I need to calculate the following integral, $$\int_{1}^{10}\frac{x-\lfloor x \rfloor }{x^2} dx$$.

So I did this-

$$\int_{1}^{10}\frac{x-\lfloor x \rfloor }{x^2} dx=\int_{1}^{10}\frac{1 }{x} dx-\int_{1}^{10}\frac{\lfloor x \rfloor}{x^2}dx.$$ Now my problem is with the second integral. How can I calculate this? I have tried to looking for examples here but in all those example there only the floor function is involved not a denominator part as I have. So how can I do this? Please help.

  • 3
    $\begingroup$ Cut it into 9 integrals: $\int_1^2+\int_2^3+\ldots+\int_9^{10}$ $\endgroup$ – Nathaniel B Jul 6 '16 at 1:48
  • $\begingroup$ Why I need to do that? Can you elaborate a bit more? $\endgroup$ – Harry Potter Jul 6 '16 at 1:49
  • $\begingroup$ For $x\in[1,2)$, $\lfloor x\rfloor=1$. $\endgroup$ – vadim123 Jul 6 '16 at 1:50
  • $\begingroup$ Over each of the nine intervals, the floor function reduces to a constant term. i.e. $\int_1^2\lfloor x\rfloor /x^2~dx=\int_1^21/x^2~dx$, $\int_2^3\lfloor x\rfloor /x^2~dx=\int_2^32/x^2~dx$, $\int_3^4\lfloor x\rfloor /x^2~dx=\int_3^43/x^2~dx$, etc. $\endgroup$ – Nathaniel B Jul 6 '16 at 1:52
  • $\begingroup$ ok got it @NathanielB $\endgroup$ – Harry Potter Jul 6 '16 at 1:53

You need to compute integrals of the form $\displaystyle \int_n^{n+1} \dfrac{\lfloor x \rfloor}{x^2} dx$. You get

\begin{align} \int_n^{n+1} \dfrac{\lfloor x \rfloor}{x^2} dx &= \int_n^{n+1} \dfrac{n}{x^2} dx \\ &= \left. \dfrac{-n}{x} \right|_n^{n+1}\\ &= \dfrac{-n}{n+1} + 1\\ &= \dfrac{1}{n+1} \end{align}


$\displaystyle \int_1^{10} \dfrac{\lfloor x \rfloor}{x^2}dx = \int_1^2 \dfrac{\lfloor x \rfloor}{x^2}dx + \int_2^3 \dfrac{\lfloor x \rfloor}{x^2}dx + \cdots + \int_9^{10} \dfrac{\lfloor x \rfloor}{x^2}dx$

and so on.

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  • $\begingroup$ I got $\int_n^{n+1} \dfrac{n}{x^2} dx= \dfrac{1}{n+1}$. Can you double check the factor "2"? $\endgroup$ – mike Jul 6 '16 at 2:12
  • $\begingroup$ @mike - got it. $\endgroup$ – steven gregory Jul 6 '16 at 3:04

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