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

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

  • $\begingroup$ what purpose, if any, is the $\pi$ serving in the question? Also please share what you have done thus far. $\endgroup$
    – Anurag A
    Jun 20, 2017 at 20:50
  • $\begingroup$ Don't get the point of $\pi$ in the fraction.. just there to cancel out? $\endgroup$
    – Nikunj
    Jun 20, 2017 at 20:51
  • $\begingroup$ removed the pi, it serves no real purpose. What I've done so far is compare the above integral to the same integral substituting $\lfloor\frac{b}{x}\rfloor$ with $\frac{b}{x} -1$ and $\frac{b}{x} +1$ $\endgroup$ Jun 20, 2017 at 20:57
  • $\begingroup$ The function appears to be bounded above by the parabola $y=2\left(x-\dfrac{1}{4b}\right)$. desmos.com/calculator/0kc645vuen $\endgroup$ Jun 20, 2017 at 21:43
  • $\begingroup$ Supposed to be $y=2\left(x-\dfrac{1}{4b}\right)^2$ $\endgroup$ Jun 20, 2017 at 21:49

1 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.


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

first image second image

  • $\begingroup$ I am interested in integer inputs for x, if that helps. I plan on using the Euler-Maclaurin formula on the function. Is there any way to tighten the bound. $\endgroup$ Jun 20, 2017 at 23:26
  • $\begingroup$ No, a look at the three graphs of equation $(1)$ shows that the bounds cannot be tightened since the middle function meets both the upper and the lower bounds. $\endgroup$ Jun 20, 2017 at 23:41

Not the answer you're looking for? Browse other questions tagged or ask your own question.