no no there are both divergent as infinity is an answer to a limit and it is not valid to have infinity - infinity as it is indeterminate. so both diverge.

I WAS JUST WATCHING and remembering my calculus classes. These types of question happens when people dont get along properly with the meaning of the symbols that makes the expressions

Thank you

Also highlights well how a limit is evaluated. In my mind it's similar to how (x+1)/(x+1) at x=1 is discontinuous. But if you were to take the limit it's equal to 1. Likewise taking the limit of that integral is similar. First you calculate the integral itself, simplify it, then solve the limit

i actually never knew that, sounds really interesting

That's what I thought, even if it's indeterminate since they are diverging at the same rate they cancel to 0

Cosine is an odd function

Someone didn't watch the video to the end before commenting...

@@thomasdalton1508 anything wrong with watching after commenting?

This is the comment that made this concept click in my brain. For a and -a to cancel out, a has to be finite. ∞ and -∞ are not finite, so they can't cancel out.

​@jimjolley it's not even really that (okay, it indirectly is) - it's that infinity deceptively can hide anything! In the original problem with the symmetrical integral, we know exactly what went into that infinity, so we can meaningfully talk about it. But if you just put infinity in, anything could be going on under the hood, and depending on what it is, the answer will differ. So when you make the statement about neg infinity to infinity, you're really making a statement about a lot of limits with different answers.

@@jimjolley not necessarily that they can't cancel, you just don't know if they do, because infinity on its own doesn't show how fast it reached infinity, which is why we need to work in the limits, on the example with a we see how they approach and that its symmetric. maybe u knew this but its important to clarify that it still could be 0, it just doesn't have to be, we just don't know its not specific.

Operator overloading really needs to be clearly defined :D

It's not really a notation problem. It's a real fact that there are many infinities, and the fact that we write "∞" to mean any one of those infinities is not really any different than writing a variable "a" to mean any number. You can fix the notation to look less like a number, but the integral will still be tricky to reason about.

@@OMGcluelessNo, fundamentally it is a problem and is NOT like writing 'a' as this video demonstrates. Because when you assign your infinity to a specific variable, then all instances of that variable are the same sized infinity. They ARE equal to eachother but not necessarily to another infinity. Which is why defining bounds as -a to a as a->infinity works but infinite bounds on their own do not.

@@Xylos144 Assigning infinity to a variable is just not an operation that makes sense. And infinity is not something that is overloaded -- if anything it is the "=" sign which is overloaded to sometimes mean assignment and sometimes mean an equality predicate and sometimes mean a definition.

@@Xylos144 In an equation like y = x sin(x), if you pick x=2 then both x's become 2. But they're not the same x as if you pick x = 3. What you're saying 100% applies to any variable, not just to infinity. If you want, put subscripts on your x's and your inf's to help tell them apart. There's no issue of notation here that isn't trivially fixed. The real solution, though, is for you to stop treating infinity like it is one fixed value. It isn't and never has been.

this would be a differnt function, what are you even saying ? its like "oh 2x-x have infinity as limit. therefore x-x cannot converge". The point literally IS that -a and a grow at same speeds, where 2a and -a dont.

@@michaelmischko847 It's a comment about the part around 0:36 where he explains how the -a,+a limit is different from the -inf to +inf integral

@@philippenachtergal6077 i see, watching the actual video would have been a good idea before becoming annoyed

also im 15, i started watching your videos when i was 12 (although i didn't really understood anything when i was 12, but it definitely made me curious) i have a question, if the infinities-bound could be a limit of any function that approaches infinity, then how most improper integrals are valuated? like the famous integral of e^-x^2, the bounds could be different from each other and would make the value different, right? maybe the reason is that no matter what function you substitute x of "e^-x^2", when it is approaching infinity, the value of "e^-infinity^2" will always be zero no matter what. but i don't know if it is correct...

