Euler would say that the Riemann integral does not exist, but he would also argue that the Riemann integral is a bad integral to use for an exercise like this one, so he would use the Cauchy principal value and tell you the answer is 0 anyway.
To respect the symmetries in math, I still want to believe that the integral is zero. After all, the areas to the left and right of zero are basically negatives of each other, so I guess that even after being infinite, they are in some ways equal, so the signed areas may add up to zero. Being divergent does seem more mathematically consistent and rigorous though TBH.
not necessarily. You can do a completely rigorous way of showing it is 0. first define L = integral from -1 to -n of 1/x + integral from n to 1 of 1/x this is also equal by the fundamental theorem of calculus to ln|-n| - ln|-1| + ln|1| - ln|n| = ln(n) - ln(n) + ln(1) - ln(1) since n is never equal to 0, ln(n) is a real number and we can do regular operations, such as subtracting therefore L = 0 for all n != 0 (not equal to 0) now take the limit of L as n-> 0 (from positive) in the limit, the integral representation of L becomes L = integral from -1 to 0 of 1/x + integral from 0 to 1 of 1/x = integral from -1 to 1 of 1/x but this is a limit, so n is never equal to 0. therefore L is still 0 FINALLY WE CONCLUDE THAT integral from -1 to 1 of 1/x = 0
one could think that i just did some mathematical trick to justify the equality, but what about calculus? if it werent for limits, when trying to take derivatives we would be saying "oh but we can't divide by 0 :(". That would be analogous to saying "the integral is divergent :(". But no, you start using something to reason around the problem and limits help. Now instead of saying "it's undefined" we can add a meaningful answer.
The issue is when you define L = , that explicitly assumes there is a solution. Since the integral diverges, its undefined. You can't assign a value to it.
These infinities are more specific than "just being infinity". If you break the original integral down to "-infinity + infinity" you have destroyed information.
So I'm going to throw a wrench in the works here: I know this is a calculus channel and not a real analysis channel, but this function just fundamentally isn't Riemann integrable over the interval [-1,1]. Isn't the best way to express that thus to approach it using the limit definition of the interval and showing the issue via calculating the Riemann sums of the function? That's the core issue here; that the "area under the curve" intuition only works for Riemann integration when the function is Riemann integrable in the first place. If it's not, then the "area under the curve" just doesn't correspond to the value of the Riemann integral. It's the same thing you get when trying to find the integral of a function with Lebesgue measure 0; the intuition of area suggests that the value ought to be 0, but the Riemann integral doesn't get you 0 because it isn't actually measuring area, it's measuring something that, given certain preconditions, is equivalent to area. That's really the reason why there's so much confusion here: because integration doesn't _really_ calculate "area under the curve", it just calculates a value that, when you have a "nice" function, is equal to the area under the curve, but when you don't have a "nice" function, calculates a garbage value that has nothing at all to do with area under the curve. I guess fundamentally what I'm saying is that the intuition that the area underneath this curve is 0 isn't _that_ bad, and it's reasonable to see why it arises. And the issue is that "area under the curve" isn't really what Riemann integration calculates when you have a function that acts up. And it'd be worthwhile to highlight that in order to illustrate the distinction between the two.
Wow. This was actually really insightful, never knew that was true for Riemann sums. Good comment! Are there any alternate definitions of integration that will actually give you 0 for this integral?
You know, I'm actually not sure. Most methods of integration that I know of at minimum require a function f to be of bounded variation over your interval. That means that for some interval [a,b], if you take every possible partition {p1, p2, p3, ..., pn} of that interval (for all integers n), and for each partition calculate the sum of f(pk) - f(p{k-1}) for all k, the supremum of the set of all these sums is finite. (It's a long-winded way to say that the function never blows up, but it's required to be long-winded because some functions blow up in really weird ways.) The issue with functions that blow up is the same issue you run into with any sum of an infinite number of terms that "should" add up to a finite value; the order that you add things matters, which means that the way in which you decompose the "infinite" portion matters. When you're calculating a finite area that should end up summing to 0, you get nice things like association and whatnot that mean that it doesn't matter how you end up pairing things off, everything will work out in the end. But like usual, infinities screw things up, because the way you pair things off _does_ suddenly matter, and so different calculations of the area, different roads that should go to the same destination, suddenly veer off in wildly different directions. That's why you get weirdness like Banach-Tarski, where you can decompose an infinite set into two copies of the original set. There might be some more obscure integral in real analysis that I'm not aware of that can handle this well - given that intuition suggests it to be 0 for fairly valid reasons, I'd suspect there ought to be some way of formalizing that intuition - but it would require more advanced analysis than I am unfortunately aware of.
Idran Completely spot on! The reason you can get zero out of thus, is because you take a simultaneous limit on both sides of the singularity. This makes it possible to do all kinds of nice cancellations, even in the Riemann sums. However, if you manipulate the limit on both sides just right (e.g. taking -epsilon on the left, and 2*epsilon on the right), it might be possible to construct even more "correct" values. That is the problem with divergent sums: you can manipulate them into being anything. Principal values (in this case being 0) have their uses, but they are more about fixing your intuition then about correctness.
I think this is a very good example why Lebesgue integral and measure theory is needed: the "area under the curve" is not always well defined without it.
@@blackpenredpen Ahh, exactly. I thought someone would bring this up. If Ramanujan can tell such insane stuffs which are not even true in the classical sense, we can easily say this is zero as well.
Indeterminate does not imply divergent! They are two different concepts. I disagree with you when you say that you need only show the left hand integral goes to minus infinity in order to conclude the overall integral is divergent. When you do both parts you get infinity minus infinity which is INDETERMINATE. Indeterminate does not mean infinite. It just means you have to go back and try to solve the problem another way. And when you do, you get zero.
you can't subtract infinities while they have different or unknown 'speed of increasing' relative to each other, but these ones behave theirselves the same way #teamzero
We can (superficially) make the integral approach -1/12 by letting the left integral's upper limit approach 0 exp(-1/12) times faster than the right integral's lower limit
By Riemann's definition, you take the limit of the sum of rectangles deltaX*f(c), where c lies in its corresponding interval [a,b]. If we choose c=a or c=b there will be a point (x=0), where the function is not defined, so the integral is divergent. But if we choose c such that a
That is not correct. The Riemann integral also includes in its definition the necessity that that it converges for *every* partition, not just some specific arbitrarily chosen partition, like in your case. You acknowledged that it is divergent for at least one partition, so the definition tells you the integral does not exist.
