1) Riemann never intended to use his definition for calculations. This is one of those typically theoretical definitions, very generell and useful for existence proofs and such. 2) In 1868 Riemann was already 2 years dead. He found his legendary definition around 20 years earlier as a young student in Göttingen.
I think the important thing about the reiman integral is it "feels" more general and natural. So proving that the Darboux integral is well founded and equivalent to the reiman integral is an important logical step, even though once the proof is done, you dont ever want to look at it again. The knowledge that they are equivalent is in itself valluable.
It's historically interesting, but I don't think it's any more natural. "The infimum of all upper sums" seems like a clean and straightforward definition, compared to a limit as the norm of some infinitely-dimensioned space of partitions and tags approaches zero. An upper sum is independent of "tags" entirely and only depends on the partition, which makes it much more natural to reason about. The fact that you often want to use some limit to bound it from above falls out naturally as you try to compute the value by sandwiching it, instead of being baked into the definition.
Riemann's definition generalizes to the Gauge Integral, where the max width δ isn't the same for all intervals, but depends on the tag, so it's really a function δ(x), called a gauge. All Lebesgue- and indefinite-Riemann-integrable functions are gauge integrable. It has apparently found use in evaluation of some path integrals in Quantum Mechanics.
when you understand the construction of the riemann integral, you understand the construction of the lebesgue integral when you yearn the basics in measure theory
We learn Riemanns definition of integrals in Calculus 1 mate... And then immideately learn Darboux integral. Then in any further analysis course, you learn about Lebesgue integrals.
Maybe highschool, but not in uni. My analysis class taught the Riemann integral first, then the darboux integral. We have some useful theorems to prove, almost never used original Riemann definition, just like you said.
There's a misprint on 15:12 - there must be "4" instead of "3" in the denominator of an expression for sum of cubes. I actually learned both integrabilities in university and can say that this is a really cool and interesting video! Keep it up 👍
Very nice video! To my mind, a strong advantage of the Darboux variant is that you need only check one specific sequence of partitions. In contrast to that, in the Riemann variant you had to keep the partitions (and tags) arbitrary. Just wanted to mention this since you used it (of course) but didn't actively point it out.
Defining integrals based on a definition would be a bit unmathematical anyway, as you always want to use an axiomatisation instead. Darboux has the first rigorous definition, equivalent to the following three/four properties. 1. The integral of a constant function is... 2. The integral of f(x) from a to b plus from b to c is the integral of f(x) from a to c. 3a. If f(x)≥0 for a≤x≤b, the the integral of f(x) from a to b is greater or equal to 0. 3b. If f(x)≤0 for a≤x≤b... The cool part about axiomatisation with inequalities is that you don't have any limits. I guess it cool, but you would want to use limits in practise anyways, since you basically only care about cases of equality.
Describing constructions as “unmathematical” is a bit silly for a variety of reasons. Tons of things in math are described via constructions, and integrals are certainly one of them. There are pros and cons to both approaches. For example, people often define the real numbers to be “the unique complete ordered field”. This is all well and good until you realize that you have no idea if such a thing exists. Indeed, in order to rigorously prove that this definition is usable, you need to A: Construct a model of the reals. B: Show that it is a complete ordered field. C: Show that all complete ordered fields are isomorphic to it. This is just straight up objectively more work than constructing the real numbers, which is the step in part A. A similar thing happens for most axiomatic definitions: To create an axiomatic definition you need to first list the axioms and then prove that they have a unique solution, but the fastest way to do that second part almost always involves constructing the thing anyway, so most mathematicians don’t bother. I realize that I said there were pros and cons but have listed only cons, so just for the record, the reason some mathematicians like axiomatic definitions is that there’s a certain conciseness to them that makes them “prettier”. Like, “the unique complete ordered field” is a much prettier definition than [insert your favorite construction of the reals], even though it takes more work to prove that the definition works.
There is flexibility in Riemann integral when integrating from the definition. For instance you can integrate sqrt(x) from 0 to 1 by choosing x_i = (i/n)^2 for i = 1, ..., n as the partition points. Now, Δx_i = (2i + 1)/n and sqrt(x_i) = i/n. So, this non-regular partition results in a sum that can be evaluated. The Darboux definition on a regular partition results in evaluations points for which the sum cannot be evaluated directly because we have no general exact formula of sum of sqrt(i).
Actually you have it backward, the Darboux integral does not require uniform partitions and indeed it is the Darboux integral that allows you to check just a single convenient partition.
Darboux integral is solvable with any partition of choice while Riemann requires you to show that all partitions work. However there are theorems that allow you to take a limit as the number of intervals goes to infinity of say for example a uniform partition.
I've actually learnt about both Riemann and Darboux integration, although the arbitrary partitioning method was attributed to Darboux, not Riemann. We've also formalized it using the Lower and Upper sums and their convergence to the same number from the get go. Never seen the arbitrary rectangle point/height method before.
Newton and Leibniz did _not_ invent the idea to approximate the area by rectangles and refining the approximation by using more and more rectangles. That idea existed centuries, if not decades, before them. E. g. Fermat, Pascal and Cavalieri, among many others, used that idea extensively.