Let's say you integrate exp(-x^2) from a to b. Because exp(-x^2) is positive, you can separate the integral as the sum of the integral from a to 0 and 0 to b and take the limit of each term separately. You can see that no matter what a and b you'd choose here, you'd still end up with the integral from -inf to 0 + the integral from 0 to +inf when taking the limit, and thus the value of the integral of exp(-x^2) from -inf to +inf doesn't depend on how you take the limit. In general, as long as you can separate the bounds of integration such that the sum of the limits of each integral is well defined, this will always work. This is possible when both integrals converge, or when they are of the same sign. The only problematic cases are +inf-inf or -inf+inf like the one in the video. When you are in one of those problematic cases, the value of the integral could indeed depend on how you take the limit, and it's a bit more tricky to even define the improper integral. You would generally use what is called the Cauchy Principal value in this case (you would integrate from -a to a and then take the limit, if it's well defined)

​@@hach1koko i was specifically talking about the -inf+inf ones, i actually never thought they would be hard to define. so what you telling is that for the *actual* integral between -inf and +inf of e^-x^2 is harder to define? and by using the cauchy principal value, it makes easily definable and thus equaling to square root of pi?

@@Tritibellum the integral of exp(-x^2) is one of the easy cases (not a -inf +inf type of situation), so you don't need to worry about cauchy principal value and can take the limit in the bounds in any way you wish. This is the integral i was talking about in the first two paragraphs, sorry if that wasn't clear. I edited my post to make it clearer

I just realized that what I meant "the -inf +inf case" maybe wasn't clear. I'm not referring to the bounds of integration when I say this, but to the value of each integral. For example in the video, you have that the integral of x from -inf to 0 is equal to -inf, and from 0 to +inf it is equal to +inf. This is a "-inf+inf case". For the exp(-x^2) integral, you would get "something positive" + "something positive" instead, so you can always add them with no issue.

That's still not working. There is no such thing as a "precise value of infinity". When you derive a function, the constant part is lost. So when you want to reverse the process and retrieve the exact primitive that was derived, you need extra info like a reference point to compute the constant that is missing. With infinity, you have a similar issue (but with a wider range) because infinity is hiding not only the constant part but also the "speed" to infinity. Even when you only consider with countable infinites. limits of x, x+1, 2x, x^2, e^x, ln(x), etc are all countable infinites. But none of them cancel out together.

I'm familiar with ordinals where ω is usually the set of natural numbers {0,1,2,...} and ω + 1 = ω \union {ω} etc. But you mention that your ω may not be countable, and I don't see how the ordinals I know would appear in limits of integration. Can you say a bit more what you are talking about?

for the "lim(a->-inf)lim(b->inf) (integral from a to b)" explaination, it wouldn't change the computation, because you are making a to be a certain function, and b to be a certain function, we don't know if it is the same function or not, we just know that a is going towards positive infinity and b is going towards negative infinity. and calculating it would lead to the same conclusion in the video, it is divergent

Due to properties of limits, we define integrals from -inf to inf that way. You can do it your way, but you end up with this: Lim(a->-inf)Lim(b->inf)[0.5b^2 - 0.5a^2] which is also divergent.

I think the explanation is meant to be intuitive.

The main issue of writing it as lim(a->-inf)lim(b->inf) (integral) is that it sort of implies an order to evaluating the limits, which would give a defined (divergent) answer of +infinity. A more precise way to write it, I think, would be the multivariate limit: lim((a,b) -> (-inf,inf)) (integral) This way you indicate that it must converge to the same value on all paths, and since this multivariate limit is undefined the integral does not exist.

@@cobalt3142 So, by this implication do you mean that the notation: lim(a->-inf)lim(b->inf) (integral) actually means: lim(a->-inf) [lim(b->inf) (integral)] ? Also, it's the first time I see a limit of an ordered pair. How is that defined? Would that be something along the lines of: [(definition of f(a) as a->-inf) and (definition of f(b) as b->inf)] ?

Yep. Or put another way - when you claim the integral from negative infinity to infinity is meaningful, you say it's meaningful (and the same) no matter how you go there. But it clearly isn't. It matters how you go there. But in the problem, that's not an issue. We know how we go there. So we're not making a statement about all integrals with endpoints that approach infinity, (which doesn't have a well defined answer) merely about a specific one, which does.

