I knew you also have a phd in integral, not just in pde.
@sofianeafra6161
5 жыл бұрын
He had a pH.D in integrals ?
@drpeyam
5 жыл бұрын
You’ll be my first PhD student, haha
@blackpenredpen
5 жыл бұрын
Dr Peyam deal!
@sofianeafra6161
5 жыл бұрын
@@drpeyam okey i will be but i want to get a pH.D in complex analysis
@sofianeafra6161
5 жыл бұрын
@@drpeyam in multivariable calculus in double integral we have x=rCos∅ and y=rSin∅ and the jacobian is r but when we have a triple integral what's z ?
@ninck8992
5 жыл бұрын
I suggested this theme in the lebesgue integral video, and i'm so glad you did it. Thanks Peyam, you're awesome!
@user-zw5jj2uf1p
2 жыл бұрын
This guy looks so excited to explain the gauge integral. I am just as excited to learn!
@loicetienne7570
2 жыл бұрын
Glad to see this kind of integral by example. It really helps understanding what is going on. Thanks.
@qubix27
5 жыл бұрын
There should not be "d", because δ(x) is already playing a role of dx - an interval on x-axis (or something that bounds this interval)
@redknight344
5 жыл бұрын
Yesssssssss! I was looking for a video of this integral! Just what I needed!!!
@fantiscious
2 жыл бұрын
2:09 "You won, you too..." Congrats to the 2 who won
@Aviationlover-belugaxl
5 жыл бұрын
Yes I’ve been wishing this vid existed so that I could Better understand this integral!
@ibrahinmenriquez3108
5 жыл бұрын
6:06 best phrase ever. It made me laugh a lot
@beaverbuoy3011
Жыл бұрын
Thank you!
@PackSciences
5 жыл бұрын
I discovered KH integrals with the book "Intégration, de Riemann à Kurzweil et Henstock", I recommend it for those who want to explore further the different definitions of the integral and jauge theory.
@imgayasheck595
5 жыл бұрын
Thank you for another video!
@Czeckie
5 жыл бұрын
nah, thanks. I'm gonna stick with Lebesgue integral since L^p spaces are well behaved. The functional analysis based on gauge integral doesn't probably exist, since the space of gauge-integrable functions is not even Banach. (or there's no known natural topology to make into one)
@tobybartels8426
5 жыл бұрын
You don't have to choose. The Lebesgue integral is recoverable from the gauge integral; a function is Lebesgue integrable if and only if its absolute value is gauge integrable (and then the value of the integral is the same). Sticking to the Lebesgue integral is like sticking to absolutely convergent series: sometimes you want to restrict your attention to those (and one reason is that the space of such functions/sequences is a nice Banach space); but at the same time, sometimes you want to integrate/sum something that isn't absolutely convergent, and it's nice to know how to do that too.
@perappelgren948
5 жыл бұрын
”...and _maybe_ take a limit of it...” 😊😊😊
@motmot2694
5 жыл бұрын
You deserve more subs!
@chasemarangu
5 жыл бұрын
neat video just to confirm all three Riemann, Lebesgue, and Gauge integrals are all rectangle sums but riemann is vertical rectangles, lebesgueis horizontal, and gauge has different widths?
@drpeyam
5 жыл бұрын
Basically
@frozenmoon998
5 жыл бұрын
Besto integral video :)
@marstruth1578
5 жыл бұрын
Great topic. Let’s not forget the Daniell Integral.
@Demki
5 жыл бұрын
28:31 isn't sin x/x riemann integrable in [0,1] ? You could define f(x)=1 if x=0 and f(x)=sin(x)/x otherwise and you'll get that f is continuous, and thus riemann integrable.
@marcelojimenezdavila7805
5 жыл бұрын
Please , extend the tetration to reals , my english IS very bad
@dgrandlapinblanc
5 жыл бұрын
Sincerely it's very hard but it seems interesting. Thanks !
@jimnewton4534
Ай бұрын
I'm confused by your definition at 11m55s. You're defining the integral of f(x)δ(x) by saying for every ε>0 there exists a δ... BUT you already have a δ. Isn't there something wrong with the definition? Do you have two different δ symbols meaning two different things?
