improper integral of x from neg inf to inf

If the improper integral of x from negative infinity to positive infinity intrigues you, then I think you will like Brilliant. Check them out here via brilliant.org/blackpenredpen/
Learn more improper integrals: brilliant.org/wiki/improper-i...
Here’s another in a similar situation • Improper integral of 1...
Here's an integral from -inf to inf that gives you 0: • improper integral of x...
0:00 we will investigate the improper integral of x from neg inf to inf
0:16 how to deal with an improper integral with both neg inf and pos inf
6:08 real analysis Cauchy principal value of this integral
7:58 check out Brilliant
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[@irinaignatova1799]

During my first semester (real analysis), I always felt uncomfortable with this improper integral because we were tought that it diverges, although it is so intuitive that the negative part and the positive part cancels outs Later that year, during a multivariable calculus, we dealt with limits in R^2. And what I learnt about limits is that in order for the limit to exist, you should show that all paths leads to the same answer. In R^2, that means that a function had a limit as (x,y)~>(0,0) if all the ways to approach (0,0) gave us the same answer for the limit. That meant that for example, •if you approach (0;0) by going from the top to the origin (for example take U_n=(0; (1/n) )) • if you approach (0;0) by going from the left to the origin (for example take U_n=(-(1/n) ; 0)) •if you approach (0;0) by going along a parabola to the origin (for example, take U_n= ( (1/n) ; (1/n)^2 ) •if you approach (0;0) by going along an oblique line (for example, take U_n = ( (1/n) ; (1/n) ) •and so on, you can image all kinds of complicated paths Then lim f(U_n) as n~>inf should all lead to the same answer in order for this limit to exist. And if that is the case, we say that this value is equal to lim f(x,y) as (x,y)~>(0,0) (it’s value is unique since all our results are the same) So if at least one of these paths leads to a different value for lim f(x;y) as (x,y) goes to (0,0), we can’t get an answer that everyone will agree with so we just say that the limit doesn’t exist. In R, the same principle exist, except it’s trickier because the only ways to approach a point is by going left on the x axis or going right on the x axis In R, only one parameter is moving But in in R^2, if there’s different parameter: •The x component can move at the same speed as the y component (line) • The x component can move at a speed that keeps increasing whereas the y component moves at a constant speed (curved graph) •and so on So in R^2, their relative speed impacts the path which leads to more interesting ways of approaching a point Our original integral (let’s call it J), is a limit of something We can express it as : J= The limit as (x;y) goes to (-inf;+inf) of integral from x to y of t * dt If we consider U_n = (-n ; n) +inf is (-inf;+inf) So U_n describes a path that leads in the input space to (-inf ; +inf) (which is what we want) Hence with this path, we should get the value for J Integral from -n to n of t * dt = 0 cause t is an odd function. Hence limit as n~> +inf of [Integral from -n to n of t * dt] = J So 0 = J We found the value for J !! But if we look using a different path... Let U_n= (-n ; 2n) +inf = (-inf ; +inf) which is the point we are interested in (input space). Integral from -n to 2n of t * dt = Integral from n to 2n of t * dt = 2n^2 - (1/2)n^2 = (3/2)n^2 And lim of (3/2)n^2 as n~>+inf = +inf So now, we find that J= +inf And if we consider another different path: U_n = (-(n +1) ; n )

[@blackpenredpen]

😮 this is amazing! Thanks for a detail comment.

[@irinaignatova1799]

@@blackpenredpen it’s far from being rigorous (and there may be mistakes in my reasoning) but I hope it can help others get the intuition behind why it’s divergent And thank you for the pin!!

[@ZipplyZane]

It doesn't really help me, as I don't see any reason why you shouldn't always assume that y and x move at the same velocity given that the equation is y=x. Sure, if we were talking about y=x+1 or y=3x, then you have things moving at different velocities. But not at y=x.

[@irinaignatova1799]

@@ZipplyZane I’m sorry, I can’t grasp what issue you are raising If y=x, then both component move at the same speed (I agree with you) But if you’re given no info, your path could be very different, so you’ll have to consider a general path instead of a precise one (like the y=x path) and show that the answer we get is the same independently of the path we choose If I’m not answering your question, please indicate me which part of my comment you were referring

[@ZipplyZane]

​@@irinaignatova1799 I suspect it's just that the math you are talking about is something I am completely unfamiliar with. I don't see intuitively why I can't assume y=x, as the problem we were given involved integrating x dx. Why can't I assume that the velocity of x and y are the same in that case? It just seems to me that, if lim [n->inf] of integral [from -n to n] f(x) = 0, then integral [from -inf to inf] f(x) should also be 0, assuming f(x) is continuous. If a function is continuous, then lim [a->x] f(a) = f(x).

[@AndrewDotsonvideos]

I’d love to hear your take on renormalization 😅

[@blackpenredpen]

I actually don’t know about it….

[@bornoflightningthunder8162]

Wow Andrew it's really you!

[@trueriver1950]

@@blackpenredpen A lot of particle physics and many cosmologies depend on it. You basically get equations that are infinite on both sides but being physicists we subtract off those pesky infinities...

[@angelmendez-rivera351]

@@trueriver1950 True, but also misleading. Renormalization in physics is sometimes necessary even when there are no singularities involved, and this can happen when accounting for self-interactions in Feynman diagrams.

[@zil1832]

@@angelmendez-rivera351 but what can be self interacting? It seems like an anomaly?

[@MonsterIsABlock]

I love how in some parts of this video, he's talking about advanced mathematical concepts, but in 0:19 he's teaching us what a definite integral is, like, we all know that definite integrals are what disintegrated our math grades.

