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

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

• I am not sure but I doubt that integral is zero. The Lebesgue measure $\mu = \lfloor x \rfloor$ is zero in $\mathbb{R}$. Jan 12, 2017 at 8:54
• 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.)
– Vim
Apr 7, 2017 at 9:08
• 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.) Apr 7, 2017 at 9:13
• 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 $\Bbb R$.
– Vim
Apr 7, 2017 at 9:34

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.

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

• How can you differentiate a function which is apparently not differentiable?
– Vim
Apr 7, 2017 at 9:36
• @Vim I have not differentiated anything.I just multiplied and divided by the differential dx Apr 7, 2017 at 9:41
• 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.) Apr 7, 2017 at 13:12
• @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? Apr 7, 2017 at 14:01
• 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. Apr 7, 2017 at 14:22