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

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• 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 $$\sum_{m\geq 1}\frac{1}{m^{n+2}(m+1)}=\int_{0}^{1}\text{Li}_{n+2}(z)\,dz = (-1)^{n+1}\left[1-\zeta(2)+\zeta(3)-\zeta(4)+\ldots\pm \zeta(n+2)\right].$$
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 $\left\lfloor\frac{1}{x} \right\rfloor$.