[@AliKhanMaths]

Nahh there's no +C with definite integrals so at least I don't lose marks for that 😂😂

[@anshumanagrawal346]

@@AliKhanMaths same lol

[@TheTalmon18]

Integrals and series are in the same course o-o It's not really more advanced. But yes, he explains things well :)

[@microscopicallysmall]

indefinite integrals give you a C if you forget the +C definite integrals give you a 0% on the test if you forget to evaluate it at 0 if one of its bounds are 0

[@MrQwefty]

Yeah, if you integrate from -a to +ma, when a approaches infinity, then you can get the integral to be +-infinity or 0 easily, when you make m between 0 and 1, exactly 1, or greater than 1. That's basically the definition of divergence - that there's no one limit.

[@thisismycoolnickname]

To be fair, this is a problem of definition. For example, if we write: lim [n->inf] of integral [from -n to n] of x dx That limit would be equal to 0, because both of those infinities are the same variable. So it really depends on how you define a double infinity under and above the integral sign. And that definition seems to be quite arbitrary. It would be very interesting to see if this problem actually pops up in any real life situation and which answer fits.

[@karansandhu4827]

I think your comment just sums up the whole thing.

[@user-nv4id1hq2t]

он так и сказал. в смысле главного значения интеграл сходится, а просто - хуюшки. чтоб отличать на письме, перед значком интеграла нас учили писать "v.p.". ну или не писать. если буковок нет, то ответ "расходится". если есть, то ответ "ноль".

[@gasun1274]

nope it won't. one approaches positive infinity and the other approaches from the opposite side. they are two different limits.

[@user-nv4id1hq2t]

@@gasun1274 вот именно, два разных.

[@Ric4562]

@@user-nv4id1hq2t Ты уверен, что тебя понимают?

[@angelmendez-rivera351]

There are a few things we need to parse out and understand here. 0. It is true that, if you integrate the identity function on [-r, r], and you let r -> +♾, then the limit is 0. 1. The above point is not actually relevant, though. The problem here is that, intuitively, people are assuuming that we are integrating on the symmetric interval [-r, r], but we are not. We are integrating on the interval [u, v], and then letting u -> -♾, v -> +♾. If the integral truly is well-defined and is 0, then evaluating the latter limit should result in 0, because it would be independent of the restriction u = -v, and limits do not exist if they are dependent on external restrictions. We do not get 0 as the result, though, as the video has shown. Thus, the integral does not exist. 2. So, with that being said, why do we tend to think that it is correct to simply have the integral on [-r, r] with r -> +♾? There are three reasons for this: A) because the notation is misleading and it visually suggests this, what with the symmetric considerations between -♾ and +♾; B) because in a calculus II course, you are not equipped to evaluate a limit in 2 dimensions, and are as such not equipped to evaluate the integral on [u, v] with u -> -♾, v -> +♾; C) because calculus courses are generally very unrigorous, and you are not much taught how to prove an integral exists, or even how an integral is defined to begin with; instead, you are taught to assume the integral exists, and symbolically compute it under that assumption, and once you have knowledge of the existence of the integral, there are no issues with using the restriction u = -v to compute it; whether the integral exists or not is an entirely different story, and attempting to compute an integral that does not exist will result in logical contradictions, as many comments in the comments section have shown by example. 3. With all of the above being said, I know there are people out there who question or reject the claim that the the limit of the integral on [u, v] as u -> -♾, v -> +♾ does not exist. We should go into more detail here. For starters, it is important to realize that the integral on [u, v] is simply (v^2 - u^2)/2. One may argue that the "infinities cancel out", but this is not the case: if u = 1 - v rather than u = -v, the limit is clearly not 0. It may also be easier to understand via a change of variables: u -> -♾ simply means 1/u -> 0, u < 0, which implies 1/u < 0, and v -> +♾ simply means 1/v -> 0, v > 0, which implies 1/v > 0. So we can let s = 1/u, t = 1/v, and so lim (v^2 - u^2)/2 (u -> -♾, v -> +♾) = lim (1/t^2 - 1/u^2)/2 (s < 0, t > 0, s -> 0, t -> 0). Now it should become obvious graphically that the limit does not exist. 4. While still on the vein of "infinities cancelling out", one can argue intuitively that the areas cancel out due to symmetry, but this fails to be a logically sound argument for the same reason point 0 is not a logically sound argument: it is not relevant. We are working with a limit of integrals, not an integral itself, so areas do not come into this, even if graphically and geometrically it may seem like they do. One should also remember that areas are defined in terms of integrals, not the other way around. 5. The Cauchy principal value is 0, and this seems to be "the answer in some sense, though not in the standard sense". But even this is not true. To understand why, one needs to look at what the Cauchy principal value is in a more general context, outside of just looking at integrals. The Cauchy principal value is a linear operator on functions, and what this operator does is it replaces the isolated singularities of a function with the "average" of the function near those singularities. To say that the Cauchy principal value "of the integral" is 0 is to say that the Cauchy principal value of the function g on [R\{0}]^2 as defined by g(s, t) := (1/t^2 - 1/s^2)/2 at the singular point (0, 0) is 0, and what this means is that the "average" value of g near (0, 0) is 0. This does not make the value of g at (0, 0) actually 0, though. g(0, 0) does not exist, it is undefined, because (0, 0) is a singularity of g. So it is inadequate to say that the Cauchy principal value gives us "the correct answer" to the "improper integral" being discussed here. It is healthier, more accurate, and more interesting, to understand the Cauchy principal value as a tool for extending functions so that they are defined everywhere, turning non-removable singularities into non-removable discontinuities in a way that is not arbitrary, but rather, natural, and useful in applications. I will go into more details about how the Cauchy principal value works in the replies to my own comment.

[@angelmendez-rivera351]

For a function f on R\{s} that is sufficiently well-behaved with a singularity at s, the Cauchy principal value of f at s is given by lim [f(s + ε) + f(s - ε)]/2 (ε > 0, ε -> 0). This is what I meant by the average near s. For a function such as g, though, it is more complicated, since g is a function on [R\{0}]^2. You must integrate g((x, y) + ε·(u, v)) over all the possible unit vectors (u, v), and then let ε > 0, ε -> 0, and divide by the length of the interval of integration that results from parametrizing the contour integral. It works the same way when working with functions in the complex plane: after all, the complex plane is isomorphic to R^2.

