# Integration with a dependence of $\lfloor x\rfloor$ or in general of a function $\lfloor g(x)\rfloor$

In mathematics, the integer part function, also known as the floor function is the function that associates with each real number $$x$$ the largest integer less than or equal to $$x$$, denoted by $$\lfloor x\rfloor$$ or $$[x]$$ being $$\lfloor x \rfloor=\max\, \{k\in\mathbb{Z} : k\le x\}$$. We suppose to consider the trivial integral $$\int \lfloor x\rfloor \, dx\tag 1$$

I have thought that since the value of the indefinite integral is independent of the integration variable could I write? $$\int \lfloor x\rfloor \, d\lfloor x\rfloor=\frac{(\lfloor x\rfloor)^2}2+\text{const.}\tag 2$$ I know that a finite function with a finite (at most numerable) number of discontinuity points is always Riemann-integrable. What would happen if the integrand function is of the kind

$$\int f(x)\lfloor g(x)\rfloor\,dx\tag 3$$

Definitively how can indefinite or definite integration be treated in the presence of integer part of functions? How should I proceed?

• Hi :-) you can also see my profile I logged in a few days ago or this morning on MSE. Please could you add a simple answer so that can I remember the procedure? Why do you use $\int_a^b [x]d[x] = b[b] - a[a] - \int_a^b [x]d[x]$? How is it possibile to prove it? In general what is the procedure to calculate an integral with the presence of $[x]$? I wait for your answer also. Jul 10, 2021 at 21:17
• Actually no, pardon me. My post was useful, but I think $[x]$ is not R-S integrable with respect to itself. So I think one will have to use Lebesgue-type notions to interpret $\int_a^b [x]d[x]$ @Sebastiano I apologize. I believe my link will not be helpful, because although it did involve integration with respect to the floor, the floor isn't RS integrable with respect to itself. So none of the formulas apply. (I'll have to leave now as well, unfortunately, so I'll come back tomorrow and we'll discuss this). Jul 10, 2021 at 21:17
• @TeresaLisbon Don't worry...I wait a complete your response to calculate a general integral (definite or indefinite) with $[x]$. Thank you very much. No apologize :-)...I done often mistakes :-) Jul 10, 2021 at 21:20
• I don't think I'm qualified enough to answer this, honestly. But I'll let you know if I give it a go for sure! Jul 10, 2021 at 21:21

## 1 Answer

For the first question: No, because indefinite integrals stands for antiderivatives, and discontinuous functions are not differentiable.

For the second question: the product of two definite integrable functions are integrable definitely. But again the indefinite integral won't make too much sense.

• Thank you very much for your contribute. Jul 10, 2021 at 21:18