Formal definition of the integral?

nilkn · 2005-09-11T18:56:14

I've just realized that often, especially is physics, integrals are used merely as syntactic sugar representing an infinite sum of the form: lim(dx->0) sum(f(x)*dx) where dx is delta-x. Take for example the simplified equation for center of mass: cm = I(p dm)/tm where I is the integral symbol, p is the position of a point mass relative to the centroid of the body, dm is the mass of the individual point masses (and so p dm is the first moment), and tm is the total mass. The integral in that equation, as far I know, isn't really calculating an area, volume, etc. It's merely a fancy way of writing an infinite sum of the above form. What I'm wondering is how the integral is formally defined. I've looked it up on the internet and found it to always be described as the area under the curve, volume, curl, etc. Is it correct to say that, in a very broad sense, the integral is just fancy syntax for infinite sums of the above form?

Math and Physics Programming

Started by nilkn September 02, 2005 05:32 PM

33 comments, last by Sheep 18 years, 7 months ago

sherifffruitfly

104

September 11, 2005 11:56 AM

Quote:
Quote:I believe there are also such functions with infinity amount of points(Countable, and in a finity fragment) with no continuity, which you can calculate their rieman integral.
The definition of a Riemann Integral requires a function to be defined at all points on an interval.

Sigh. It is the case that Riemann-integrable functions may be discontinuous at an infinite number of points. A correct statement is: A Riemann-integrable function is continuous except possibly on a set of measure zero.

Intelligent discussion of the limits of Riemann integration requires an understanding of the Lesbeque and the Stieltjes integrals (they "solve" different "problems" with Riemann's version).

NotAnAnonymousPoster

286

September 11, 2005 12:28 PM

Quote:Original post by sherifffruitfly
False. A Riemann-integrable function's domain does not have to include an interval. Consider the function f, defined only at 0, such that f(0)=0. This is Riemann-integrable (everywhere continuous and differentiable even). But its domain does not include any interval.

It's a degenerate case, where definitions often disagree. Still, I think most definitions of Riemann-integral will entail that f in this case is not Riemann-integrable.

The theorem you are thinking to justify otherwise will state that any function continuous on a closed interval is integrable on that interval, not that a function which is everywhere continuous is everywhere Riemann-integrable.

"C combines all the power of assembly language with all the ease of use of assembly language"

sherifffruitfly

104

September 11, 2005 01:28 PM

Quote:Original post by NotAnAnonymousPoster
Quote:Original post by sherifffruitfly
False. A Riemann-integrable function's domain does not have to include an interval. Consider the function f, defined only at 0, such that f(0)=0. This is Riemann-integrable (everywhere continuous and differentiable even). But its domain does not include any interval.
It's a degenerate case, where definitions often disagree. Still, I think most definitions of Riemann-integral will entail that f in this case is not Riemann-integrable.

The theorem you are thinking to justify otherwise will state that any function continuous on a closed interval is integrable on that interval, not that a function which is everywhere continuous is everywhere Riemann-integrable.

You can say that all you want - it's just false. Definitions do not vary on this to any substantial degree (and hence neither to the theorems). ROFL - you think there's substantial variance in exactly *what* the Riemann integral is? LMAO.

And there's nothing "degenerate" about my example. It's a perfectly fine continuous, differentiable, and Riemann-integrable function.

NotAnAnonymousPoster

286

September 11, 2005 02:11 PM

Quote:Original post by sherifffruitfly
You can say that all you want - it's just false. Definitions do not vary on this to any substantial degree (and hence neither to the theorems). ROFL - you think there's substantial variance in exactly *what* the Riemann integral is? LMAO.

Not for the Riemann integral, no. Which is why I said:

Quote:Original post by Me
I think most definitions of Riemann-integral will entail that f in this case is not Riemann-integrable.

Quote:Original post by sherifffruitfly
And there's nothing "degenerate" about my example. It's a perfectly fine continuous, differentiable, and Riemann-integrable function.

I wasn't clear there. I meant that your function f is defined on the degenerate interval [0,0].

But we won't get anywhere here without actually supplying some definitions. Try this one at PlanetMath:

Riemann Integral

The definition requires that the function in question be defined on a non-degenerate interval.