[@robertveith6383]

*@ blackpenredpen* --- You left off the negative sign on the third line for what was supposed to be the "lim x ---> - oo" part. You corrected the absence iin the next line.

[@blackpenredpen]

Ah, yes, thank you.

[@elithanathileoathbound3772]

I think this reflects a general misunderstanding about infinity, namely that it is a number. In math Infinity eq Infinity so while it may look like an integral from -a to a it really isn't, it is a integral from a to b.

[@Magis-rt1lw]

It depends on the definition but the most "appropriate" one is Lebesgue. In this sense, the integral doesn't exists. But if you insist for the existence of this integral, then you have to let go of some of the familiar properties. For example, you can no longer split the integrals in any way you want, otherwise, you'll treat infinity as if it were a finite real number. Its a trade but for practical purposes and consistency, such as those in probability theory, the Lebesgue integral is used. A similar question to this is: how do we add all of the integers? If you'll define its "n'th Partial sum" as the sum from -n to n, then it would make sense but we are no longer allowed to interchange the order of the terms in the sum. Otherwise, the sum of all the integers would be 0 and at the same time 2. If you'll accept the strict ordering of the sum, then it will be fine. If a "mathematical definition or formulation" is consistent with a (physical) theory you are working, then you are only limited by the experimental results.

[@user-tn4qe9ne9j]

To be honest, If they gave me this question, I would use the second way to give an answer without hesitation. It seems the most intuitive.

[@erynn9770]

I would probably do the same, but this implies that you move left at the same "Speed" als you move the limit right. Setting the integration limits to -t and +2t would bei valid, top, since you also end up with infinity, but twice as fast to the right. The integral then would resolve to 1/2 (2t)^2 - 1/2 (-t)^2 = 3/2 t^2 and Infinity as limit...

[@angelmendez-rivera351]

The reason it is intuitive is because this actually correct to do for most integrals you have encountered. Why? Because those integrals can be proven to converge, and so for computational purposes, you can apply the second method without creating logical contradictions. But since f in R with f(x) = x is not Lebesgue integrable, the integral does not converges, and the second method does lead to logical contradictions.

[@thatkindcoder7510]

@@angelmendez-rivera351 This maths rabbit hole just ends no-where damn

[@angelmendez-rivera351]

@@thatkindcoder7510 Well, if we are talking about integrals, that is not entirely true. I would argue it ends with topology, measure theory, and differential geometry. It is in those contexts that we get the most generalized idea we can have on the subject and why things work the way they do in calculus I and II.

[@thatkindcoder7510]

@@angelmendez-rivera351 Aye thanks

[@TreeCube]

I think a more rigorous approach to showing that this integral diverges is to use a two dimensional limit: ∫[-∞,∞] x dx = lim (a,b)->(-∞,∞) ∫[a,b] x dx Then from there, you can show that the limit does not exist, therefore implying that the original integral diverges. Also, according to mathworld, the most common designation for the cauchy principal value seems to be writing "PV" before the integral, not above the equal sign, like this: PV ∫[-∞,∞] x dx = 0

[@angelmendez-rivera351]

Even more robust would be to understand that r -> +♾ is just shorthand notation for 1/r -> 0, r > 0, and r -> -♾ is similarly just shorthand notation for 1/r -> 0, r < 0. So in this case, we have a = 1/s, b = 1/t, with s < 0, t > 0, and (s, t) -> (0, 0). Now it becomes visually obvious, if you plot the Riemann integral on [1/s, 1/t] of the identity function, as a function of (s, t), that the limit as (s, t) -> (0, 0) does not exist, even with the restriction that s < 0, t > 0, so the improper integral diverges.

[@wassimtouyar7916]

The methode in the video is rigorous. If a part of the integral diverges then the whole integral diverges. There is no need to over complicate things and call it "more rigorous"

[@mtaur4113]

The truest safe integral is one where int |f| converges, whatever the domain. Weird things can happen with infinite cancellation. sin(x^2) has finite improper integral from 0 to infty, but only in the sense of integrating from 0 to M and letting M approach infty. If you integrated segments in a different order, you could get large positive or negative amounts. In theory you could devise a u-sub that would amount to doing the same thing. Any time you do improper integrals mixed with u-sub, there is an opportunity to flip the order of limit and integral, which can be invalid. Calc can be mean sometimes.

[@wiesorix]

The integral has to be undefined, otherwise we run into problems. For example, consider the function g(x)=x+1. Since this is just a horizontal translation of f(x)=x, the integral from -infinity to infinity of both functions has to be the same (if it exists). So the integral of g(x)-f(x) over R has to be zero, but on the other hand g(x)-f(x)=1, and the integral of the constant function 1 diverges to infinity. Or, if you try to calculate the integral of g(x) over R by taking the limit of t->infinity of the integral of g(x) from -t to t, you get infinity. But again, this is just a translation of f(x), so the integrals have to be the same (if they exist). Since we've found two contradictions, the integral cannot exist i.e. it's undefined.

[@Anthos147]

A beautiful word from Romania.Thank you for all your math classes.It's great to know so many mathematical expression,you really are a genius.Excuse my English.Good luck!

[@MathTutor1]

Well done. Great explanation!

[@danibarack552]

With finite ranges you can consider 0 the middle of the range, and the integrals to the left and right cancel each other out. With an infinite range ANY number can be the 'middle' since there are still an infinity if numbers to the left and right, so the same argument doesn't work

[@pladselsker8340]

Ahh, I see. Intuitively, I was thinking that infinity and minus infinity must have equal and opposite sizes, but that's not what the first written integral enables us to suppose. For example, you could integrate between -t and 2t, and you would end up with positive infinity as the answer. So I guess 0 is one answer to the first integral, but there are infinitely many. You could also integrate between -t and t+k for example, and you would end up with a real value that's different than 0 for k different than 0. Pretty cool, thanks for the insight.