Your question is wrong, since it is undefined in zero you can define the integral in multiple ways, the way you broke it into to limits is undefined (you cant add/subtract divergent limits we "do" it just to get a general "feel" of the limit. A proper definition would be lim b -> -0 [integral from -1 to b of (dx/x) + integral from -b to 1 of (dx/x)] and it is clear that it is zero. I repeat, you can't break it into to limits because each of them is divergent and if you define it as such you can't add them, you can't add or subtract infinities, that's just scratchwork before we actually calculate the limits.
I think a geometric proof would be needed to solve this completely. Something along the lines of proof by contradiction using geometry. We know the graph is anti-symmetric about zero. An integral like this is the area under a curve. Anything other than zero would contradict the fact that integrals are the area under a curve.
Actually both are wrong. OP is wrong for the reason you explained very well. But you are also wrong because you can't just pick symmetric bounaries when you feel like it just to make it converge. If you have any doubt when using these technical methods, go down a level and look at the the Riemann sum directly or use one of the criteria for convergence (e.g. Cauchy criterion, Darboux criterion..)
It would be a contradiction to assume it was non-zero due to symmetry. It could be an infinitely large square, but you can always divide the square in half.
Your limit is actually two limits a and b, tending to zero from + and -. Ony when a equals b do the two areas cancel, therefore for any particular a, a decreasing b can give you a divergent area - the integral is divergent.
This MUST be zero, right?? If the integral is interpreted as an area in the algebraic manner, these two pieces are EXACTLY THE SAME. It doesn't matter that they diverge, as they diverge in the same way. I mean, a rotation of one set, gives you the other set, so they have the SAME cardinality. If you add the fact that you associate a sign to one of the pieces, then they cancellate each other. I see this as a notation/procedure limitation rather than a actual conclusion. Please, BPRP, bring some abstract algebra to solve this problema. Where is Dr. Peyam? This has to have a solution! #teamzero
Amahury Diaz I know that, it's just that these sets are essentially the same but with different sign, so it feels that the divergent solution is due to a limitation of the approach. I have that feeling that an abstract algebra approach could solve this to be zero.
Amahury Diaz but that’s because when you evaluate those limits they converge to different values. in this case as we evaluate the limit ∞ - ∞ goes to zero for all values
Amahury Diaz that’s kind of like saying the lim x-> ∞ of 1^x =/= 1. of course we know that 1^∞ is an indeterminate case, but if we evaluate the limit for this specific function 1^∞ does equal 1 for that case
I remember getting into an argument with my calculus teacher over a cancellation I did in one integral problem that was considerably less trivial than this, but still had the basic idea of a fraction with a polynomial denomination resulting in a vertical asymptote, and I flipped the result on one side of the asymptote onto the other side and added the two to make the infinities cancel each other out and leave a finite result. I believe they told me that they understood what I did, but this type of cancellation was not allowed because it could lead to nonsense answers if applied to less well-behaved functions, so the correct answer was that the integral diverged, and I wasn’t given credit.
BPRP, initially I thought this was a silly video but after reviewing Cauchy Principal value, I'm beginning to realize just how well thought-out your videos are. This integral has to do with the strength of the conditions for the theorem. According to my complex analysis textbook (John M. Howie, Springer SUMS, pg. 14 section 1.7 Infinite integrals), The weaker theorem is exactly what you just showed in the video, the limit of the sum of the two integrals, but the stronger version of the theorem says that the integral exists ONLY if the two separate finite limits lim(e->0+) integral(a to (c - e)) f(x)dx AND lim(d->0+)integral( (c+d) to b) f(x)dx exist (sorry for shitty notation, also a = -1, b = 1, c is some value in between, f(x) = 1/x) on an interval a
Umm, Who? Wow, what a compliment!!! Thank you so much! I did show both in this video and the rest is left to the comment section, which I AM having tons of enjoyment of reading them.
The function f(x) = 1/x is the textbook example given to introduce divergent improper integrals and Cauchy principal value. It's a nice presentation to be sure.
Sorry yeah you did, now I realize that's why you put the "gap"! My bad ahaha! See, just another reason why your videos are brilliant, you incorporated the "epsilon"-proximity (arbitrary gap "near" infinity) without scaring people by using the actual hardcore rigorous definition for the argument. Please keep doing videos, they are really helpful! You're the best! ^_^
You aren't dealing with an arbitrary 0- and 0+. The same X applies to both by definition of the initial integral. This absolutely links both sides of that resulting equation. It wouldn't if both sides of the equation were derived from unrelated variables or functions, but that isn't the case here. If there is no X for which evaluating the area to the left of the origin for that X is not equal to the area to the right of the origin for the same X, why does that not allow us to conclude that both infinities, in this special case, are absolutely related and inverse of each other for any evaluation of points along their graph sharing the same X, and therefore accept that the limit of this process conforms? This feels like saying the limit of the function y = 1 at infinity isn't 1 even though the value was 1 at every single point along the graph. I know you've shown why it's dangerous to rely on intuition when dealing with infinities before, but this case just defies intuition so heavily and without an actual application or example to contradict that I can't accept the conclusion. If we were to assume it is 0 like all intuition says we should, what breaks down? What proof by contradiction arises from such an assumption? If you could show me where assuming this is 0 begins to lead to conclusions that violate other proven mathematical structure, then I would be convinced.
Let J = ∫[-1 to 1] 1/x dx Take n ∈ ℕ fixed ( n ∈ {1,2,3,…} ) Then we can express J as: J = ∫[-1 to 0-] 1/x dx + ∫[0+ to 1] 1/x dx = lim k->0+ { ∫[-1 to k-] 1/x dx + ∫[nk to 1] 1/x dx } = lim k->0+ [ln(k)-ln(1)+ln(1)-ln(nk)] = lim k->0+ [ln(k/nk)] ==> J = ln(1/n) for any n ∈ ℕ Conclusion: u said the integral is 0, and I’ve said the integral can be made to be whatever value u desire. Contradiction - the integral is well defined and should only output 1 value, not infinitely many And the limits need not be tied together as at the limit, my construction above still covers all the area under the graph over [-1,1] and is therefore valid
All calculus rules aside, we *know* that the positive area and the negative area are exactly equal but opposite, thus their sum is 0. This simply indicates that there are some problems that the tools of calculus aren't sufficient for.
