1. ## Integral Notation

One of the thing that bothered me for a long time with the $\displaystyle \bold{d}x$ appearing in the end of the integral. What is it for? And why was it but there? People told me that it is because we want to show what we variable we are integrating, but it is clear without it even being there. Behold! I have dreamt a dream and I have a revelation! Behold! It was Riemann he told me the answer. Behold! It was the Riemann-Stieltjes Integral.

I want to explain what it is, it will make the notion of why we put a $\displaystyle \bold{d}x$ in the integral a lot more clearer.

First let us begin with a simpler question. What is a Riemann Integral? If you taken a basic course in analysis you would know there are two ways to define it. The classical Riemann definition which is a little discussed in a Calculus course also, and another one (which is exactly the same thing) developed by Darboux. Since Riemann's definition is more elementary (but not as neat) let us do that definition.

Definition: Let $\displaystyle f$ be a bounded function on a closed interval $\displaystyle [a,b]$. We say that $\displaystyle f$ is integrable on this interval when there exists a real number $\displaystyle I$ such that: for any $\displaystyle \epsilon > 0$ there exists a $\displaystyle \delta >0$ so that for any partition $\displaystyle P= \{ a=x_0 < x_1< ... < x_{n-1}<x_n=b \}$ satisfing $\displaystyle \text{mesh}(P) = \max_{1\leq k\leq n} \ \{x_{k} - x_{k-1} \} < \delta$ we have that $\displaystyle \left| I - \sum_{i=1}^n f(t_k)(x_k-x_{k-1}) \right| < \epsilon$ where $\displaystyle t_k$ is any point chosen on $\displaystyle [x_{k-1},x_k]$-subinterval.

Basically the definition is saying that we can make the finite sums (approximating areas) as close as we want to the true value which we call $\displaystyle I$* as long as the partition $\displaystyle P$ of the interval is fine (or thin) enough. And note how much freedom we have, it says for any partition and there are infinitely many, and it says any point in sub-interval and again there are infinitely many. So there is so much freedom with these finite Riemann sums.

So we know that if $\displaystyle f(x) = x \mbox{ on }[0,1]$ then to show that $\displaystyle \int_0^1 f = \frac{1}{2}$ we need to show that the number $\displaystyle I = \frac{1}{2}$ is this number we need to satisfy the definition given above.

Note, that is what the fundamental theorem of Calculus is doing. Instead of going through all that difficult definition, it says that if we can find the anti-derivative it is that value $\displaystyle I$ that we are looking for.

If you think the Riemann integral definition is complicated just look at the Riemann-Steiljes integral definition. Now the Riemann-Steiljes integral is more general. It is an integral with respect to another function. Before stating their definition there is just one technical detail.

Definition: Let $\displaystyle g:[a,b]\mapsto \mathbb{R}$ is a bounded variation when there exist a constant $\displaystyle M>0$ so that if for any partition $\displaystyle P = \{ a=x_0<...<x_n = b\}$ we have that $\displaystyle \sum_{k=1}^n |g(x_k)-g(x_{k-1})| \leq M$.

Now we can state the definition (which might look monstrous in the beginning).

Definition: Let $\displaystyle f$ be a bounded function on $\displaystyle [a,b]$ and $\displaystyle g$ be a bounded variation on $\displaystyle [a,b]$. We say $\displaystyle f$ is Riemann-Steiljes integration with respect to $\displaystyle g$ when there exists a real number $\displaystyle I$ such that: for any $\displaystyle \epsilon > 0$ there exists $\displaystyle \delta > 0$ so that for any partion $\displaystyle P = \{a = x_0<x_1<...<x_n = b\}$ satisfing $\displaystyle \text{mesh}(P) < \delta$ we have that $\displaystyle \left| I - \sum_{k=1}^n f(t_k)[g(x_{k}) - g(x_{k-1})] \right| < \epsilon$ where $\displaystyle t_k$ are any points in the $\displaystyle [x_k-x_{k-1}]$ sub-interval. We call this distinguished number $\displaystyle I$ to be the Riemann-Steiljes integral of $\displaystyle f$ on $\displaystyle [a,b]$ with respect to $\displaystyle g$. And write $\displaystyle I = \int_a^b f \bold{d}g$.

Now why is this a generalization? Because if $\displaystyle g(x) = x$ then it is the standard Riemann integral! And that means with respect to $\displaystyle x$ we would write $\displaystyle \int_a^b f \bold{d}x$. And that is where the $\displaystyle \bold{d}x$ comes from.

In fact it turns out that if $\displaystyle f$ is continous and $\displaystyle g$ is smooth (continously differenciable) then it is a bounded variation and:
$\displaystyle \int_a^b f \bold{d}g = \int_a^b fg'$. Where the RHS is the standard Riemann Integral.

So not only does this explain the $\displaystyle \bold{d}x$ part it also explain the differencial part of a function.
For example,
$\displaystyle \int_0^\pi \sin x d(x^2+x) = \int_0^{\pi} \sin x (2x+1) dx$.
By the Riemann-Steiljes Integral.

Maybe you find that interesting, that is why I posted it.

*)It can be easily show that if $\displaystyle I_1,I_2$ are any possible real values for then Riemann integral then $\displaystyle I_1 = I_2$. Meaning there is only one such possible value $\displaystyle I$ and we define it to be the integral $\displaystyle \int_a^b f$.

2. If you would like to explore more of the topics you have introduced then that are two classics that are standards: INTRODUCTION TO THE THEORY OF INTERGRATION by T. H. Hildebrandt and THEORY OF THE INTEGRAL by Stanislaw Saks. I think that Hildebrandt is still the best on Riemann and Riemann-Stieltjes integrals there is. He also has a good discussion of content of a set which was the subject of another one of your postings.

If you are interested in modern work in integration theory I would suggest Robert McLeod’s book THE GENERALIZIED RIEMANM INTEGRAL. It explores a relatively new definition of the integral that makes any derivative, F’(x), integrable on [a,b] to F(b)-F(a). That is not true of earlier attempts. That book is an MAA publication: Carus#20.

3. Thank you. I do like theory of integration, one of my favorite in an analysis course but since I am doing so much stuff it seems I have to wait until I have time to explore integration in more detail.

Whose is Stieltjes pronounced?

4. Originally Posted by ThePerfectHacker
Whose is Stieltjes pronounced?
Thomas Joannes Stieltjes (pronounced 'sti:ltʃəs).