[@arekkrolak6320]

I like the image at 5:55 with two infinities - a big one and a small one - this is some original maths indeed! :)

[@fun_engineering5710]

Great video :) but... which one is more correct? And which one we should use by default ?

[@blackpenredpen]

Depends on what class you are in. For example, before algebra 2, we answer sqrt(-9) as “no real value”. But after we learn the complex numbers in algebra 2, we answer sqrt(-9) as 3i

[@AliKhanMaths]

I really love how you don't just talk about advanced topics, but you make sure to explain them so that they're accessible for everyone. Your videos really inspire me to share my own maths content!

[@klementhajrullaj1222]

YOU HAVE MADE US DIFFICULTY WITH THIS BEARD THAT KEEPS THESE LAST TIMES, WHEN EXPLAINING THE EXERCISES! ...

[@zitianwang1205]

this is so damn enlightening thx！even as a phd student i still have that doubt since yrs ago but now its clear！

[@michaelspence2508]

The most intuitive way to understand this (which you touched on but only VERY briefly) is that there are many different infinities. Thus the problem itself is like saying, "exactly how tall is a human" or "which letter of the alphabet is the only one that's a vowel"

[@angelmendez-rivera351]

This is one way to put it, but far from the most intuitive. The issue here is in the symbol ♾. Personally, I take issue with the fact that the symbol ♾ is used in mathematical analysis at all, due to the fact that it is used inconsistently and is always ambiguous. Because the improper integral is presented with having the bounds -♾ and +♾, we visually get the suggestion that there is some symmetry here, and so the integral is 0. But this is inaccurate: 0) -♾ and +♾ are not even well-defined mathematical symbols; 1) in whatever sense they used, it is not valid to place them as bounds of an integral, even though we tend to not care about this in practice, and we abuse notation, because a mathematician would know what is being meant in context. So there is no symmetry here; 2) it is because of the symbol ♾ that the impression that there is "one infinity that is a unique, well-defined object" occurs. The answer to this problem becomes obvious once you become transparent with the viewers and stop exploiting mathematically incorrect notation to blow their minds. Peeling back one layer of this, we realize that the expression being discussed is not an integral itself, but actually, a limit: we evaluate the integral on [u, v] of the identity function. From the fundamental theorem of calculus, we get (v^2 - u^2)/2, and what we are doing is evaluating lim (v^2 - u^2)/2 (u -> -♾, v -> +♾). So far, everyone can agree that this is the case. The disagreement and confusion is in the fact this limit seems to be equal to 0, but it is not 0. Here is where your explanation ceases to be intuitive: even if a reader realizes that "there are many infinities", it does not change the fact that the symbol ♾, as used in this limit, suggests that, by symmetry, the limit should be 0. Also, the suggestion that there are many infinities implies that ♾ stands for many different mathematical objects, but this is inaccurate: in actuality, it stands for nothing at all. It is merely a symbol used for a few notational shorthands in analysis. So ultimately, this could easily mislead people. And many people in the comments section did indeed acknowledge your comment, but have said "there may be different infinities, but in this problem, we are dealing with the same infinity". So this fails to be the most intuitive way to understand the problem. How exactly does one come to realize that lim (v^2 - u^2)/2 (u -> -♾, v -> ♾) does not exist? Well, what you really need to understand is that if v -> +♾, then a·v -> +♾ as well, as long as a > 0. So if I replace v with a·v, the result I get should still be 0. But is that true? lim (a^2·v^2 - u^2)/2 (u -> -♾, v -> +♾) = lim [(a^2 - 1)·v^2 + v^2 - u^2)/2 (u -> -♾, v -> +♾) = lim (a^2 - 1)/2·v^2 + (v^2 - u^2)/2 (u -> -♾, v -> +♾), but notice how lim (a^2 - 1)/2·v^2 (u -> -♾, v -> +♾) is not 0, unless a = 1. This should not be the case: as discussed, it should not matter what real number a we choose if a > 0. But it does matter. So the limit does not exist. Another way to put it is that, directly in the original expression, if I replace +♾ with 1 + ♾ or 2·(+♾), the value should not change, since +♾ = 1 + ♾ and +♾ = 2·(+♾) is true in the context of calculus. But it does change. Contrast this to the integral of exp(-x^2) from -♾ to +♾. Making these replacements will never change the value of the integral, because the limit actually does exist, and it is indeed equal to sqrt(π).

[@Chalisque]

If int_infty^infty xdx converged to a value, say A, then consider the change of variables u = x + B for some constant B. Essentially we could make the integral sum to anything.

[@wojtek43]

I've never found this convincing. I'm trained in the art of physics first and mathematics second. If this were quantum mechanics, we would say that if you take a symmetrical integral of an odd function, it's zero, and the way the universe behaves will almost always agree with this. It seems like this is just defined as divergent to keep our mathematical semantics consistent.

[@nicksmith4507]

It seemed the most obvious approach was to notice the symmetry to me too. I get you can't treat all infinities as the same, but this just seemed long-winded.

[@angelmendez-rivera351]

Yes, but in quantum mechanics, you are not actually dealing with an improper integral, you are dealing with the Cauchy principal value.

[@danibarack552]

I think it's also that 0 doesn't have to be the 'middle' of the real axis, any number you pick as the 'middle' there are still an infinity of numbers to the left and right of it, so you can't really use the symmetry argument

[@lexus_bkl]

Bro, Wolfram alpha says this integral is equal to 0.

[@angelmendez-rivera351]

@@lexus_bkl WolframAlpha makes many mistakes. It also says 3^(3^(3^(3^3))) = ♾. Calculators are a tool for computation, not an omniscient infallible god of mathematics.

[@W-HealthPianoExercises]

yes, you are in fact mirroring the notion of symmetry of the function within the actual operation of taking the limit. This reflects the fact that the result may depend on how you define the improper integration. I would take it more as a demonstration of the correct way to define it. The initial splitting into a sum of 2 improper integrals was already something not really justifiable when dealing with infinite, which yields the consequent inconsistencies.

