Integral of ceiling I have tried to bound this definite integral:
$f(r,i)=\int_{0}^{1}\left (\lceil r^{i+x}\right \rceil - r^{i+x}) dx$
$r,i \in \mathbb{R}$, $r>1$, $i≥0$
I want to prove upper bound for all i when $r∈[3,5]$. In practice, this upper bound is 0.65 but we want to prove it in a theoretic way. Could you help me with that?
p.s. Firstly , I have got the below bound, but it didn't help:
$\frac{r^{i}+r^{\left(i+1\right)}}{2}-\frac{r^{\left(i+1\right)}-r^{i}}{\ln r}$
 A: Splitting the integral into two yields $$\int_0^1 \left\lceil r^{i+x} \right\rceil dx - \int_0^1 r^{i+x} dx$$
The second integral can be evaluated in closed form as $\frac{r^{i+1}-r^i}{\log(r)}$, and for the first, you can sum over $n = \left\lceil r^{i+x} \right\rceil$. Specifically, the integral would be $$\sum_{n=1}^{\infty} \int_{R_n} n dx $$
where $R_n$ is the region such that $0<x<1$ and $\left\lceil r^{i+x} \right\rceil=n \to n-1<r^{i+x} \le n$. Solving the inequality for $x$ yields $\log_r(n-1)-i < x \le \log_r(n)-i$. For $\log_r(n)-i < 0$ or $\log_r(n-1)-i>1$, the inequality $0 < x < 1$ wouldn't be satisfied at all, so we can just sum for $r^i < n < r^{i+1}+1$. The integral would then be equal to $$\int_{0}^{\log_r(\lceil r^i \rceil)-i}\lceil r^i \rceil dx + \sum_{n=\lceil r^i \rceil+1}^{\lfloor r^{i+1} \rfloor} \int_{\log_r(n-1)-i}^{\log_r(n)-i} n dx + \int_{\log_r(\lfloor r^{i+1}\rfloor)-i}^{1}\lfloor r^{i+1}+1 \rfloor dx$$
Evaluating the integrals in closed form yields
$$\lceil r^i \rceil\left(\log_r(\lceil r^i \rceil)-i\right) + \sum_{n=\lceil r^i \rceil+1}^{\lfloor r^{i+1} \rfloor} n(\log_r(n)-\log_r(n-1)) + \lfloor r^{i+1}+1 \rfloor\left( 1-\log_r(\lfloor r^{i+1}\rfloor)+i  \right)$$