If you think about it, the infinity on the left side of the integral is the same as the infinity in the right side of the integral even though it is infinitely many digits, but it is still the same thing, therefore it cancels out. #TeamZero
Usually, when this appears in physics problems, it is zero by symmetry -- for every point on the +x axis, there is a point on the -x axis with the opposite value. However, as demonstrated, there are ways to take the limit in which it becomes +/- infinity, but in physics problems that would correspond to an obviously distinct physical situation if it happened. When doing physics, one has to think carefully about the physical situation in a problem when potentially ambiguous limits are encountered, in order to know what the true answer should be. Stuff like this happens a lot in Quantum Field Theory, for instance.
Got asked this one on my PhD oral final. I simply discussed the issue of what happens at zero. Since I had used the Cauchy Principle Value in several parts of the work, which involved both analytic and computer solutions to the time dependent transport equations, I passed with a few chuckles from my committee. By the way, I returned the favor on several other students later.
Why can't you use algebra to find that it's 0? You'll get -ln|0|+ln|0|=ln|0|-ln|0| Which obviously equals 0 The only thing that stops this is to make a difference between 0plus and 0minus but in regards to ln|x| they are symmetrical and should be regarded as equals
It diverges. You can prove this with e.g. showing it doesn't meet Cauchy criterion or by comparing to a suitable series. In fact, if I remember correctly: The integral of 1/x^a converges in (0,1] for a1, which is why it never converges on the entire (0,inf) for any value. I may have got it mixed up though.. :-P
I'm an year late but the controversy seems to be on whether it is indeterminant or divergent. Well the simple explanation is that it's indeterminant only if we are dealing with the same variable in the limit on both sides but it depends on the problem solver to choose different variables approach 0 from either side(i.e. a, b) and you don't have any relation between the two. And therefore he claimed that the integral diverges. After all we are including a point that is not in domain of an unbounded function.
The limit is undefined. When you end up with "infinity-infinity", you are trying to combine two separate limits of two variables into one. This cannot be done as the rate at which each limit approaches the singularity may be different. It is only indeterminate if that result happens within the same limit. Because it's with two different limits, the answer is instead undefined.
@@KokomiClan Just because the rates *can* be different doesn't mean they have to be. In this case the solver gets to choose their variables, and makes a mistake in choosing two distinct arbitrary variables with no relation to one another rather than one variable that forces the integration to be symmetric (or at least an assertion along the lines of let b = -a or something similar). Let a -> 0+ and then integrate over -1 to -a and a to 1. The value of a is still arbitrary, but by now we get to conclude that the resulting infinities are indeed symmetric and therefore cancellable.
It's not two limits. It's one limit, and the same limit with a -. A thing - the same thing is zero. If the definition does not work, the definition is wrong. It's not two different apples. It's an apple, remove it, no apples. It's literally a mirrored image, point by point. It's not subtracting two infinites, it's subtracting an infinite from itself. From some definition, lately, women may have a penis. Definitions may change language, not reality.
What is exactly the definition of 0+ and 0-? I say it would make sense to define it as: lim d-->0 of d and lim d-->0 of -d. And then you have: area = lim d-->0 ln(d)-ln(1))+ln(1)-ln(d) = lim d-->0 0 = 0
Really late to post here but I just want to add that you don't know if 0+ or 0- is bigger because you defined it that way. If you did it such that one integral goes to (0 - b) while the other goes to (0 + b) and then take the limit as b approaches 0 you preserve the symmetry of the integrals. It still means that b can get arbitrarily small and both halves will cancel each other out
While that's true it doesn't really help, because it's not true that lim x->a (f(x)+g(x)) = lim x->a f(x) + lim x->a g(x) in general, and the existence of an improper integral requires the latter.
For me the divergent explanation doesn't make sense. You can operate the ln before subtitute it, obtaining that ln|0-|-ln|0+| its in fact 0. Its like saying that the lim x->inf of x-x diverges because you substitute before operate the "x" and you obtain inf-inf that is inditerminated form. No sense for me.
#TeamDivergent The theorem states: the integral of any odd function on an interval, which is symmetric with respect to the origin, equals zero, if the function is continuous on that interval. The function 1/x is not continuous on [-1, 1], thus the theorem cannot be applied.
Who cares about the theorem when both of the functions behave the same whether x is positive or negative. The infinites are symmetric and are the same value. This integral must be 0.
You're all wrong. The area under the curve is complex. EDIT: The explanation: The integral of 1/x is lnx, so we have ln(1)-(ln(-1)) = 0 - ln(e^iπ) = -iπ. Hence proved. EDIT 2: This essentially violates the fundamental theorom of calculus, since the function isn't continuos. But it Is always fun to venture into the complex world.
His graphical explanation of the 0+ and 0- corresponds to using two separate variables for the limits, such as ln|b|-ln|a| (rather than ln|b|-ln|b|). While intuitively it would be nice to call both limits the same variable as you do, I'm not sure if that can be rigorously justified.
If you want to define the integral as the area under the curve, then I think it's not rigorous, why you should integrate from -1 to -b and from b to 1 instead of from -1 to -b/2 and from b to 1? At the limit b->0 the two options define the same area, so the result must be the same for both, but it's not.
math.stackexchange.com/questions/2665679/cauchy-principal-value-and-divergent-integrals/2665708#2665708 this action(of setting both sides to have the same limit) is called Cauchy PV
Roger Balsach why not b/2? Because that's inconsistent. The goal behind the limit is to capture everything except the point zero. If you're already as close as you can possibly get, or taking a limit, it doesn't matter if you divide by 2.
It is very simple. It is a matter of definition. Unfortunately, the integral of a discontinuous function with a single discontinuous point in the middle of the integration interval is defined to be the sum of the of the limit of the 2 integrals to and from the discontinuity point provided that each of the 2 limits converges. If anyone of the limit diverges, the whole thing diverges. Of course, one can change the definition from the "sum of the limits" to "limit of the sum", in that case, we can show that this integral is 0. But this changed definition is not the standard textbook definition. If we really like 0 to be the answer. We will need to use a non-textbook definition.