"C combines all the power of assembly language with all the ease of use of assembly language"

sherifffruitfly

104

September 11, 2005 02:55 PM

Quote:Original post by NotAnAnonymousPoster
Quote:Original post by sherifffruitfly
False. A Riemann-integrable function's domain does not have to include an interval. Consider the function f, defined only at 0, such that f(0)=0. This is Riemann-integrable (everywhere continuous and differentiable even). But its domain does not include any interval.
It's a degenerate case, where definitions often disagree. Still, I think most definitions of Riemann-integral will entail that f in this case is not Riemann-integrable.

The theorem you are thinking to justify otherwise will state that any function continuous on a closed interval is integrable on that interval, not that a function which is everywhere continuous is everywhere Riemann-integrable.

Moreover, the theorem I was thinking of was not the one you tried to putin my mouth. The one I was after was the one I actually said - that the Riemann-integrable functions are those which are continuous except on a set of measure zero.

sherifffruitfly

104

September 11, 2005 03:01 PM

Quote:
But we won't get anywhere here without actually supplying some definitions. Try this one at PlanetMath:

Riemann Integral

The definition requires that the function in question be defined on a non-degenerate interval.

Kindly point out *exactly*, by quoting, where in the PM definition it is "required" (required for what?) for an interval (what interval?) to be "non-degenerate" (what does non-degenerate mean?).

NotAnAnonymousPoster

286

September 11, 2005 03:29 PM

Quote:Original post by sherifffruitfly
Quote:
But we won't get anywhere here without actually supplying some definitions. Try this one at PlanetMath:

Riemann Integral

The definition requires that the function in question be defined on a non-degenerate interval.

Kindly point out *exactly*, by quoting, where in the PM definition it is "required" (required for what?) for an interval (what interval?) to be "non-degenerate" (what does non-degenerate mean?).

Looking again, it probably doesn't require it. A degenerate interval is an interval [a,a]={a}. A non-degenerate interval is not a degenerate interval. But this is moot anyway.

The PM definition defines Riemann-integration on intervals. You said that your function f is not defined on an interval, so its nowhere Riemann integrable by the PM definition.

"C combines all the power of assembly language with all the ease of use of assembly language"

sherifffruitfly

104

September 11, 2005 03:42 PM

Quote:
The PM definition defines Riemann-integration on intervals. You said that your function f is not defined on an interval, so its nowhere Riemann integrable by the PM definition.

Apparently you're under the massive misconception that [a,b] is not an interval when a=b. Well, it is. It is a closed, bounded, compact, finite, perfect interval. I'm sure there are a number of other qualities such intervals have as well.

Moreover, it pretty much *has* to be this way - if [a,a] weren't an interval, then all sorts of topological and lattice-theoretic things would f^ck up (closure under arbitrary intersections being the most obvious).

There's absolutely nothing degenerate about my example, and my theorem was correct as I originally stated: f Riemann-integrable = f continuous but at a set of measure 0.

sherifffruitfly

104

September 11, 2005 03:48 PM

Quote:ervals

. You said that your function f is not defined on an interval, so its nowhere Riemann integrable by the PM definition.

rofl - and although it must be blasphemous to someone who thinks as you do, the empty set is also an interval.

NotAnAnonymousPoster

286

September 11, 2005 03:49 PM

Quote:Original post by sherifffruitfly
Apparently you're under the massive misconception that [a,b] is not an interval when a=b. Well, it is. It is a closed, bounded, compact, finite, perfect interval. I'm sure there are a number of other qualities such intervals have as well.

Moreover, it pretty much *has* to be this way - if [a,a] weren't an interval, then all sorts of topological and lattice-theoretic things would f^ck up (closure under arbitrary intersections being the most obvious).

There's absolutely nothing degenerate about my example, and my theorem was correct as I originally stated: f Riemann-integrable = f continuous but at a set of measure 0.

Great. Now can you retract your original objection, and accept that your new position is in contradiction with it:

Quote:Original post by sherifffruitfly
A Riemann-integrable function's domain does not have to include an interval. Consider the function f, defined only at 0, such that f(0)=0. This is Riemann-integrable (everywhere continuous and differentiable even). But its domain does not include any interval.

Your function's domain is and includes the interval [0,0].

"C combines all the power of assembly language with all the ease of use of assembly language"

Formal definition of the integral?

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