You can make it go from 0 to a. Then for every finite value of a, the integral is defined. Yet it diverges. So that alone is not enough to make it a limit.

You can but it still gives you a divergent limit

Which is exactly what the first integral is saying.

Yes you're right and the value of the limit not being equal to the value of the function is precisely the definition of the function being discontinuous at that point

The limit can be different from the function value yes. The definition of a limit (epsilon-delta) requires the distance between x and the target point to be strictly positive. So what happens when x is exactly at the target point does not affect the limit.

No!

Cardinality is for sets. We have different infinities for functions. For example, e^inf is “way bigger” than ln(inf). A bit more rigorously, if you do limit as x goes to inf of (e^x-ln(x)) you will get inf.

@@bprpcalculusbasics Gotcha. Infinity is weird, man. Anyway, thanks for the clarification.

yes, a and b can grow at different rates and therefore it diverges

It's not just a matter of rate, though. For example, consider the limit as a goes to infinity of the integral from -a to a+1. Both bounds are now changing at the same rate, but this limit also diverges. The interpretation of the integral as the signed area under the infinite curve fails to be invariant under translation.

That isn't the definition that is generally used, no. (Edited to include the following) The standard definition is that the improper integral from -∞ to ∞ represents the sum of the improper integral from -∞ to some finite value c and the improper integral from that finite c to ∞. This definition turns out to be more well-behaved than the alternatives.

@@TJStellmach so then how are you supposed to evaluate improper integrals? That's what I learned in my calc 2 class at penn state

​@@mitchratka3661 for improper integrals that approaches a certain value, thats because of how limits are defined lets use for example, the famous gaussian integral between -inf to inf. no matter what function you can plug in x in e^-x^2, incredibly slow or fast, as long as the function approaches infinity or negative infinity, it will always be e^-(inf)^2 and e^-(-inf)^2, which evaluates to 0, which means the integral **could** converge to a value (doesn't work for the integral of 1/x unfortunately) you can plug the famous harmonic series as x in e^-x^2, and it will still evaluate to 0 because we know it approaches infinity, slowly. and this is how limits are defined, they are meant to approach a certain value L with no matter what rate of change you can plug, like for example of x approaching infinity, you can plug f(10), f(100), f(1000), ... or you can plug f(pi + 5), f(2pi + 10), f(3pi + 15), ..., or you can even plug the harmonic series into it. the reason why it doesn't work for integrals like this one in the video, is because it doesn't evaluate to 0 in the limit test, it just diverges off to infinity with this case, we don't know what the area could be, because we don't know how the rates of change are acting, they could be the same, or one of them could be growing bigger, or smaller, we don't know! and for that reason, we know that it could possibly be 0, 2, 25, 106, pi, e, any value you can think of, because we have an inf minus inf case.

i thought so too, but i think i see a problem there - in the first problem, they define the bounds as -a and a. we know that those are necessarily symmetrical, when lim a -> any number at all in the second one, i think we’d be introducing our own new constraint that’s not actually given if we insisted that the bounds were specifically -a to 0 and then 0 to a. there’s nothing in the bounds of the improper -inf to inf integral that declares that those two infinities are the same

The issue of an improper integral with an asymptote within its interval is different from that of one whose interval is unbounded.

Its the same scenario here. You would have to break the integral into two parts from [-2,b] and [a,2] and let b->0- and a->0+. The first integral gives -inf and the second is +inf. But these two infinities are entirely unrelated to each other. They come from two different limits and theres no connection between them. So we cannot say it is 0 nor can we continue any further. So we conclude it diverges since we cannot get a single finite answer.

You can't "really" use infinity as an integral bound, it serves a stand-in for an arbitrary large value (which can and does always get larger). That's what limits do, right? The key idea is that when there's a negative and a positive infinity outside of the limit, there's nothing allowing you to assume they're based on the same arbitrary value- they could be two unrelated values each moving away from 0 at different speeds, and thus two "different" infinities.

