How to integrate the following floor function?

Question

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.

Answer 1

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}

Then

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

and so on.