This is a great example of why Riemann integration on open intervals doesn't work. However, in the video the problem is represented as a sum of two integrals: from -1 to 0 and from 0 to 1. Riemann integration works with closed intervals, so that corresponds to [-1, 0] and [0, 1], and the intervals actually overlap. Another mistake is using a limit to calculate an integral at the point where the original function is not actually defined. We're actually trying to calculate a sum of integrals over 3 intervals: [-1, 0), [0], (0, 1]. We know that the function 1/x we're trying to integrate is smooth and finite on the intervals (-∞, 0) and (0, ∞), therefore the integral of the function must also be smooth and finite on any subset of those intervals, so for any real b > b' > 0, integral over [-b, -b') has to be finite, and equal to - integral over (b', b]. This can be applied recursively - no matter how small b is, the integral over (0, b] can be written as a sum of an integral over (b', b] plus an integral over (0, b']. This gives us two infinite sums a = a_1 + a_2 + a_3 + ... and a' = a'_1 + a'_2 + a'_3 + ... such that a_1 = -a'_1, a_2 = -a'_2, a_3 = -a'_3, ... This means a + a' = 0, therefore the integral over [-1, 1] equals 0 and does not diverge.
I'm going to give you the classic response from a mathematician: It depends on your definition of an integral. The most useful definition of an integral is probably the Lebesque integral. By this definition, the integral diverges.
8:06 notice how you’re “subtracting” infinity from infinity in one step, but the step before, you’re subtracting ln|0-| - ln(0+). but that’s just a shorthand for the limit as x->0+ of (ln|-x| - ln(x)). but we know that |-x|=x whenever x is positive so you get the limit as x->0+ of (ln x - ln x). ln x subtracted from itself is 0 for all positive x. therefore the integral is 0 and i strongly disagree with anyone who says it shouldn’t be 0.
Your arguement is not convincing when you chose to arbitrarily give different values to "0" on the negative and positive sides of x-axis. You don't have the luxury of treating the two parts of Y=1/X equation differently! The two parts of this equation approach Y-axis asymptotically (divergently.)
That is an odd function from -a to a therefore it’s zero right? These comments are great btw and also I realised that the theorem above doesn’t hold in this case because the function isn’t continuous on [-1,1]
There is a continuous one-to-one correspondence between all points left of zero and right of zero, so everything must cancel. Imagine an excluded band inside of -1 to +1 centered around zero that is shrinking. If we add the left-side integral from -1 to -f to the right-side integral from +f to +1 (where 0
The issue is, the way improper integrals are classically defined, the "f" used as the limit for the right-side integral doesn't relate to the "f" used as the limit for the left-side integral. It is not defined to be a symmetric band. There is a way to make it symmetric though, (en.wikipedia.org/wiki/Cauchy_principal_value), but again, that differs from how improper integrals are usually defined.
#TeamZero This is where I'd favor the sensible answer (0) and say that improper integral as a technique just breaks down. Diverges is just so unsatisfactory of an answer.
Define F(c) = lim b -> c Integral from -1 of -b of 1/x dx + Integral from b to 1 of 1/x dx, c in (0, 1). Then F(c) is zero everywhere. Notably, as c -> 1, F(c) = 0, so F(0) is a removable discontinuity whose limiting value is 0.
"If one is piece is divergent then the whole thing diverges." Let's try that, ln|x| diverges as x approaches infinity. So I do the limit of ln|x| - ln|x| as x approaches infinity. Both pieces diverge but the whole thing converges, so no, your reasoning doesn't quite work out. But, yes I agree it does diverge.
Btw, when I said that, that was meant to be a quick way to draw the conclusion. I did provide another reason why it diverges at the end with the graph.
I have heard the infinity minus infinity argument and the graphical argument, but neither helped me understand it, because the symbolic answer does not make sense intuitively. My reasoning therefore is that if you were to numerically integrate this on a computer with infinite memory, a Turing Machine, then it would return different answers each time you ran the calculation, i.e. one time you get 0, another time you get 243534574, and another time you get 12. Its's a little wishy washy since we don't have true turing machines, but when I learned this, I needed the numerical argument for me to get it.
MuffinsAPlenty That seems correct to me, but it's not really intuitive, and you certainly wouldn't have convinced me with that a year ago. BPRP was trying to answer this in terms that would make sense to students, but this is something that will never make sense. So I would say go for a rigorous proof instead, considering that will make just as much sense, and it's not like BPRP hasn't done rigorous proofs and such before.
What if you take the following limit of integrals: Limit as c goes to 0 from the positive side. First integral: -1 to -c. Second integral: c^2 to 1. You'll get 0 - 2 ln (c) + ln (c) - 0 Simplifies to - ln (c) for any small positive c. In the limit as c goes to 0, this is infinity.
I think you should've evaluated the integral on "b" and left thrlimit for the end, so you couldve saved the indetermination. Limit b->0 [Ln (b) - Ln (b)] = Ln (b/b) = Ln 1 = 0 You can do this in this way because the function is symmetrical by turning the axis 180°
Approaching zero from both sizes (positive and negative) in an absolute value is equal, therefore we can simplify to one limit, either positive or negative, and therefore simplify it to 0.
#TeamDivergent. I did the limits on both, but on one I used epsilon, and on the other k*epsilon. So, if the answer is finite, it must not depend on k. Then, I ended with lim epsilon -> 0 ln|k*epsilon|-ln|epsilon|, which, by the log properties, throws ln|k|, which contradicts the suposotion to the integral being finite. So, it does diverge.
usuario0002hotmail I am not at high math level,,, only took calc 1 ,,, but seems divergent because both are sides sum to same, (one side neg Infinit , other side pos Infinit,),, , yet never zero. Because for he fact that they never actually touch. The true origin / 0 , they diverge. ( I'd argue in a real world In real world I'd be 0, both are infinite and cancel out) But !,,, Since They're is a gap between both 0- and 0 and also gap 0 to 0+. #DiVERgEnT 🔥💯🙏🙏🙏🙏
You did not show, that it cannot be finite. Just that the value can be made into any real number ln (k) by choosing any positive k you like. "Real" divergence is also possible, eg by using nonlinear limit variables (b, b^2). I don't really like the word divergent for indeterminant expressions ^^ sounds too much like "approaching infinity". I know that's stupid :D Edit: i think the original commentator also took "divergent" as "going to infinity" and tried to show that ;p
You can make the left hand integral ln(0+), which is the same as the right hand integral. You can then say it is the limit as t-> 0+ of ln(t)-ln(t). This is simplified as limit as t->0+ ln(t/t) which is ln(1), so the integral does equal 0
Man, you're incredible, you explain so good... With you I'm learning all the calculus because in school I haven't learnt it. Thanks, man. #TeamDivergent #YAY Salute from Spain.
