Riemann-Stieltjes Integral geometric intepretation

1. Jul 27, 2011

Zeato

Hi all,

I would like to ask for the geometric interpretation of the riemann-stieltjies integral.

Suppose we have an integral, (integrate f dg) over the interval [a,b], where g is monotonically increasing.

Can i interpret it as the area between f and the g function?

Moreover, i am a little confused on what do they mean when they say f is integrable with respect to the g function. E.g. (integrate f dg). Do they treat g(x) as a function, g(x)= y and then (integrate f (y) dy) ?

2. Jul 27, 2011

pwsnafu

Igor Podlubny wrote a paper on finding http://arxiv.org/abs/math/0110241" [Broken] This paper relies on a previous result about finding a geometrical interpretation for Riemann-Stieltjies. You can read Podlubny's paper for a short synopsis on that result (it should have a citation in there).

Last edited by a moderator: May 5, 2017
3. Jul 27, 2011

Zeato

Hi pwsnafu,

Thanks for the article.
Although i am able to understand intuitively from the article that the geometric interpretation for the riemann-stieltjes integral is actually the area under a vector function (E.g. K( f(x), g(x), x) for ( integrate f dg ) )
I am still unable to comprehend how they actually come to such a conclusion. I mean for Riemann, Darboux integral, we often look at the the partition over an interval x, but the article mention the use of projection of vectors instead. Also, where do they take the partition in this case? Do they take the partition from the plane that contain f, g, and x vectors? How do they come to such a conclusion then?

Also, when they say f is integrable with respect to g, how do we interpret this geometrically, with the shadow and the fence in the mentioned article?

Last edited: Jul 27, 2011
4. Jul 27, 2011

HallsofIvy

Note that in the Riemann-Stieljes integral, $\int f(x)dg$, g only has to be "monotone increasing". It does not have to be differentiable or even continuous. Taking g(x) to be the "greatest integer less than x" gives
$$\int_0^N f(x)dg= \sum_{n=0}^N f(n)$$

If g is, in fact, differentiable, then $\int f(x)dg= \int f(x)g'(x)dx$

5. Jul 27, 2011

Zeato

Hi HallsofIvy,