end{align*}

Let

$$F(n)= \sum_{k=2}^{\infty}\frac{1}{(k-1)k^{n+1}}$$

Hence:

\begin{align*}
F(n) &=\sum_{k=2}^{\infty} \frac{1}{k^n} \left ( \frac{1}{k-1} - \frac{1}{k} \right )\\
 &=\sum_{k=2}^{\infty}\frac{1}{(k-1)k^n} - \sum_{k=2}^{\infty}\frac{1}{k^{n+1}} \\
 &=F(n-1) -\zeta(n+1) +1 \\
 &=\cdots \\
 &=F(0) - \left ( \zeta(n+1) + \zeta(n)+\cdots + \zeta(2) \right )+n 
\end{align*}

Hence

$$F(n)=n+1 - \sum_{k=2}^{n+1}\zeta(k)$$

The result follows.