Riemann-Stieltjes integral (floor function) 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.
 A: 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.
A: 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:
$$
\begin{eqnarray*}
\int_{-1.2}^{3.9}x\,d\left\lfloor x\right\rfloor 
&=&\int_{-1.2}^{-1}x\,d\left\lfloor x\right\rfloor
+\int_{-1}^{3}x\,d\left\lfloor x\right\rfloor
+\int_{3}^{3.9}x\,d\left\lfloor x\right\rfloor , \\ 
&=&-1+6+0=5.
\end{eqnarray*}
$$
