# Asking for an analytical proof

Calculate the following integral for $n\in\mathbb {N}$" $$\int_{1}^{n+1} \frac {\{x\}^{[x]}}{[x]} dx$$ :lol: Where $[x]$ is the largest integer $\leq x$ and $\{x\}=x-[x]$

Well my attempt was using induction... Let $I_n$ denotes the following integral; It is quite easy get $I_1$ and $I_2$ Obsevrving $I_1=\frac 12$ and $I_2=\frac 23$ Let me set an inductive hypothesis, $I_n=\frac {n}{n+1}\forall n\in\mathbb{N}$

Base case is no doubt true.. Let it be true for any arbnitar $m\in\mathbb{N}$ We have, $I_{m+1}=\int_{1}^{m+2}\frac {\{x\}^{[x]}}{[x]} dx$ $=I_m+\int_{m+1}^{m+2}\frac {\{x\}^{[x]}}{[x]} dx$ $=\frac {m}{m+1}+\frac {1}{m+1}\int_{0}^{1} \{t\}^{m+1}dt$ Where $t=x+m+1\implies dx=dt$ But obviously for $t\in(0,1)$ we have$d\{t\}=dt$ So $I_{m+1}$ can be written as $I_{m+1}=\frac {m}{m+1}+\frac {1}{m+1}\int_{0}^{1} t^{m+1}dt=\frac {m}{m+1}+\frac{1}{(m+1)(m+2)}=\frac {m+1}{m+2}$; hence the claim Butihave tried so far to obtain it analyticallybut didn't get it! I am very keento see it's anal;yticalsolution if there's one, and any other interesting method that make me aware!

• You have an uncommon notation. Usually, your $\{\cdot\}$ is denoted by $[\cdot]$ (Gaussian brackets) and your $[\cdot]$ is denoted by $\lfloor \cdot \rfloor$ (the floor function) – AlexR Oct 30 '13 at 15:18
• @AlexR No, the Gauß brackets are floor, and $\{\cdot\}$ is the common notation for the fractional part $x - \lfloor x\rfloor$. – Daniel Fischer Oct 30 '13 at 15:31
• @DanielFischer At least that's what I learned and what I see more often (maybe it prevents ambiguous notation in corelation with integrals over sets... $\int_{\{\alpha\}} \ldots d\#$). Not working too much with this though. - If that's the case, I'll adapt my answer. – AlexR Oct 30 '13 at 15:32

I will use the notation I described in my comment. \begin{align*} I_n & = \int_1^{n+1} \frac{\{x\}^{\lfloor x \rfloor}}{\lfloor x \rfloor} dx \\ & = \sum_{j=1}^n \int_j^{j+1} \frac{(x-j)^j}{j} dx \\ & = \sum_{j=1}^n \int_0^1 \frac{u^j}{j} du = \sum_{j=1}^n \frac{1}{j(j+1)} \\ & = \frac{n}{n+1} \end{align*}