@Vladimir_Pavlov
5 жыл бұрын
t=1/x^3 ........ integal 1/3*(sint/t) from 1 to ∞
@nds142
5 жыл бұрын
Can't I do the first integration by the means of substitution followed by byparts integration? As answer is coming from that too.
@eliyasne9695
5 жыл бұрын
What if you made an integral that would sum infinitly many curved sections each of which is rimman integrable instead of summing infinitly many rectangels?
@azzteke
2 жыл бұрын
infinitEly & RIEMANN please!
@xy9439
5 жыл бұрын
Why do you set F(0)=π/2 and not something else?
@BrainsOverGains
5 жыл бұрын
Maybe it's the limit as x goes to zero? Im not sure myself
@xy9439
5 жыл бұрын
@@BrainsOverGains It must be that bjt since he didn't check...
@drpeyam
5 жыл бұрын
It’s the limit as x goes to 0
@mildlyacidic
5 жыл бұрын
Would this be more useful than using residues to compute some complex real integrals, or would that just depend on the situation?
@ListentoGallegos
5 жыл бұрын
It looks like you're making a marked partition.
@Ricocossa1
5 жыл бұрын
Very beautiful! Thank you very much. Do you know if there's a way to express this in the language of differential forms? Perhaps defining the gauge as some volume form, and demanding convergence? How do you know the gauge integral is unique?
@drpeyam
5 жыл бұрын
Not sure :) Given delta, it’s unique, but different deltas give different integrals
@Strikingwolf2012
5 жыл бұрын
@@drpeyam Different deltas actually do not give different integrals because the particular gauge is not mentioned in the definition! I will give a short proof here that the gauge integral is unique, justifying the use of the notation int f(x) delta(x). So. Suppose I1 and I2 are gauge integrals of f. I.E. For every e > 0 we have gauges d1 and d2 such that for all d1-fine partitions P1 and for all d2 fine partitions P2 we have |Sum(P1,f) - I1| < e and |Sum(P2,f)| < e Now let e > 0, we know that d1 and d2 exist for e/2. Let d(x) = min(d1(x), d2(x)). Clearly if P is a d-fine partition it is also a d1 and d2 fine partition. Such a partition P is guaranteed by Cousin's theorem (en.wikipedia.org/wiki/Cousin%27s_theorem). | I1 - I2 | 0 Thus I1 = I2. This is made a bit confusing by the fact that your notation in the video uses the same notation as that for a gauge!
@douro20
4 жыл бұрын
BTW Denjoy was French...
@FunctionalIntegral
5 жыл бұрын
Ok, the next step is multidimensional gauge integral haha
@curtiswfranks
5 жыл бұрын
Oh, I went to school with Boming!
@drpeyam
5 жыл бұрын
Wow!!! What a coincidence! His notes are excellent!!!
@iabervon
5 жыл бұрын
Is the definition implicitly excluding partitions where the gauge sum doesn't exist? I don't see how you can pick a gauge function that keeps partitions from picking an undefined point as a t_i.
@redknight344
5 жыл бұрын
www.math.uchicago.edu/~may/VIGRE/VIGRE2006/PAPERS/Herschlag.pdf you choose those tags t_i.
@md2perpe
3 жыл бұрын
When will you introduce the gay integral?
@drpeyam
3 жыл бұрын
LOL
@tobybartels8426
5 жыл бұрын
The main example here is _improperly_ Riemann integrable, so it's not that crazy. (ETA: And of course, the other example is Lebesgue integrable.) If you want an example that's neither Lebesgue integrable _nor_ improperly Riemann integrable, then let f(x) be 1/x sin(1/x^3) when x is irrational or 0 when x is rational. This is not Riemann integrable on any nontrivial interval, so not improperly Riemann integrable either, but it's also not Lebesgue integrable on any interval with 0 (as an endpoint or as an interior point), since the integral of the absolute value is infinite. Yet it _is_ gauge integrable on every interval (with the same values as the function in the video).
@drpeyam
5 жыл бұрын
Nice!
@tobybartels8426
5 жыл бұрын
As for sin(x)/x, that's _properly_ Riemann integrable (and so also Lebesgue and gauge integrable), since the discontinuity is removable (the limit is 1).
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