I got a similar question on a test and said 0. It was the only question I missed and I got into a debate with my teacher about it and got the points back lol
If you approached this problem in the manner of Riemann sums, for every x value on one side of the y-axis, the opposite x value will return the same "area" of its corresponding rectangle, just negative. This is true for any x value on either side of the curve, no matter how small the rectangles get, so, logically, it'd make sense for the integral from - 1 to 0- and 0+ to 1 to be 0, as whatever 0- or 0+ is, its counterpart can be just as small, just opposite.
you can't do Riemann sums because for a function to be Riemann integrable it has to be defined over an interval (this function is not defined on all [-1,1]) and it must be bounded on this interval (this function is not bounded) so the only way to approach this problem without using more advanced theories of integration is to use the limit, which gives 2 integrals that MUST individually converge, but they do not.
Zone Goku Essentially with principal value you're taking the integral from - 1 to - z and from z to 1 then taking the limit as z goes to zero, but there's no reason it couldn't be say taking the integral from - 1 to - b*z and from a*z to 1 then taking the limit as z goes to zero, the should give us a value of ln(a)-ln(b).
But what if you calculate the indefinite integral first and then evaluate from -1 to 1? You always get 0. Only when splitting the integral at the asymptote do you get an indeterminate value. Also, using the limit definition of an integral, even when you do split the integral at the asymptote, you still get a limit of 0. Because infinity minus infinity still means something in limits.
you cannot integrate between -1 and 1 because for a function to be Riemann integrable (the most well know and common integral) it must be defined in the whole interval (this function is not), bounded (again this function is not) and also another condition that i can't really type on here but already the first two aren't satisfied. There exists a natural way of dealing with non bounded functions, and it's splitting up the integral in two distinct integrals and to take a limit. Both of these integrals must converge for the original integral to converge. Maths doesn't really care about intuitive results, but only rigorous definitions and what follows from them, so talking about "symmetry" or "intuitive results" doesn't work very well
#TeamZero is, sadly, wrong, thought it feels so right... Here's more proof Let's apply the same reasoning to the the integral from -1 to 1 of 1/x^2 1 ∫ 1/x^2 dx = ? -1 If we don't pay attention, we just use the anti derivative and get 1 -1/x ] = -1/1 - -1/-1 = 0 -1 We can already see that this is wrong, but let's keep going Trying to integrate over dx, split up the improper integral into two parts, one from x=-1 to x approaches 0-, another from 0+ to 1 a 1 lim ∫ 1/x^2 dx + ∫ 1/x^2 dx a-->0- -1 b b--->0+ The anti-derivative of 1/x^2 is -1/x, so when we plug in 0 like you did at 6:28, we get that the first integral diverges: 0- -1/x ] = -1/0 - -1 = undefined -1 Likewise, the second integral diverges, and as a result: ***Using **#TeamDivergent**'s reasoning, the integral from -1 to 1 of 1/x^2 is undefined. Which is correct*** Now, lets approach the integral a different way. Instead, we'll integrate over dy. I can't draw a picture here, but the are bound by x=-1, x=1, y=0, y=1 is within the area of this integral and has an area of 2 square units. So, we only have to integrate above y=1. y = 1/x^2 -------> x = ±√1/y ∞ ∞ ∞ ∞ ∫ xdy = ∫ ±√1/y dy ===> ∫ +√1/y dy + ∫ -√1/y dy 1 1 1 1 The anti-derivative of √1/y is 2√y. So, we get ∞ ∞ 2√y ] + -2√y ] 1 1 Which we can already see diverges, because we have √∞ The same logic applies to any 1/x^n. Therefore, it's important that we always split the integral at any break in differentiability, and #TeamDivergent is right
+Oleksiy Bazhenov Just a minor correction which does not affect your argument, quite the opposite: If we don't pay attention, we just use the anti derivative and get 1 -1/x ] = -1/1 - -1/-1 = 0 -1 Or we don't pay attention to the numbers :). The first fraction (-1/1) equals -1. From that, we subtract 1 (-1/-1). -1 - 1 = -2, not 0. This makes things even weirder, because we are adding positive numbers and get a negative result (Zeta function anyone?). Since it contradicts your second solution using #TeamZero logic (integrating using dy), it further proves that #TeamDivergent is correct. Unless we might miss something. Sincerely, ZfE
The problem with 1/x^2 is that both halves of the integral are positive. Adding +inf to +inf is always going to diverge. In the case of an odd reciprocal power we are adding -inf to +inf which only potentially converges - to show it actually converges we need to show both sides approach the limit at the same rate. One way to do so was shown in my separate post for #TeamZero
Wow, great explanation! :) And I have similar example. I have to integrate -e^(-x^2/2)/(x^2) from -Inf to Inf, but this function is undefined at x=0. And with integration by parts(u=e^(-x^2/2), v=-1/(x^2)), we get sqrt(2Pi). However, if we integrate from 0 to Inf, it should be half of sqrt(2Pi), but in this time it diverge! So I'm really confused about this. I need to calculate this integral to solve my problem, and if this integral diverge, then I have to throw away this integral and find the other way. lol In fact, I have some sequence, and my integral is -1th term of that sequence. Also, 0th, 1th, 2th, ... term are all sqrt(2Pi). But -1th, -2th, ... term are all "undefined integration" like my example. So,, I hope that (integral of -e^(-x^2/2)/(x^2) from -Inf to Inf) is sqrt(2Pi), but after I read your comment, I think it may diverge. ...... What is correct answer?
@@trueriver1950 You cannot actually prove that the limit on both sides converge to their respective values at the same rate. What you are doing is simply calculating the Cauchy principal value.
What if you took the integral of the function (x / (x^2 +a)) with respect to x, for positive constant values of a. Then, taking the limit as a approaches 0+ should give you a consistant value equivalent to the cauchy principle value of 0 for the - 1 to 1 bounds. Of course, since the limit a approaches 0 - does not exist, you can still argue that the integral diverges as well.