[@ethanjerram4209]

At 3:00, can you replace the first limit with a --> pos inf, and make all the a's in the equation negative. And then you can combine both limits to get lim a->inf (1/2(a^2) - 1/2(a^2) = lim a->inf (0) = 0, or is that not allowed

[@ssusp]

You're videos are so helpful because I'm just curious how you describe math in Egnlish --from Japan

[@Tom-xq7ys]

Sepling worgn

[@user-uj8ts7pi9y]

どういうこと？

[@Tomohiko_JPN_1868]

@@user-uj8ts7pi9y "Spelling Wrong" English のスペルを間違えている …ってさ。

[@user-uj8ts7pi9y]

@@Tomohiko_JPN_1868 それぐらいわかってるけどそもそも文章の意味がわからなかったんだよ

[@Necrobin]

Never said it but I love your pen swaps.

[@wiwaxiasilver827]

Ah, the Cauchy Principal Value. Nice 😊👍

[@aconfusedhuman8439]

I have a question: how do you define fractional tetrations? Are you able to differentiate the x-th tetration of x?

[@angelmendez-rivera351]

The analytic continuation of tetration exists for fixed base, but there is essentially no concept of what a moving base looks like.

[@curtiswfranks]

4:55 Replace 'a' xor b with their respective squares and it becomes pretty obvious on an intuitive level, I think. If we have a² and just b, and we treat each 'a' and b as linear functions going to ∞, then we can intuit that a² gets there faster, so the -∞ answer would dominate. But if insteas we have 'a' and b², then +∞ would dominate. And, of course, we can do any mix-matching so that we can arrive at other sums. This is not the best mathematical justification, but I think that it would give a new student some intuition for why the infinities do not just cancel.

[@curtiswfranks]

Ah, you kinda do do that. Hahaha!

[@awesokestephen3494]

saying this before i finish the video... why dont we do lim of b as b approaches infinity of the integral from -b to b? would this not resolve the infinity - infinity indeterminate form problem?

[@angelmendez-rivera351]

It does not. The video explained why.

[@briemann4124]

Great video, bprp!

[@emmanuelpaygar3736]

Thanks very much sir.

[@ricomenez5362]

The way he flawlessly switched between red and blue and black and blue was nice

[@Jeon1958able]

The term infinity does not confer any exact idea of magnitude, thus, we don't know if the two infinities are of equal magnitudes. Therefore, you can't say the result of that integral is 0.

[@idrissimonssef9978]

you make maths funny thanks

[@Martin-qb2mw]

I'm in my masters, passed all calc stuff several years ago. Just watching these for fun.

[@gabrielaudi100]

Excellent job man! Greetings from a Brazilian's physicist! You know how we treat those things... kkkkkkk

[@rafectennis]

What if you switched the upper and lower bounds of the first integral and then wouldn’t you have a positive and negative of the same integral?

[@angelmendez-rivera351]

No, because the bounds are just a and b, arbitrary real numbers, and you are just taking a -> -♾, b -> +♾.

[@BalaMurugan-hk2gd]

Love you dude

[@MathAdam]

The CPV identifies the approach that many of us assumed, but unneccessarily. Infinity is not a number.

[@mauricioalarcon4284]

How to graph?

[@santiagoparera5531]

aight quick question integration is about the area of a certain function. And theres not really such a thing as negative area. the area from -inf to 0 us equal to the area from 0 to inf and they dont really cancel each other out

[@joluju2375]

Is there a similar "CPV answer" for a sine from minus to plus infinity ?

[@angelmendez-rivera351]

Yes, there is, and it is also 0.

[@AlgyCuber]

A question related to this, what is the sum of all integers?

[@Grassmpl]

Does brilliant have a course on invertible sheaves and line bundles?

[@bappalaswamybonu6793]

Hey bro I like your vedios on maths.I am a math lover too. Could suggest me some books to develop caluculus to solve problems and apply it 🙏

[@Peter_1986]

Every time I run into an integral with infinite lower and upper bounds, I try to turn it into 2 integrals with one infinite bound per integral. It always feel as if that should work, but it usually gives me some hopeless integrals that cannot be solved for some other reason, like "∞ - ∞" or something like that.

[@nationalstudyacademykim5030]

Actually, the answer is to infinity and beyond!

[@alberteinstein3612]

Could you find some integrals that do not diverge from -infinity to positive infinity, like 1/(1+x^2)? And then make a video about them?

[@blackpenredpen]

Sure, the classic Gaussian integral : ) Or this one, integral of x*e^(-x^2) from -inf to inf, of kzbin.info/www/bejne/aF7ZaqJrdqtjfaM

[@alberteinstein3612]

@@blackpenredpen thanks 😁

[@triton62674]

What if we kept both parts of the integral under the same limit where the lower end is -a to 0 and the upper end is 0 to a, wouldn't that evaluate to 0?

[@officiallyaninja]

yes, but for different expressions for the limits of integration you get different value. say instead of taking -a to a, you took -a to a+3, in this case the limit of the integral is not 0.

[@lelouch1722]

You can do that, but this is not the definition of the integral between -inf and +inf.

[@chrisjillings]

He did that at the end. That is the Cauchy Prinicple Value.

[@user-qi8ic2rc8y]

Can you find continuous, differentiable function f(x) when f(f(x))=2^x for all real x? What about g(x) when g(g(g(x)))=2^x?

[@dansheppard2965]

Surely lim(t->inf)integral([-t,2t] x dx) or any other multiple also "equals" the indefinite integral in the limit, and is non-zero? Why is the symmetrical case chosen other than because it's the intuitive one?

[@alex_marr]

No reason, they ARE growing in an equal way and cancel each other.

[@funktorial]

yeah, the real point is that the indefinite integral as written is just *a shorthand* for some particular limits. this one: lim [a->inf] lim [b -> inf] int [-b, a] x dx diverges, and since that's the limit we assigned as the meaning of int [-inf, inf] x dx, the integral, as written, diverges. there are obviously related limits we can write down that we may want to write using the same shorthand int [-inf, inf] x dx, but of course there is no unique choice here (there are infinitely many; compare, for instance lim [a -> inf] int [-a, e^a] x dx. this one diverges)! the Cauchy principal value he mentions is just one way of making a principled choice of some limit to disambiguate and assign some convergent limit to the expression we wrote down.

