Indefinite Integral of Floor Function Integration by Substitution

Question

From knowing the anti-derivative of floor function to be x*floor(x), is it possible to find the derivative of a function contained within a floor function?

The particular question I had in mind is floor(y(x)/17) and I believe in the same way y in an equation can be treated as y function of x, I tried using integration by substitution.

Nevertheless, the integral appears to be too complex to integrate; as by using the rule for substitution, every integration opens up a new iteration of another integral, sometimes even equating to the L.H.S. but unable to be simplified. It is a shame my rough work is too unorganized to be shown on here.

It is hoped that someone may shed some light how to integrate implicitly, an example of a similar iterated equation may be fine, but it would be best if the integral of the equation prior can be solved.

Answer 1

First, the antiderivative for $\;\lfloor x \rfloor$ is $x\lfloor x \rfloor - \frac12 \lfloor x \rfloor(\lfloor x \rfloor + 1)\;$, not $x \lfloor x\rfloor$.

For any continuous function $g(x)$ defined on $(0^{-}, L^{+})$ for some $L > 0$ which satisfies:

• The points $\lambda_i \in [0,L]$ with integral value $g(\lambda_i)$ are all isolated.
• For any such $\lambda_i$, $g(x)$ is strictly monotonic in some neighborhood of $\lambda_i$.
  Define $\epsilon_i = \pm 1$ depends on whether $g(x)$ is increasing or decreasing there.

If one interpret all integrals involved as Riemann Stieltjes integral, one can integrate by part the integrand $\lfloor g(t)\rfloor$ over an interval $[0,x ] \subset [0,L]$ and get: $$\int_0^x \lfloor g(t) \rfloor dt = \int_{0^{-}}^{x^{+}} \lfloor g(t) \rfloor dt = x \lfloor g(x) \rfloor - \int_{0^{-}}^{x^{-}} t\,d \lfloor g(t)\rfloor = x \lfloor g(x) \rfloor - \sum_{i : \lambda_i \in [0,x]} \epsilon_i \lambda_i $$ The problem of computing the anti-derivative reduces to a summation over the position of discontinuity of $g(x)$.

Please look up the wiki entry of Riemann-Stieltjes integral, once you understand under what condition you can integration by part, a lot of integral over functions with only jump discontinuous can be converted to and from corresponding sums easily.

Answer 2

$\lfloor f(t) \rfloor = \int_0^t \sum_{i=0}^{t+\epsilon}\left(f(a_i+\epsilon) - f(a_i-\epsilon)\right) \delta(x-a_i) dx$ where the $a_i$'s are the points where $f(t) \in \mathbb{N}$. the integration by parts with Dirac works if : it's an integral on a finite interval $[b,c]$, there is no Dirac at the limits, and the Dirac are integrated times a $C^1$ function on $[b,c]$. if the interval is $[b;\infty[$ then the limit $c\to \infty$ must be investigated. your problem is to reverse that integration by parts thus it works. note that $\delta(x^2-a)$ is non-sense : only $\delta(x-a)$ is allowed and that every time you get a Dirac. So integrating a $\delta$ is nether a problem, and thus derivating the obtained function neither. this means don't try to make a change of variable when you integrate a Dirac ! (or at least respect the weird rules for it)