Alexander Nenninger In the second case we have S(n)= [-1;1/n] u[exp(-x);1] but 1/n and exp(x) are always >0 so I think that exists an interval (0;ε),ε>0, that it's considered two times; in other words the intersection between [-1;1/n] and [exp(-x);1] ≠0
My problem with this is that the integral from -1 to 0-E plus the integral from 0+E to 1 equals zero, even for arbitrarily small consistent E. The reason that the integral fails to converge is only the point x=0, a fact which I feel like this video glosses over. If you take the limit as E approaches zero but remove that one point, the integral converges, but that’s not an equivalent integral. The indeterminate form comes from the fact that the area of the “rectangle” with zero width centered at x=0 is not necessarily zero, as the rectangle has infinite height (0*infinity is undefined)
(Also, I realize I was vague as to why the altered integral converges. If you take the limit of the Riemann sum that represents the full integral, separated into its two parts, with the terms grouped so as to cancel one another, only one term in the sum is undefined: the infinity minus infinity term. Removing the x=0 point changes these from rectangles of infinite height to rectangles of arbitrarily large finite height, and thus cancelling one another out.
Another way to understand (using definition): If we divide [-1, 1] into N pieces equal in length and N is an even number, Ci is the midpoint of each piece. The sum of all f(ξi)·Δxi = Σ(1/Ci)·(2/N) will always be 0. So the same is true when N goes to infinity. HOWEVER, if N is an odd number, for every piece which doesn't contain 0 we choose their central point. For the central piece (the only piece contains 0) we choose one point bigger than 0 from it. We call this point Q. The sum of all f(ξi)·Δxi will be (1/Q)·(2/N) = 2/(Q·N). One thing we can know is Q can be any value in (0, 1/N]. Thus (Q·N) can be any value in (0, 1]. When N goes to infinity the sum can be any value in [2, +infinity). Thus by definition the integral of 1/x from -1 to 1 cannot exist. I hope this is more explanatory.
I like this answer a lot! And of course, the definition of an integral doesn't require your sample points to be chosen in the same location for each piece. So, playing around with that, with careful selection, you can probably get _anything_ you want, including in (-infinity,2) as well.
My interpretation is that the integral is certainly divergent, but the area the integral is supposed to represent is 0. Thinking about the way the integral defines the area underneath a function as many tiny rectangles, we see that as the number of rectangles goes to infinity there is only a single rectangle which doesn't cancel with a mirrored one: the one at x=0. Now for the integral this has disastrous consequences but in terms of intuition about area it is an infinitesimally small strip that can be disregarded. Therefore the area is 0, but the integral is divergent.
Alright, as a proud member of #TeamZero, here is my rebuttal. When we evaluate ln(|x|) [-1,0-], we get 0-lim[x->0-] ln(|x|). When we evaluate ln(|x|) [0+,1], we get lim[x->0+] ln(|x|). If we dont evaluate the limits yet and add them we get integral from -1 to 1 of 1/x = lim[x->0-] ln(|x|) - lim[x->0+] ln(|x|). Since it was established that ln(|0-|)=ln(|0+|), then this can be rewritten as lim[x->0+] ln(|x|)-lim[x->0+] ln(|x|)=lim[x->0+] ln(|x|)-ln(|x|) = lim[x->0+] 0=0. Therefore, the integral from -1 to 1 of 1/x dx is 0
Int(1/x)=ln(x) And since ln(-1)=ln(1) It becomes ln(1)-ln(1)=0 Your way ends up with not defined, therefore you can't say if it diverges or not. Just like you are able to create a not-defined almost anywhere. For example before using l"hopital.
blackpenredpen said that the two areas (left side area and right side area of the curve) are not equal because -------> +(0)[very little right side of zero] is not equal to -(0)[very little left side of zero] can someone tell me why? i think this is the only reason why the area is not zero
Both sides grow/shrink to ±∞ at the same rate, so personally I wouldn’t have two separate limits for the two integrals but rather: lim_{b→0⁺}(∫₋₁⁻ᵃ + ∫ₐ¹), where both integrals are of 1/x dx. If we computed it like this, it would indeed turn out to be 0, and since infinity is reach from both sides at the same rate I think this is how we should do it.
But what if you, before you work out those 'ln' things at the end, leave those and simplify it using those. That way you'd get ln|0-| - ln|-1| + ln|1| + ln|0+| = ln|-1| + ln|1| = 0. That way you don't have to deal with an 'infinity' and it is simply 0
However, in the explanation that leads to a divergent series, the ending result after solving the logarithms is infinity-infinity. But infinity-infinity is an indeterminate form like 0/0 and can therefore take any value. The trick, as with the indeterminate form 0/0 generated by computing derivatives, is to find which value is the expression really approaching, it's the essence of calculus. And since both infinities seem to grow at the same rate and both infinities seem to get closer to zero at the same rate, one could then argue that the expression actually reaches zero. This is the reason why I chose to believe that the integral is equal to zero.
#TeamZero taking the integral of this by not equal spaced intervals doesn't work, or at least to me, if they are not equally spaced, then we are always missing a bit "almost zero" here but not there, or an infinitismal. An integral is taking the ENTIRE function, not almost this function, but it's still okay to do equal intervals to get all values except the singularity. 0 is the Cauchy Principal Value of this integral, the main sensible value.
The theorem for odd functions holds whenever the integral exists. Continuity is not necessary but sufficient (continuous images of compact sets are compact, compact sets are bounded, bounded piecewise-continuous functions on compact sets have an integral). The theorem already holds for, e.g., the integral of x^-3.
Assume deltaX as a distance to 0. Use -deltaX and deltaX respectively as the boundaries of the integrals. Solve for the area as deltaX goes to 0. Wouldn't this yield a proper answer? You still approach 0 but with an equal distance from 0 at both sides.
Here's proof that it diverges , we can first of all devide the integrals into two parts one from -1 to 1/n and the other from 1/n to 1 under these conditions we get ofc 0 , however if we change 1/n by 1/2n if it were convergent we would still get 0 as n goes to infinity but infact we get infinity and so we have shown the existance of two possible values for the integral wich is ofc impossible and therefore it is divergent
Destroctive Blade But if we integrate 1/z over the upper half of the unit sphere in C, since 1/z is Holomorphic in C\{0}: integral_0^1 exp(-i (π t)) d/(dt) exp(-i (π t)) dt = 0 #teamzero 😁 Under some other comment I have shown that it's (infinity)
If you let b go to 0 *simultaneously* for the two limits, it will actually be 0, for instance A = \lim_{b\to 0} \ln(1) - \ln(b) + \ln(b) - \ln(1) = \lim_{b\to 0} \ln(b/b) = 0.
