8

Assume $f$ is continuous on $[1,n]$. How would you go about taking the integral $$\int_1^n f(x)\,d\lfloor x\rfloor$$ where $\lfloor x\rfloor$ represents the greatest integer function?

user127.0.0.1
  • 7,435
user10311389
  • 209
  • Just some thoughts, $d\lfloor x \rfloor$ is $dx$ multiplied by an impulse train, and integrating an impulse train times $f(x)\ dx$ from $1^-$ to $n^+$ gives $\sum_{t=1}^{n}f(t)$, if $n$ is a natural number $>1$. – peterwhy Jan 23 '14 at 04:46
  • You can use the theorem in my answer. – Mhenni Benghorbal Jan 23 '14 at 07:26
  • $[x]$ isn't derivable in integer values of x; and where it is derivable, it's $0$. Unless you're willing to improvise a bit, and consider that its derivative for integer values is a Dirac impulse. In which case you get a discrete sum, I think. – Lucian Jan 23 '14 at 11:58

2 Answers2

1

This is a Stieltjes integral. By definition it is a limit under refinement of sums of the following kind: Consider (tagged) partitions $${\cal P}:\quad 1=x_0<x_1<x_2<\ldots<x_N=n$$ of the interval $[1,n]$, and put $$S_{\cal P}:=\sum_{i=1}^N f(\xi_i)\bigl(\lfloor x_i\rfloor-\lfloor x_{i-1}\rfloor\bigl)\ .$$ Note that $\lfloor x_1\rfloor-\lfloor x_0\rfloor=0$ and $\lfloor x_N\rfloor-\lfloor x_{N-1}\rfloor=1$ for all sufficiently fine partitions. It is then easily seen that $$\lim_{{\cal P}\to?}\> S_{\cal P}=\sum_{k=2}^n f(k)\ .$$

Christian Blatter
  • 232,144
1

$$\begin{align*} \int_1^n f(x)\ d\lfloor x \rfloor =& \left[f(x)\lfloor x \rfloor\right]_1^n - \int_1^n \lfloor x \rfloor df(x)\\ \end{align*}$$ Since $f$ is continuous, we use the additivity of integration on intervals property for the steps $$\begin{align*} =& nf(n)-f(1) - \sum_{i=1}^{n-1} \int_i^{i+1}\lfloor x\rfloor df(x)\\ =& nf(n) - f(1) - \sum_{i=1}^{n-1} i\int_i^{i+1}df(x)\\ =& nf(n) - f(1) - \sum_{i=1}^{n-1} i [f(i+1)-f(i)]\\ =& nf(n) - f(1) - \sum_{i=2}^{n} (i-1)f(i) + \sum_{i=1}^{n-1} if(i)\\ =& \sum_{i=2}^{n} f(i) \end{align*}$$

Leslie Brenes
  • 25
peterwhy
  • 22,930