That is not what they said though, "(...) and the basic idea of partitioning random shapes into rectangles and triangles *has existed since ancient times* , however *the integral as we know today* was invented by Newton and Leibniz, which uses rectangles to approximate the area (...)".
Also, I remember learning that Cauchy did it before Riemann, but his idea was to have arbitrary partitions but with a fixed tag on the leftmost point of each subinterval. Later, someone proved Cauchy's and Riemann's definition are equivalent, so the tags aren't adding anything of value. Very interesting. All in all, Darboux will usually be the best one for working with proofs.
@@LinaWainwright Huh? So it says that the idea to approximate the area with rectangles was invented by Newton and Leibniz, just as I said. And as I said, that's wrong, the idea had been used by others before.
@@bjornfeuerbacher5514 the idea did exist but those people did not apply in the same vein. It's no wonder squaring a circle wa seen as impossible until they went and did it, because of said ideas.
I don't know where you're coming from saying we probably never learned the Riemann integral in college. Every elementary calculus textbook I've ever seen teaches some variation of the Riemann integral. By contrast, I've never seen Darboux's definition in anything short of an upper-level text.
Quite right 👍 the Darboux version is technically easier and more natural to prove results with. For first year calculus it won't matter but later on it's preferable to use Darboux.
At around 12:20, you switch from max/min to sup/inf. This is good to handle the problem you pointed out, but opens a new problem: sup and inf can be infinite, so what if there is a vertical asymptote? For example, if you want to find the integral of log(x) between 0 and 1, how would you find the limit of L(f,P) as ||P|| tends to 0? Since one of the intervals will be sure to contain the asymptote, then for that partition inf[f(xi*)] will always be -infinity, which would make every L(f,P) undefined, no matter how you partition it.
Riemann/Darboux's original notion of integrability was not meant to tackle such issues. Those are resolved using the definition of improper Riemann/Darboux Integrals, which is taking the limit of the integral as a or b approaches such asymptote
So in my Analysis class we did start with the upper/lower integral versions, but also saw the Riemann version and proved it's equivalence, so the title did confuse me
yeah i dont know what you are talking about, we learned this at the community college in Calc2 and had to use regular Reimann and the trapezoidal method.
Nah, mate, I actually had the Riemann definition in my Calculus 2 course. You are just tripping. The sad thing is that there was no time to go into the other ways of integrating.
@@HaramGuys All he did to make the Riemann integral harder is not take a uniform partition like he did in the second example. That simplification has nothing to do with the differing definitions and works perfectly well with the Riemann integral too. He even mentions taking the left or right endpoints which is all he did in the second example for L(f,P) and R(f,P) as x^3 is monotone increasing.
@@abebuckingham8198 no, because you need a general proof for the riemann definition. It was okay in the second part because that was explicitly for the upper and lower bounds. But the original riemann definition requires *for all partitions with norm ||P||* . Hence the Darboux definition is better because it lets you simplify the Riemann definition
@@icodestuff6241 You just prove that if it converges for one that it converges for all of them. This is very easy to do in full generality. That's what motivates the U(f,P)-L(f,P) definition in the first place.
@abebuckingham8198 Sorry, but your statement is just false. Take, for example, the integral of 1/sqrt(x) from 0 to 1. The function is not integrable, yet the right-riemann sum converges. The proof is as follows. Taking N partitions we have, we get dx = 1/N, and x_n = n * dx = n/N. Thus the right riemann sum is S[ 1/sqrt(x_n) * dx ] = S[ 1/sqrt(n/N) * 1 / N] = S[ 1 / sqrt(N * n)] where S represents summing from n = 1 to n = N. Since N is a constant, we can pull it out, giving 1/sqrt(N) * S[ 1 / sqrt(n)]. S[ 1 / sqrt(n)] is bounded by 1 + 2sqrt(N) which can be proven by integrating 1/sqrt(x) from 1 to N. Thus, 1/sqrt(N) * S[ 1 / sqrt(n)] < ( 1 + 2sqrt(N) ) / sqrt(N) = 2 + 1/sqrtN Therefore, as N goes to infinity, the right-riemann sum converges to a value less than or equal to 2.
For Riemann, it is not true that you have to know the result in advance. In case of convergence, every single sequence of tagged partitions converges to the same value, so you can just choose one that's convenient to determine the value. Note that you can do that *before* proving convergence because all you do at this step is to say *if* the integral exists, *then* it has to have this value; you then in the second step use that value to prove that the integral actually exists. I'm also not sure that Darboux integration should be considered as a separate type of integral rather than as a method of evaluating Riemann integrals.
The failure of the Riemann integral for functions like the one showed that had one constant value for rationals and another for irrationals is a classic and fun one. There’s an interesting pedagogical/thought experiment that points toward more advanced treatment of the integral (Lebesque, eg) by way of “rescuing” the Riemann integral in that function by taking the sampling step of selecting tags in a “statistical” way. So in each sub interval, you _could_ pick a rational. But if you were to choose randomly, you never would, since rationals are infinitesimally rare compared to irrationals. The probability of selecting a rational from a finite real interval is zero. So that tagged partition with all rational tags can be ignored since😊 it basically would never ever be picked by random selection. The probability of picking that from all the possible samples is zero. It’s a handwavy, “physics class math” sort of explanation. And to make it rigorous you’d have to reinvent the same ideas of measure theory anyway. But pedagogically I think it’s an interesting bridge.