#TeamZero - I like so much your videos. Actually, there are infinite that diverges faster than others. For example, the function y = x^x and the function y = x. The both functions diverges when x comes to infinity, but y = x^x diverges faster. Also, we know that there are more real numbers than natural numbers, even when there are infinity real and natural numbers. In this case, it makes sense that the both sides comes to infinity in the "same velocity", what lead this integral to 0. Well, that is just my positioning, but I am just a boy in math, I hope learn even more if your videos. Thanks!
The definition of a convergent integral states : int from a to b of f converge if and only if for all [c,d] in [a,b], the integral over [c,d] converges. However, int from 0 to 1 of 1/x diverges. Thus int from -1 to 1 of 1/x diverges. Simple as that.
Does it make a difference if you write one limit as epsilon->0 of the sum of the two integrals (i.e. from [-1,-epsilon] and [epsilon, 1]. In the end you have the difference of two logs which is the log of a ratio which equals 1 hence the answer is 0?
Yes I know. But as you say yourself this is something else, namely the cauchy principal value of the integral (or whatever you would call it in english) and not the value of the 'normal' integral.
OK. So this strikes me as something like the alternating harmonic series in that we have a conditionally convergent function as long as you evaluate it in a particular way. If you add the terms of the ahs out of order, you can make it equal anything. If you evaluate the given integral as a left-to-right Riemann sum then you'll soon get to infinity and all bets are off.
How about the next best thing? Define a function f(a) = integral from -1 to -a of 1/x plus integral from a to 1 of 1/x. Then f can be continuously extended as f(0) = 0. That way you can tell people they are wrong about the integral while also satisfying their intuition.
The first integral you wrote does not exist. The reason is that 1/x is not continuous at x=0, so it's not Riemann-integrable at that point. As for the split integral with lateral limits, it diverges if you separate the limits, but converges to 0 if you consider a single limit of (int - int) #TeamDivergent
I have a suggestion for strengthening your argument at the end: by letting 0^+ be +a and 0^- be - a*b, with arbitrary positive b>0 and letting a>0 approach zero, you can show that the combination of the limits approaches ln(a*b) - ln(a) = ln b. That shows the indeterminacy in its pure form: by being able to choose b>0 arbitrarily, the integral can take any real value. The number b can be interpreted as the relative rate at which 0^+ and 0^- approach 0. Taking b = 1 would give you the Cauchy Principal Value of the integral
Here is my take... when working the improper integral, we get (before you simplify ln|0| as -oo) ln|0| - 0 + 0 -ln|0|... we know that ln|x| converges to -oo at 0 from both sides (absolute value functions are symmetric), so lim x->0- ln|x| = lim x->0+ ln|x|. Now we get into semantics. I argue calling ln|0-| = -oo is a generalization - that is, we lose information by representing it this way. However, if the above limits are true, then there is no need to generalize them this way. ln|0| - ln|0| = 0 because ln|0| is always = ln|0|. This would mean we could simplify your equation to 0. Note that oo - oo is still undefined because many expressions “evaluate” to infinity, like 1/0 and 2/0. That said, by my argument, 1/0 = 1/0 ≠ 2/0 because if we had the means of solving this expression we would get the same result using the same numbers. So, in short, I claim by semantics that calling ln|0| = -oo is a generalization (we lose information), and that ln|0| - ln|0| = 0 because ln|0| would consistently map to the same answer, if we had the means of calculating it, because the input is the same every time. Thoughts?
I would note that the idea of integration is to accumulate all of the values of a function as an infinite sum, the notion of infinity is not a value rather a symbol that represents the continuing of numbers to no end. In a similar fassion to the problem here, we can, for example, integrate the gaussian from "negative infinity" to "positive infinity" which really just represents the total value the function has as a sum, if we were able to evaluate infinitely many terms. An integral is more so asking, what does this infinite sum APROACH as I use smaller and smaller portions of the domain 'dx' to accumulate a function, in a similar way, we know that by nature of 1/x, all finite inputs cause it to act as an odd function, hence, as we choose smaller and smaller portions of the domain to examine and accumulate, the sum doesn't exist, as 1/x as x approaches 0 leaves a figuratively thin sliver of infinite height in the center of the integral, which even as dx gets smaller and smaller, still gives an infinite accumulation. #teamdiverge
FYI: limx->0+ doesn't mean approaching to a number a little bit more than 0, since for that matter you can also approach that from the left. However it means the x approaches (but never equals) exactly 0 from the positive side. Although I do agree that the integral diverges because the bottom and top limits of integration are independent of each other
#teamzero I mean just look at the first graph he made. Like, it's obvious the area cancels out. Just to be clear I haven't studied integrals thoroughly so what I say might not be mathematically rigorous, but based on my intuition the integral should be 0
That might be true for all the other values either side, but what happens at 0? Both sides are trying to claim that piece of area to cancel the other one out, and they can't both have it, and both cancel out.
Horep I'm not quite sure I understood what you mean there. If you are talking about the slice at x=0, then the areas can't split it as it's not in the function's domain.
This improper integral is also the solution of -ln(-1). This is because the natural logarithm function is the integral from 1 to n of the function 1/x respect to x. The negative natural logarithm is formed when you switch the upper and lower bounds. Using this rule, ∫_(-1, 1) dx/x = -∫_(1, -1) dx/x = -ln(-1). Since e^(pi*i) = -1, ln(-1)=pi*i, -ln(-1)=-pi*i, and the solution to this integral is -pi*i. This is because this integral can be broken into 2 parts: The positive side, and the negative side. ∫_(0, 1) dx/x = infinity, and ∫_(-1, 0) dx/x = -infinity. Therefore, the integral brings us to an infinity-infinity paradox as now we can see that infinity-infinity can create imaginary numbers as well. This is why infinity-infinity is indeterminate without using 0/0, 0*infinity, or infinity/infinity.
Exact. You can prove the integral( 1/x ) over 0, 0 is 0 ! :D But Zero is a more comfortable value to accept. The shame is that the mentor ends the video with "take a stand", like if we could choose the result we prefer in Math !
At 0, the function approaches infinity and infinity-infinity is undefined. It gives a feel that due to symmetry, the areas below and above x-axis cancels out each other to give 0 but actually its not. It is undefined.
If you instead take a complex integral, using any path not containing the origin, then the integral from a to b of 1/z dz ln|b/a| + (angle(b) - angle(a))*i. This means that the real value is always zero, but the imaginary component is (2*k+1)π, depending on the direction and number of times the curve loops around the origin.