Infinity is not a number and can’t be used like one, it has its own rules. I think of it as a sign post on the road saying “go that way forever”rather than a destination (like a number). Knowing the direction doesn’t tell you either the path you follow or the speed (rate) you are going there. Unless you add additional information (like using the Cauchy principal value or using something like a->∞) to agree on a path and rate, infinities can’t be compared. BTW, the cardinality of the infinites doesn’t factor into this case. Even if you said that both infinities had the same cardinality (in this case, the Cardinality of the continuum, |R|) you still couldn’t compare them. I hope this helps.

@@zombie_pigdragon I kind of understand what you're saying, like ln(x) as x -> infinity is smaller at every x when compared to x as x -> infinity even though ln(infinity) = infinity. Thus they "approach" infinity at different speeds. My hold up is that the infinities would still be the same size irrespective of the speed they took to get to infinity no? Even if at every point along, one area was bigger than the other, I thought that at infinity they would still be the same size because where would the difference in cardinality even come from?

@@jhonnyrock No, infinities are not the same, take the two limits at infinity, e^x - x and x - e^x . Both are infinity - infinity superficially but taking the functions as a whole you see that the e^x factor decides the limit. It's not about cardinality you are mixing up terms, simply think of it as which function takes the wheel.

@@jhonnyrock BPRP was trying to give an intuitive explanation which don't always align with a more detailed logical explanation. The two limits are not connected in any way so their individual outcomes matter. Infinite area is not a value so neither limit exists. Had the limits shared the same variable, then we could potentially use l'Hôpital's Rule to find a finite value.

"Going with whichever makes sense in context" is possibly the least rigorous thing I've heard in my life. Things have definitions. You don't get to pick and choose.

​@@sylowlover Yes, but if the question doesn't stop at such an integral and there's more to the question to solve, then I'm not writing Divergent and leaving the problem there. I'll use the more definite answer(0 in this case) if there is one and solve further. Unless the question is specifically asking me to check convergence, I'm using 0 because that makes the most sense. It's wrong, but at least I go further.

@@sylowlover in maths you do. When a problem has various answers you choose the one that accommodates to your necessities. take for example a physics problem. if you get that the answer to the speed of a body is sqrt(300) you know the answer is only the positive value because the negative answer has no logical sense. Indeterminations are full of situations in which you have to choose what that indetermination means

@@kraklos don't "in maths" me. I have a bachelor's in maths. Of course you can come up with an example where it's applicable, and there being multiple roots to a polynomial with one valid solution makes sense, but it's rather clear that I'm implying that convergence of this integral is not conditional on your problem. It is neither Riemann or Lebesgue integrable. No problem changes that fact.

@@sylowlover It isn't Riemann or Lebesgue integrable, but it does have a well-defined Cauchy principal value, as explained in the video. If you are trying to solve a problem where the Cauchy principal value is what makes sense, then the correct answer is zero. Infinities in mathematics usually arise from some kind of limiting process. You can examine that process to see how you should treat the infinities. If that integral has arisen during your attempt to solve some problem then you can look at where it has come from and determine whether it is approaching infinity symmetrically or not. If it is, you can say the integral equals zero. If it isn't, then you have to say it isn't defined and find another way to solve your problem.

Probably because it evaluates it as if it was formulated in the first way. This seems more like a point about notational precision.

For me wolfram alpha says the integral does not converge, though it does say the Cauchy principal value is 0.

As good as Wolfram Alpha is Prof. Chow has proved it wrong many times.

Nvm you didn't say they are equivalent. That other expression is a different beast with a different evaluation.

The first integral is not how you define the second one. You define the second one as a limit with two variables, not one.

It's not nitpicky or "just notation". The first limit is 0. The second doesn't exist. They are not equal.

​@@herbie_the_hillbillie_goatwriting infinities on bounds of an integral does not make any sense unless it is notation, it's a short for something,, and that something usually is a limit of a variable tending to infinity so that the bounds have the same magnitude of infinity, I have already stated that context matters did I not? If the infinities are generated by 2 different functions things change

@@Manu-se5tx You made the false statement that the second integral is defined by the first. It's not, so it's not a notational issue. It's also not bound to context. They're not equivilant regardless of context or notational conventions. You're just wrong. You can argue with me further or just rewatch the video.