Did the formal definition of the integral not come before that of the derivative? Jacob Bernoulli created the definitions and even the notation, namely the large S standing for sum, by himself in the late 1600s. Then the derivative was later created or formalized by the likes of Euler of Cauchy. I'm not sure if I am entirely correct, but I do believe the integral came first.
Newton and Leibniz definitely used differentiation as well, since their notation has survived to the present day. (Leibniz used the familiar df/dx, while Newton put a dot over the dependent variable, which I can't get on my phone's keyboard.)
The large S was invented by Leibniz, not by Bernoulli. Bernoulli invented the _name_ integral. But both did not really give a _formal_ definition. After all, the whole concept of limits was only rigorously established long after Bernoulli.
@@tomkerruish2982 Yes, they used it - but they never rigorously _defined_ what this actually means. They kept talking about "infinitesimals" without specifying what that is actually supposed to mean.
@@icodestuff6241 Well, Euler's approach was not really "formal" and/or "rigorous" in the modern sense - he did his calculations by freely using an "infinitely large" and an "infinitely small" number... But I agree on the others.
Interesting. I actually learned the Riemann Integral in University at 2001. And I wouldn't say it's useless, if the calculation may contain a limited amount of errors. You see, if you have a sampling (a value recorded for timestamps), you might want some sort of "integral" of that, but that input is not even continuous. What's useful is not the formulation of the Riemann Integral itself, but the idea behind it (take those rectangles and their area up, and you'll have a sum with some error estimation).
When you say that you already need to know the answer, I don't think that is an impediment. At least in the example you showed you can find the integral of polynomials just using the summations formulas.
You need Riemanns Definition for the proof of the Linearity of the Integral. Or for the Minkowski Inequality. Or even much more important for the Fundamental Theorem of Calculus. This should be enough to state the need of Riemanns view.
This is great and interesting man.. But can you make a video or two on Lebesgue integral? Since it is subtle and blends sets, measures and calculus ideas unlike others.. I am still looking for good videos on these topics!?
The notation ∫f(x)dx = ∫f(t)dt = ∫f(u)du clarifies that we are choosing to integrate the function over some specific choice of independent variable. ∫f is 1800s mathematicians saying it doesn't matter what letter we use, it's all the same anyway, so lets be lazy
In analysis, it's sometimes conventional to just write the integral without a "dx", since for single-variable functions, it not only makes no difference, but also can't be rigorously defined. You only get to a rigorous definition with the theory of "differential forms", which comes much later!
As many others I also did learn both definitions in school so the title does not apply to me. I also do not agree with the Darboux integral being overall better - try proving linearity... for Riemann it is a piece of cake.
You say we often tag the rightmost or the leftmost point on the interval, but Darboux integrals, which you claim is most often the taught variation, tags the minimum and maximum of the function on each interval, which is a variation you don't really mention in the intro.
The Darboux integral is the one that's most often taught *in analysis courses*. The version where you tag the rightmost/leftmost point is the one that's most often taught *in introductory calculus courses*. Hopefully that clarifies things...
Okay. Let's make something clear. Sure, int_(a)^(b)(f(x)dx) finds the area under the curve of y=f(x) from a to b, but it also finds the length of the curve y=int(sqrt(f(x)²-1)dx) from a to b
Hi EpsilonDelta, appreciate the effort put in to this video. Would you be interested in new video topic suggestions? What would be the best email address to contact you at?
Many people appraised the video but I kind of disagree. For constructive criticism, I think you should not delve too much into "deeper" topics without explaining concepts or ideas. For example, you have mentioned tat irrational numbers are a dense subset of real numbers at 5:48 . Yes, for someone who knows some higher level math this might be a better clarification but for the general viewer I think this video is intended to, I think it just causes a ton of confusion.
All you did was use an arbitrary partition for the first example and a uniform partition for the second because as you said, it's easier. It's an artificial complication.
Not an artificial complication. Riemann's definition is the way it is because he wanted to make a definition (any partition and any tags) that is as general as possible to be able to tackle all pathological fringe cases. Darboux's improved definition requires you to only provide any reasonable upper bound on the difference between upper and lower integrals, and the regular partition just happen to be a convenient choice for a bound for a specified function like x^3. but definition itself still assumes arbitrary partition It was much later proven that even if you relax the condition of the definition of riemann integrability to sequence of regular partition (but still arbitrary tag) to converge, then it is equivalent to the Riemann's definition. This is a highly nontrivial theorem that takes quite a bit of work to prove, but guess how its proven?? using Darboux's definition
@@HaramGuys All he did was apply the squeeze theorem to the Riemann definition. The fact that if it converges for one sequence of refined partitions it converges for all of them is very easy to prove in full generality. It's what motivates the U(f,P)-L(f,P) definition in the first place.
