# Evaluate $\int_0^1 x^n\left\lfloor\frac{1}{x}\right\rfloor^{-1}dx$ [closed]

I'd like help with calculating: $$\int_0^1 x^n\left\lfloor\frac{1}{x}\right\rfloor^{-1}dx$$ where $⌊\cdot⌋$ is the floor function. Any suggestions?

Thanks!

## closed as off-topic by Xander Henderson, B. Mehta, Namaste, user296602, Andrew LiApr 19 '18 at 23:59

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, B. Mehta, Namaste, Community, Andrew Li
If this question can be reworded to fit the rules in the help center, please edit the question.

• That is $$\int_{1}^{+\infty}\frac{dx}{\lfloor x\rfloor x^{n+2}} = \sum_{m\geq 1}\frac{1}{m}\int_{m}^{m+1}\frac{dx}{x^{n+2}} = \ldots$$ – Jack D'Aurizio Apr 19 '18 at 23:31
• @Steven Please include your efforts/workings toward solving this question. – Namaste Apr 19 '18 at 23:35
• By chance, is the original question from the American Mathematical Monthly or Cornel Ioan Valean? – Jack D'Aurizio Apr 19 '18 at 23:52

That is $$\int_{1}^{+\infty}\frac{dx}{\lfloor x\rfloor x^{n+2}} = \sum_{m\geq 1}\frac{1}{m}\int_{m}^{m+1}\frac{dx}{x^{n+2}} = \frac{1}{n+1}\sum_{m\geq 1}\frac{1}{m}\left(\frac{1}{m^{n+1}}-\frac{1}{(m+1)^{n+1}}\right)$$ and by summation by parts this boils down to computing $$\sum_{m\geq 1}\frac{H_m}{m^{n+1}}$$ which can be done through Euler sums. See this related question.
The first instances are $$\sum_{m\geq 1}\frac{H_m}{m^2}=2\,\zeta(3), \qquad \sum_{m\geq 1}\frac{H_m}{m^3}=\frac{\pi^4}{72},\qquad \sum_{m\geq 1}\frac{H_m}{m^4}=3\,\zeta(5)-\zeta(2)\zeta(3).$$ As an alternative $$\sum_{m\geq 1}\frac{1}{m^{n+2}(m+1)}=\int_{0}^{1}\text{Li}_{n+2}(z)\,dz = (-1)^{n+1}\left[1-\zeta(2)+\zeta(3)-\zeta(4)+\ldots\pm \zeta(n+2)\right].$$

\begin{align}\int_0^1 \frac{x^n}{⌊\frac{1}{x}⌋}\ \mathrm{d}x = &\;\int_{\frac{1}{2}}^1 \frac{x^n}{1}\ \mathrm{d}x+ \int_{\frac{1}{3}}^{\frac{1}{2}} \frac{x^n}{2}\ \mathrm{d}x + \int_{\frac{1}{4}}^{\frac{1}{3}} \frac{x^n}{3}\ \mathrm{d}x\ +\cdots\\ \end{align}

HINT: It is $$\int_{1/2}^{1}x^n\,dx+2\int_{1/3}^{1/2}x^n\,dx+3\int_{1/4}^{1/3}x^n\,dx+\ldots$$ It is equal to $$\sum_{k=1}^{\infty}k\frac{1}{n+1}((1/k)^{n+1}-(1/(k+1))^{n+1})$$

Hint: Split the interval up into regions depending on the value of $\left\lfloor\frac{1}{x} \right\rfloor$.

• I tried but it was hard for me – Steven Apr 19 '18 at 23:30