[@MrAskGamer]

Does that mean integral -infinite to infinite of x^3 also ? it equals to zero using CPV?

[@jimallysonnevado3973]

This is why infinity in math is not a number but a concept. It is very subtle. Numbers are concrete things that you approach into. While, infinity is a process that goes indefinitely. Because of that, there are many ways to go indefinitely. In the example about CPV, we are assuming that we are going indefinitely to the right with the same rate as the thing goint to the left. However, it is not the only reasonable way. We instead, get the same area atleast from the figure sake if we go twice as fast to right than to left. Say 2t for upper bound and -t for lower bound. Based from the figure we still cover same area but the numbers we get now is different. So, in dealing with infinities the rates on how it goes to infinity matter.

[@angelmendez-rivera351]

I should point out that the rate at which stuff is being approached _always_ matters, regardless of whether you are working with the symbol ♾ or not. Just to make it clear how this is true, consider the integral on [u, v] of the identity function. This is (v^2 - u^2)/2. What the video is proposing is that the limit as u -> -♾, v -> +♾ does not exist, but if you restrict the limit function to u = -v, then it does exist, and it is equal to 0, which is what the Cauchy principal value is. But we can say the exact same thing without using any "infinities" whatsoever. Let u = 1/s and v = 1/t. Then the above limit can be rewritten as lim (1/t^2 - 1/s^2)/2 (s < 0, t > 0, s -> 0, t -> 0), and this still does not exist, but now, it is irrelevant whether you consider ♾ a number or not.

[@hotlatte1222]

Div!!! I thought it was 0 equally. I love this question. It implies how diversely unfair a social system can be even if the appearance shows a equal form. Maybe in infinity, like George Orwell said, someone is more equal.

[@genocider5868]

Okay, this might be a stupid question but if we have integral from -inf to +inf, can we just assume the positive and negative infinities are "equally as big"? Like can we really introduce a limit of [t->inf] and just call both sides of the integral -t and t? Or do we have to use a two dimensional limit?

[@zanti4132]

That would be my way of looking at it. If I draw any closed shape that fits in the area above the x axis and to the right of the y axis, then an identical shape (just upside-down) can be created in the area below the x axis and to the left of the y axis. From that I conclude these are equal infinities - I would agree that infinities don't have to be equal, but these two are. I'll add that f(x) = x is an odd function, and we know that for an odd function, the value of its integral from -a to +a is zero. So, shouldn't its value be zero from -∞ to +∞? Why should this change? I don't think it is illogical to think two divergent limits can cancel each other out and give a convergent result.

[@angelmendez-rivera351]

@@zanti4132 That is just not how that works. If you look at the integral of the identity function on [-a, a], then it is 0, and if you let a -> +♾, then this does not change. But if you look at the integral of the identity function on [-a, a + 1], then it is not 0, and if you let a -> +♾ then, it is not 0, yet you still have an improper integral from -♾ to +♾ in the latter. Which path you choose should not matter, yet it does. So the improper integral simply does not exist. The identity function being odd is irrelevant, because we are not looking at specifically the limit as a -> +♾ of the Riemann integral on [-a, a], and there is no reason to assume we are.

[@angelmendez-rivera351]

@Blazer Warrior: We have to use a 2-dimensional limit, since this a function of (a, b), since we are looking at (a, b) -> (-♾, +♾) of the Riemann integral of the identity function on [-a, b].

[@zanti4132]

@@angelmendez-rivera351 If I randomly select a point {a,b} in the xy coordinate plane and I ask you, what is the probability this point is in quadrant 1, would your answer be 1/4?

[@angelmendez-rivera351]

@@zanti4132 No, I would say the answer is undefined. It is not possible to define a uniform probability distribution on "the xy plane". This is actually a big reason behind why measure theory exists.

[@ashotdjrbashian9606]

The situation here is much simpler than the author and many commenters think. The question of divergence or convergence depends completely on the definition of what convergent integral is. If we look at the "standard" definition, then obviously this integral is divergent, as it was explained in this video. At the same time (and this is mentioned in video too), if we accept Cauchy's principal value definition, then it's convergent. Similar situations exist in the theory of infinite series too. For example, the series 1-1+1-1+1-1+... is clearly divergent under the classical definition, but if we use Abel summation, then it will become convergent and the sum of the series is 1/2. Another familiar example is the difference between Riemann and Lebesgue integrals. The function defined on [0,1] as 0 for rational numbers and 1 for irrational numbers is not Riemann integrable but has Lebesgue integral equal to 1.

[@angelmendez-rivera351]

*At the same time, if we accept Cauchy's principal value definition, then it's convergent.* No, it is not, because the Cauchy principal value is not an "alternative definition" of the integral like you claim it is. The Cauchy principal value is actually an averaging operator, and it is not specific to integrals, since you can apply it to arbitrary functions on R. For example, the Cauchy principal value of the Riemann Zeta function at 1 is equal γ. This is to say, lim [ζ(1 - ε) + ζ(1 + ε)]/2 (ε -> 0) = γ. This is true, and yet it would be incorrect to say that ζ(1) = γ, since 1 is not in the domain of ζ, which is defined on C\{1} as a holomorphic function. *For example, the series 1 - 1 + 1 - 1 + ••• is clearly divergent under the classical definition, but if we use Abel summation, then it will become convergent and the sum of the series is 1/2.* No, this is demonstrably not how that works. When we speak of the symbol f(0) + f(1) + ••• formally, we are talking about the sequence f : N -> G, where G forms a topological group, and we define the partial summation transform operator S as the discrete integral operator on sequence in G^N, so that S[f] is the partial summation transform of f, a sequence of the partial sums of f. We say the series converges iff S[f] converges, and the convergence of a sequence g is simply defined as follows: there exists some L in G such that for every neighborhood M of L, there exists some natural number N, such that for every natural number n > N, g(n) is in M. That is what convergence is, and there is only one definition. The convergence of S[f] does not depend on "the type of summation" we choose. With that being said, Abel summation is well-defined, but the Abel sum existing does not make f(0) + f(1) + •••, since S[f] continuous to diverge. The Abel summation being 1/2 implies that the sequence h : N -> {1, -1} with h(n) = (-1)^n has a generating function F : (-1, 1) -> R with F(x) = 1/(1 + x), and that lim F(x) (x < 1, x -> 1) = 1/2. This is all it is.