@@herbie_the_hillbillie_goat I said that usually the way of writing the first integral is by writing the second integral, I'm sorry but it's a fact writing limits is too verbose and it's always implied when the infinities have the same magnitude, if they don't then using limits is mandatory. I know it's a fact because everyone I know does it, be it professors or peers, and of course it's all notation you think those symbols mean anything? We are the ones giving it meaning

"We don't know" because, for example lim(a->inf) integral(-a, 2a) x dx is also technically integral(-inf, inf) x dx, but give +inf as the answer. The same way lim(a->inf) integral(-2a, a) x dx gives -inf. They all are integrals (-inf, inf) x dx, but each of them gives different answer.

@@MacronCPP like i get that, but it isn't the situation. like i would argue that 1/0 < 2/0 is true. clearly it's "wrong" or "nonsense" but it's clear to see that when you take their limits the 2/0 increases faster than the 1/0 situation. In the example of this video it's also clear to see they diverge at the /same/ rate.

@@Jalae When you get into analysis many things that seem 'obvious' or 'clear' start to break down when you actually drill into them, counter examples etc and many things that seem counter-intuitive can actually be shown to be true. This is why it's useful to be able to write/prove things rigorously. For example feels like ∞ - ∞ should equal 0 but it can equal almost any value you want. It's a similar thing here. You always hear "Why can't we just assume this?!?", the answer usually is "Well we tried and it doesn't work and/or breaks down." .

@@thomaspickin9376 i mean, i didn't use ∞ - ∞ because that doesn't say anything about how infinity is achieved. x^2 and x can be pushed to infinity, but they aren't the same infinity. That's in fact why im making the claim I am, we do know the rate of getting there. Is there some place i can see rigorous example of -x + x to infinity doesn't equal 0? or where limit x->0: 1/x < 2/x isn't true? like i said initially if the pure rigor/procedural approach just throws up it's hands at these kinds of constructs, it seems to me the procedure is lacking. These aren't nuanced cases of unsimilar infinities, how we got them is right there, and they are symmetrical. like the B in the video is purely a result of the limits of that kind of analysis as far as I can see. maybe this example is just too simple to really understand the problem, because i am on board if the -inf to zero part was in any way different from the 0 to inf part.

the problem is how we define limits and infinities, there are countless amount of functions which, by taking the limit to approaching a certain value, will make them diverge towards infinity, but each infinity could be different or equal from others. in this example, we don't know what happens with ∞ - ∞, all we can know is that the result could be any value, which also includes 0.

Did you even read the original question? It wasn't asking about an arbitrary function.

@@Arycke of course I read the question. What I am saying is that the question is the same if we replace x by any odd function F(x). Maybe you don't know odd functions. We say F is an odd function if F(-x) = -F(x) If F is integrable, the integral over a symmetrical interval is always zero, simply because int_{-a}^a F(x) dx = int_{-a}^0 F(x) dx + int_0^a F(x) dx = int_{-a}^0 -F(-x) dx + int_0^a F(x) dx = int_a^0 F(x) dx + int_0^a F(x) dx = - int_0^a F(x) dx + int_0^a F(x) dx = 0 If the function is differentiable, its derivative is an even function and this reasoning is even easier. We say g is an even function if g(-x) = g(x) If F is odd, meaning F(-x)=-F(x), and F is differentiable, then F'(-x) = - (-1) F'(-x) = - (-x)' F'(-x) = - F(-x)' = F(x)' = F'(x) So, the derivative of odd functions is even, and the derivative of even functions is odd. The same thing happens with primitives. So, if F is odd and it has a primitive P, then int_{-a}^a F(x) dx = P|_{-a}^a = P(a) - P(-a) = 0 That's what he did. But this only works in the case F has a primitive. But the exercise is general for odd functions. Because of that, I said it before and will say it again: there is no need to invoke x²/2. Now, did you even read my comment?

@@samueldeandrade8535 nope. tl;dr

@@Arycke haha. I imagined. No problem. Good people will read it.

It is a definite integral

bro did not watch the video