Challenge Question: Integral involving floor and reciprocal of natural numbers

Question

$$\int_m^{n}\frac{1}{\frac{b}{x -\frac{1}{2b}}-\lfloor{b/x}\rfloor} dx = I$$

Where $b$, $m$, $n$, and $x$ are natural numbers $s.t.$ $m <n<b$

Answer 1

Since $m<n<b$ and all three are natural numbers it follows that $b\ge3$.

An investigation of the graph of

$$ y=\frac{1}{\frac{b}{x -\frac{1}{2b}}-\lfloor{b/x}\rfloor} $$

for $b\ge3$ indicates that

$$ \frac{2\left(x-\dfrac{1}{4b}\right)^2}{1+2\left(x-\dfrac{1}{4b}\right)^2} \le \frac{1}{\dfrac{b}{x -\dfrac{1}{2b}}-\lfloor{b/x}\rfloor}\le2\left(x-\dfrac{1}{4b}\right)^2\tag{1}$$

A substitution of $x=\dfrac{u+1}{4b}$ turns $(1)$ into the inequality

$$ \frac{u^2}{u^2+8b^2}\le\frac{1}{\dfrac{4b^2}{u-1}-\left\lfloor\dfrac{4b^2}{u+1}\right\rfloor}\le\frac{u^2}{8b^2}\tag{2} $$

Perhaps this could be of use in approaching the problem.

https://www.desmos.com/calculator/h5qdnimgm2

Here are graphs of the three function in $(1)$ for $b=3$ and $b=40$.