Riemann-Stieltjes integral (floor function)

Question

Please, I am having problem with this function.

$$ \int_{-1.2}^{3.9} xd[x] $$

Here is what I have done $$\int_{-1.2}^{1}xd[x] + \int_{0.8}^{2}xd[x] + \int_{1.8}^{3}xd[x] + \int_{2.8}^{3.9}xd[x] $$

I don't know if I am on the right track, help.

Answer 1

I'm not quite sure how you're breaking up the integral - if you look at it, there is clearly overlap between the limits of integration in the different parts.

Let me give you an extended hint. The Riemann-Stieltjes integral is defined as the limit of Riemann sums of the type $$ \sum_{i=0}^{n-1}x_i([x_{i+1}]-[x_i]) $$ It should be clear, then, that the only pieces of any partition $x_0=a,x_1,x_2,\ldots,x_n=b$ for this integral will cancel out all intervals $[x_i,x_{i+1}]$ where $[x_i]=[x_i+1]$... in other words, we only care about the sub-intervals where we cross an integer value.

So, if we write $$ \int_{-1.2}^{3.9}x\,d[x]=\int_{-1.2}^{-1-\epsilon}x\,d[x]+\int_{-1-\epsilon}^{-1+\epsilon}x\,d[x]+\int_{-1+\epsilon}^{0-\epsilon}x\,d[x]+\cdots, $$ we can cancel out all all intervals that are not of the form $[a-\epsilon,a+\epsilon]$ for an integer $a$. Hence for any $\epsilon$, we can write $$ \int_{-1.2}^{3.9}x\,d[x]=\sum_{a=-1}^{3}\int_{a-\epsilon}^{a+\epsilon}x\,d[x]. $$ So, what's left is to figure out what each of these integrals must be. Try thinking about how these work from the definition, noting that you can make $\epsilon$ as small as you like.

Answer 2

Let $f$ be a continuous function and $g$ a constant function $g(x)=c$ in the interval $[a,b[$ and $g(b)$ its value at $x=b$. Then the Riemann-Stieltjes integral

$$ \begin{equation*} \int_{a}^{b}f(x)\,dg(x)=f(b)\underset{\text{jump of }g(x)\text{ at }x=b}{\underbrace{\left( g(b)-c\right) }}.\tag{$\ast$} \end{equation*} $$

In your case $f(x)=x$ and $g(x)$ is the floor function $g(x)=\left\lfloor x\right\rfloor $. Then for $n\in\mathbb{Z}$ $$ \begin{eqnarray*} g(x)&=&\left\lfloor x\right\rfloor =n,\quad\text{for } x\in [n,n+1[,\\ g(n+1)&=&\left\lfloor n+1\right\rfloor =n+1. \end{eqnarray*} $$

The jump of $g(x)$ at $x=n+1$ is equal to $1$:

$$g(n+1)-n=\left\lfloor n+1\right\rfloor - n =1.$$

By $(\ast)$ we have $$ \begin{eqnarray*} \int_{n}^{n+1}x\,d\left\lfloor x\right\rfloor &=&(n+1)(1)=n+1, \\ \int_{n}^{n+k}x\,d\left\lfloor x\right\rfloor &=&\sum_{m=1}^{k}n+m=nk+\frac{ k(k+1)}{2}, \\ \int_{-1}^{3}x\,d\left\lfloor x\right\rfloor &=&\sum_{m=1}^{4}-1+m=6, \end{eqnarray*} $$

and $$ \begin{eqnarray*} \int_{-1.2}^{-1}x\,d\left\lfloor x\right\rfloor &=&(-1)(1)=-1, \\ && \\ \int_{3}^{3.9}x\,d\left\lfloor x\right\rfloor &=&3.9\left( 3-3\right) =0,\text{ the jump of $g(x)=\left\lfloor x\right\rfloor$ at $x=3.9$ is $0$}. \end{eqnarray*} $$ Therefore the given integral evaluates as: $$ \begin{eqnarray*} \int_{-1.2}^{3.9}x\,d\left\lfloor x\right\rfloor &=&\int_{-1.2}^{-1}x\,d\left\lfloor x\right\rfloor +\int_{-1}^{3}x\,d\left\lfloor x\right\rfloor +\int_{3}^{3.9}x\,d\left\lfloor x\right\rfloor , \\ &=&-1+6+0=5. \end{eqnarray*} $$