finding $\displaystyle \int^{4}_{0}(x^2+1)d(\lfloor x \rfloor),$ given $\lfloor x \rfloor $ is a floor function of $x$

Assume $\displaystyle I = (x^2+1)\lfloor x \rfloor \bigg|^{4}_{0}-2\int^{4}_{0}x\lfloor x \rfloor dx$ ( integration by parts )

i have a doubt about limit part , did not understand whether the limit corresponding to $x$

or corrosponding to $\lfloor x \rfloor$

because when we take $\displaystyle \int^{b}_{a}f(x)dx,$ then limits are corrosponding to $x$

but when we take $\displaystyle \int^{b}_{a}f(x)d(\lfloor x \rfloor ),$ then limit corrosonding to $\lfloor x \rfloor$

please clearfy my doubt and also explain me whats wrong with my method above , thanks

  • $\begingroup$ I am not sure but I doubt that integral is zero. The Lebesgue measure $\mu = \lfloor x \rfloor$ is zero in $\mathbb{R}$. $\endgroup$
    – Alex Silva
    Jan 12, 2017 at 8:54
  • $\begingroup$ The floor function, being right continuous and of bounded variation on $[a,b]$, corresponds uniquely to a signed measure $\mu$ on $[a,b]$, and the integral should be understood as $\int_{[a,b]}f(x)\mathrm d\mu.$ (wait a minute, I just realised the floor function is increasing, so the integral can be alternatively treated as a Riemann-Stieljes integral, in a much easier way. However, for the general case $\int_A f\mathrm d g$ where $g$ isn't increasing, we have to resort to Lebesgue theory.) $\endgroup$
    – Vim
    Apr 7, 2017 at 9:08
  • 1
    $\begingroup$ Limits always correspond to the variable w.r.t. which you are integrating. And in both cases, it is simply $x$. Limits change only when you make a substitution or split the integral into parts, and neither is done in your case! (Using the "differential" $d\lfloor x \rfloor$ means that are changing the way how the length is measured, but it does not mean that you are deforming the domain of integration.) $\endgroup$ Apr 7, 2017 at 9:13
  • $\begingroup$ Also, it seems that there's ambiguity in your formulation of this integral. You should specify what $\int_0^4$ means, is it over $[0,4]$ or $(0,4)$ or $[0,4)$ etc? As said in my answer below, singletons can no longer be neglected because the floor function assigns positive measure (or intuitively, "jumps") to a lot of singletons (all integral points) in $\Bbb R$. $\endgroup$
    – Vim
    Apr 7, 2017 at 9:34

2 Answers 2


About your attempt to integrate by parts: instead of doubting about the legitimacy of the change of the "corresponding limits", I advise you take the following steps:
1). Try to understand the basic theory of Riemann Stieljes integration.
2). Find a proof of integration by parts for R-S integrals, be careful about the assumptions, and then try to fully understand the proof. (A quick search on this site gives many, for example.)

A helpful answer would have to be based on the definition you adopt for this integral. Here I give two treatments, which should lead to the same result.

1). From a measure theoretic perspective, we have to find the measure $\mu$ on $\Bbb R$ induced by $g(x)=\lfloor x\rfloor$, in a way such that $\mu((c,d])=g(d)-g(c)$ for $c<d$. Clearly, we can find that $\mu$ is an atomic measure which assigns mass $1$ to each integral point in $\Bbb R$ and $0$ elsewhere, which basically says that you need only to care about integral points, to which you should assign mass $1$, over your integration domain, i.e. $$\int_A f(x)\mathrm dg(x)=\sum_{i\in \Bbb Z\cap A} f(i).$$

2) Yet another simpler to understand treatment is the Riemann-Stieljes integral. For details of the computation, they bear much resemblance to computing a Riemann integral by definition, the only change being replace $\Delta x_i$ by $\Delta g_i$. For a fuller explanation, you should be able to check the definitions and relevant properties online (where resources should be plenty) by yourself.


you can rewrite the integral as $$\int_0^4 x^2+1 \frac{d[x]}{dx}dx$$ where $[x]$ is greatest integer function.Now apply IBP taking $\frac{d[x]}{dx} $ as function to be first integrated.

$$ (x^2+1) \int (\frac{d[x]}{dx} dx)\text{from x=0 to x=4}-\int_0^4 2x ( \int \frac{d[x]}{dx} dx)dx=0$$

  • $\begingroup$ How can you differentiate a function which is apparently not differentiable? $\endgroup$
    – Vim
    Apr 7, 2017 at 9:36
  • $\begingroup$ @Vim I have not differentiated anything.I just multiplied and divided by the differential dx $\endgroup$
    – catechol
    Apr 7, 2017 at 9:41
  • $\begingroup$ Technically, "quotient of two differentials" is a syntactic sugar for differentiation. The issue of non-differentiability still needs to be resolved. If you want to regard it other than the classical differentiation-thingly, then you have to supply all the relevant theory so that the integral makes sense. (Both measure theory and distribution theory are capable of serving this purpose.) $\endgroup$ Apr 7, 2017 at 13:12
  • $\begingroup$ @SangchulLee I follow what you said but I am just a begineer in calculus and don't know about the theories but I have one thing to say.even if we notice that my method is not correct we could perhaps split the integral into 0,1;1,2;2,3;3,4 and then apply my method.There should not be any issue of differentiability then? $\endgroup$
    – catechol
    Apr 7, 2017 at 14:01
  • $\begingroup$ The issue that the "differential" $d\lfloor x\rfloor$ has mass concentrated at integer points. So splitting the interval into parts (especially at integer points) does not improve the situation. But your solution can be savlaged by proving a version of IBP for Riemann-Stieltjes integral. $\endgroup$ Apr 7, 2017 at 14:22

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