[@harshitjuneja9462]

The second method, even though is intuitive but it is also like saying that 0/0 equals 1 as lim (x -> 0) for x/x is 1. however, we know that x/x is not the only way to reach the fore-mentioned indeterminate form.

[@gigachad6844]

exactly my view, some people say 0/0 is 1, I mean they're actually telling their IQ score lol.

[@divyanshudevesh6016]

Sir are you teachs mathematics for bsc

[@alien3805]

I still have my high school calculus papers gotta restudy them soon for uni

[@hankseda]

Here's another way to understand the indeterminate nature of the problem. We need to be more specific about what something ---> infinity means. If b---> infinity like n and a---> - infinity like -√n then the difference between the two halves of the integral is (1/2)n^2 - (1/2)n = (1/2)n(n-1). This quantity gets infinitely large as n---> infinity.

[@you8sand]

I'm wondering why you can't do a u substitution with `-x` letting you update the bounds of integration resulting in \int_0^\inf x - \int_0^\inf x ..

[@angelmendez-rivera351]

Because they are different limits, not the same limit

[@gigachad6844]

Good way to making people understand infinity, infinity is not a number it just mean too far right ( don't know how far), and similar thing for negative infinity.

[@ProCoderIO]

You can’t have it approach both ends at the same rate?

[@orionmartoridouriet6834]

You can, but you are not required to. If 0 were the correct answer, *all* of the limit manipulations would give you 0, but because not all of them do, we say it diverges

[@blackpenredpen]

That’s what the CPV is for. So we can have an “answer” to a divergent integral.

[@vikraal6974]

Is the answer 'zero' if we consider a different integral: Limit (n->inf) Integral (x) dx from -n to n If yes then why? Thanks

[@ericsmith1801]

Try to integrate this: |x|. This will avoid negative areas under the graph by flipping the graph into the positive areas of the graph.

[@angelmendez-rivera351]

Not necessary. The integral of the identity function on [s, t] is equal to (t^2 - s^2)/2. lim (t^2 - s^2)/2 (s -> -♾, t -> +♾) does not exist. Therefore, the improper integral does not exist. Q. E. D.

[@giorgibliadze1151]

what about symmetry sir?

[@mizu7284]

inf - inf is like (-∞, x] - (-∞, y] where x, y can be anything. So basically we can't really define it as zero since we don't know which scale of infinity it is.

[@imbreakingdown2683]

HAPPY TEACHER'S DAY!!!!!

[@divyanshudevesh6016]

Sir how to learn calculus from basic

[@AnakinSkywalker-zq6lm]

What if it’s inside of a limit?

[@alexandresoaresdasilva1966]

Couldn’t a cardinality argument be made? Like there’s a one-to-one mapping between (-∞,0] and [0,∞) by means of f(x)=-x; therefore the cardinalities are the same and thus it makes sense that the areas would be equal?

[@ttercheng]

I think this make use of the ambiguity of Inf sign, where infinity isn’t a finite value or state such that it can be cancelled out like in the example. If I write {lim m -> inf [ m Integrate -m ( xdx)]} then it shall be zero, as this doesn’t remove the rotational symmetry property of the function. 😌

[@angelmendez-rivera351]

The ♾ sign is not ambiguous. The improper integral, in every calculus textbook, is implicitly defined as the limit of the Riemann integral on [a, b] as a -> -♾, b -> +♾. This limit does not exist. Hence the integral does not exist. There is exactly no ambiguity at all.

[@brainsofmathematics999]

You told that integration of xdx limit from -1to 1 is zero,,, but actually the area is not zero. What is actually correct ?? Pls ans Mee??

[@mighty8357]

I love you!

[@gianlucadegliesposti7241]

Integrals like this pop up in theoretical physics, and people always assume that they're zero. Another popular one I see often is the integral from 0 to infinity of e^(iax) for real a. It is assumed that the integral is carried out over an imaginary direction or Im(a) =/= 0 to make it converge (and Im(a) = 0 is the analytical continuation)

[@angelmendez-rivera351]

People do not "assume" the integral is 0. Physicists use the Cauchy principal value in replacement of these integrals, though often not denoting the fact that they are, because the correct notation is cumbersome, and it is contextually unnecessary. Functions with singularities are problematic within physics, so functions are repalced by their Cauchy principal valued counterparts. Integrals are no exception. This implies nothing regarding the integral itself.

[@gianlucadegliesposti7241]

@@angelmendez-rivera351 using the Cauchy principal is an assumption. There's no physical reason to do so, it just works.

[@angelmendez-rivera351]

@@gianlucadegliesposti7241 No, there are physical reasons to do so. I would know. I studied physics.

[@gianlucadegliesposti7241]

@@angelmendez-rivera351 can you provide examples?

[@angelmendez-rivera351]

@@gianlucadegliesposti7241 Regularization in quantum field theory comes to mind. Infinite quantities appear where they obviously should not. Hence by physical necessity, one must take regularization into consideration before taking any mathematical statement to be a hypothesis regarding the universe we live in. There are also physical phenomena with boundary conditions that naturally lead those boundary conditions to modify singularities in such a way as to eliminate them.

