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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!

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

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  • 2
    $\begingroup$ That is $$ \int_{1}^{+\infty}\frac{dx}{\lfloor x\rfloor x^{n+2}} = \sum_{m\geq 1}\frac{1}{m}\int_{m}^{m+1}\frac{dx}{x^{n+2}} = \ldots $$ $\endgroup$ – Jack D'Aurizio Apr 19 '18 at 23:31
  • $\begingroup$ @Steven Please include your efforts/workings toward solving this question. $\endgroup$ – Namaste Apr 19 '18 at 23:35
  • $\begingroup$ By chance, is the original question from the American Mathematical Monthly or Cornel Ioan Valean? $\endgroup$ – Jack D'Aurizio Apr 19 '18 at 23:52
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That is $$ \int_{1}^{+\infty}\frac{dx}{\lfloor x\rfloor x^{n+2}} = \sum_{m\geq 1}\frac{1}{m}\int_{m}^{m+1}\frac{dx}{x^{n+2}} = \frac{1}{n+1}\sum_{m\geq 1}\frac{1}{m}\left(\frac{1}{m^{n+1}}-\frac{1}{(m+1)^{n+1}}\right) $$ 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 $$ \sum_{m\geq 1}\frac{1}{m^{n+2}(m+1)}=\int_{0}^{1}\text{Li}_{n+2}(z)\,dz = (-1)^{n+1}\left[1-\zeta(2)+\zeta(3)-\zeta(4)+\ldots\pm \zeta(n+2)\right]. $$

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\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}

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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 $$ \sum_{k=1}^{\infty}k\frac{1}{n+1}((1/k)^{n+1}-(1/(k+1))^{n+1}) $$

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

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

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