[@iitjee5765]

Thx sir am clear 😄

[@oyunuzmanlar8726]

The limit as a aproaches infinty of the improper integral of x from neg a to a = the limit as a aproaches infinity of(1/2)a^2-(1/2)(-a)^2 = the limit as a aproaches infinity of (1/2)a^2-(1/2)a^2 = the limit as a aproaches infinity of 0 = 0

[@kaan4472]

Which one is correct in math?

[@angelmendez-rivera351]

The former

[@meltingzero3853]

I think a good way to demonstrate that inf - inf = 0 can't simply be stated is as follows: if inf - inf = 0, with inf + 1 = inf, goes that inf + 1 - inf = 0 which would mean that 1 = 0. In this sense, having two infinities disallows this algebra.

[@lexus_bkl]

bruh, this is calculas not algebra. In calculas, algebra wouldn't makes sense. If algebra would make sense in calculas, then derivative of everything will be 0/0.

[@lexus_bkl]

Also, according to wolfram alpha, that integral is 0.

[@meltingzero3853]

@@lexus_bkl I just typed it in WolframAlpha to check, and it disagrees with you. The result is that the integral does not converge, and it adds underneath that that the Cauchy principal value is 0. All of this was talked about in the video. By your line of reasoning, results in geometry don't have to make sense in an equation because "this is geometry, not algebra".

[@ThePeterDislikeShow]

What's the application of the CPV? I didn't learn that in calculus!

[@angelmendez-rivera351]

The Cauchy principal value takes the isolated singularities of a function and replaces them with the "average" of the function near those singularities. For example, if f(x) == 1/x on R\{0}, then the Cauchy principal value of f at 0 is 0, because lim [f(0 + ε) + f(0 - ε)]/2 (ε > 0, ε -> 0) = lim (1/ε - 1/ε)/2 (ε > 0, ε -> 0) = lim 0 (ε > 0, ε -> 0) = 0. So in many applications, it is useful to replace f with f*, where f*(x) = 1/x if x is nonzero, f*(0) = 0. Integrals are functions of their bounds, and so the concept is equally applicable.

[@aakh3500]

Before 3:23 you fogot minus at lim in line #3, a to inf but on other lines a to -inf

[@Lmno274]

Assalam o alaikum dear, kindly make a video about the functions who's integral can not be found and plz explain the condition for integration, 🙏🙏

[@manastoews6679]

The GOAT

[@purim_sakamoto]

なるほど！

[@derpsalot6853]

What would happen if we were to say that the cardinality of both infinities were equal? Could we then say it was zero, or prove it was zero?

[@angelmendez-rivera351]

There are no actual infinities involved in this problem. This is a problem with limits. Infinite quanties never actually appear in this problem.

[@averagecornenjoyer6348]

but i thought you could split the integral only under the condition that the two converge?

[@patrickpablo217]

Isn't the issue not that you don't know "how far to the left and right" the two integrals are? Isn't it more like the issue with infinite series that are not absolutely convergent or divergent? Like, you *can* add them up in certain ways to get them to give a certain value - but the problem is that *that* value is not the only value they can give. Similarly with the two integrals: if you march away from 0 at the same rate in both directions, then you get the CPV of 0, but you don't actually have to do that and doing something different can give a different value. However with integrals that behave nicely, it doesn't matter *how* you "march away from 0 in each direction" (or whatever the appropriate phrasing is for that hypothetical problem). For those well behaved integrals, they will give the same answer no matter how you get there. This seems like more the flavor of what's going on, but I might be wrong.

[@angelmendez-rivera351]

You are correct.

[@Tomohiko_JPN_1868]

Only within CPV. ∫[-∞, 1] e^x dx - ∫[-1, +∞] e^(-x) dx = 0 ... right ?

[@pradyumnakumarnayak9384]

Namaste sirji.

[@shadmanhasan4205]

So... basically the integral of X by +/- infinity is really 1/2(b^2 - a^2)? And outcome of ♾ depends on if a=b?

[@yannik1635]

not exactly, you need both of the limits and you can't just combine them, because you don't know how big these infinities are. So i think that your thought is correct, but you can't just write it like this, without the limits and say that it depends on if a=b. Sadly limits and infinities are a bit more complicated.

[@angelmendez-rivera351]

The integral of the identity function on [a, b] is indeed (b^2 - a^2)/2, provided a = b or a < b, at any rate, if we ignore orientatuon. The limit of (b^2 - a^2)/2 as b -> +♾, a -> -♾ does not exist. Hence the improper integral does not exist.

[@jarikosonen4079]

What means the Cauchy principal limit? I had asked earlier if infinity is multiple of π. It could be. Maybe the infinity is multiple of every number, including π and all trancendentals, etc. Wouldn't that be nice property of infinity?

[@angelmendez-rivera351]

This is nonsensical. Infinity is not a mathematical object, and it also has nothing to do with the Cauchy principal value.

[@jarikosonen4079]

I dont know

[@joeyoest1105]

If you don’t see the reason the improper integral doesn’t clearly converge to 0, here’s something that may help... Considering the integral from -t to t, then letting t-> inf, it seems we’re taking the limit of ‘0’. However, we could just as easily have integrated from, say, -t to 2t then let t->inf. For real t, this integral would be (2t)^2/2 - (-t)^2/2 = 3/2*t^2 -> inf as t-> inf. There’s no reason the (-t,t) interval is a better method than the (-t,2t) interval (or (-2t,t) for that matter). This isn’t intended to be a careful proof - just a way to intuitively make sense of why we have to be extra careful here.

[@danibarack552]

So you split a limit into two limits that diverge, that shouldn't mean that the original limit diverges. It's like splitting lim(n-n) into lim(n)-lim(n) which is inf-inf and so is indeterminate, but the original limit is clearly zero

[@seegeeaye]

the first part of integral (from negative infinitive to zero), it's integrand function is (-- x), not x.

[@gamepplayker-of6in7do8to9nice]

X tO sIn&cOs rIt？ Are there any mathematical formulas that can draw triangles? I only knew that there are circles and squares, please

[@pabitraghadei6857]

Sir plz solve integration of cos(e^t) dt