Maybe we should accept that it is unknown rather than be on team zero or team divergent. My intuitive opinion is but I don't know the mathematical proof is if both sides diverge at the same rate in opposite directions then they would cancel each other out. If one side diverges more or quicker than the other than the whole thing diverges.
Ayman Abdellatief Believe me We integrate over the upper half of the complex unit sphere: integral_0^1 exp(-i (π t)) d/(dt) exp(-i (π t)) dt = 0 😁 Edit: BS calc. It's already way too late
It does not have an improper Riemann or Lebesgue integral, but it does have a Cauchy principal value of 0. The Cauchy principal value splits the integral into left and right (proper) integrals, then takes the limit as both of them approach the singularity at the same rate. In this integral, the value is 0 for all δ>0, so of course the limit is too.
Look up Cauchy's principal value. Instead of using 0-minus and 0-plus, use -a and +a (same magnitudes). Integrate, combine results, and let a->0. The limit is zero.
I wish Euler was still around, he'd know what to do
He would win dota 2 the international
Euler would say that the Riemann integral does not exist, but he would also argue that the Riemann integral is a bad integral to use for an exercise like this one, so he would use the Cauchy principal value and tell you the answer is 0 anyway.
@Asyam Abyan damn I’d get that link ASAP
@Asyam Abyan this is the way.
Euler deserved immortality if anyone did
To respect the symmetries in math, I still want to believe that the integral is zero.
After all, the areas to the left and right of zero are basically negatives of each other, so I guess that even after being infinite, they are in some ways equal, so the signed areas may add up to zero.
Being divergent does seem more mathematically consistent and rigorous though TBH.
Reeta Singh same my thinking 😉😊
not necessarily. You can do a completely rigorous way of showing it is 0.
first define L = integral from -1 to -n of 1/x + integral from n to 1 of 1/x
this is also equal by the fundamental theorem of calculus to ln|-n| - ln|-1| + ln|1| - ln|n|
= ln(n) - ln(n) + ln(1) - ln(1)
since n is never equal to 0, ln(n) is a real number and we can do regular operations, such as subtracting
therefore L = 0 for all n != 0 (not equal to 0)
now take the limit of L as n-> 0 (from positive)
in the limit, the integral representation of L becomes
L = integral from -1 to 0 of 1/x + integral from 0 to 1 of 1/x = integral from -1 to 1 of 1/x
but this is a limit, so n is never equal to 0. therefore L is still 0
FINALLY WE CONCLUDE THAT
integral from -1 to 1 of 1/x = 0
one could think that i just did some mathematical trick to justify the equality, but what about calculus?
if it werent for limits, when trying to take derivatives we would be saying "oh but we can't divide by 0 :(". That would be analogous to saying "the integral is divergent :(". But no, you start using something to reason around the problem and limits help. Now instead of saying "it's undefined" we can add a meaningful answer.
The issue is when you define L = , that explicitly assumes there is a solution. Since the integral diverges, its undefined.
You can't assign a value to it.
you said "since the integral diverges"
you can't argue with the proposed conclusion
-infinity + infinity = whatever #TeamWhatever
hmmmyes very well said
infinity - infinity = pi (mathologer video)
#TeamPi
These infinities are more specific than "just being infinity". If you break the original integral down to "-infinity + infinity" you have destroyed information.
It's actually not infinity
I actually agree, because it's indeterminate. Even by summing up the area piecewise you can get any old answer.
So I'm going to throw a wrench in the works here: I know this is a calculus channel and not a real analysis channel, but this function just fundamentally isn't Riemann integrable over the interval [-1,1]. Isn't the best way to express that thus to approach it using the limit definition of the interval and showing the issue via calculating the Riemann sums of the function? That's the core issue here; that the "area under the curve" intuition only works for Riemann integration when the function is Riemann integrable in the first place. If it's not, then the "area under the curve" just doesn't correspond to the value of the Riemann integral. It's the same thing you get when trying to find the integral of a function with Lebesgue measure 0; the intuition of area suggests that the value ought to be 0, but the Riemann integral doesn't get you 0 because it isn't actually measuring area, it's measuring something that, given certain preconditions, is equivalent to area.
That's really the reason why there's so much confusion here: because integration doesn't _really_ calculate "area under the curve", it just calculates a value that, when you have a "nice" function, is equal to the area under the curve, but when you don't have a "nice" function, calculates a garbage value that has nothing at all to do with area under the curve.
I guess fundamentally what I'm saying is that the intuition that the area underneath this curve is 0 isn't _that_ bad, and it's reasonable to see why it arises. And the issue is that "area under the curve" isn't really what Riemann integration calculates when you have a function that acts up. And it'd be worthwhile to highlight that in order to illustrate the distinction between the two.
Wow. This was actually really insightful, never knew that was true for Riemann sums. Good comment!
Are there any alternate definitions of integration that will actually give you 0 for this integral?
You know, I'm actually not sure. Most methods of integration that I know of at minimum require a function f to be of bounded variation over your interval. That means that for some interval [a,b], if you take every possible partition {p1, p2, p3, ..., pn} of that interval (for all integers n), and for each partition calculate the sum of f(pk) - f(p{k-1}) for all k, the supremum of the set of all these sums is finite. (It's a long-winded way to say that the function never blows up, but it's required to be long-winded because some functions blow up in really weird ways.)
The issue with functions that blow up is the same issue you run into with any sum of an infinite number of terms that "should" add up to a finite value; the order that you add things matters, which means that the way in which you decompose the "infinite" portion matters. When you're calculating a finite area that should end up summing to 0, you get nice things like association and whatnot that mean that it doesn't matter how you end up pairing things off, everything will work out in the end. But like usual, infinities screw things up, because the way you pair things off _does_ suddenly matter, and so different calculations of the area, different roads that should go to the same destination, suddenly veer off in wildly different directions. That's why you get weirdness like Banach-Tarski, where you can decompose an infinite set into two copies of the original set.
There might be some more obscure integral in real analysis that I'm not aware of that can handle this well - given that intuition suggests it to be 0 for fairly valid reasons, I'd suspect there ought to be some way of formalizing that intuition - but it would require more advanced analysis than I am unfortunately aware of.
Idran best comment here
Idran Completely spot on! The reason you can get zero out of thus, is because you take a simultaneous limit on both sides of the singularity. This makes it possible to do all kinds of nice cancellations, even in the Riemann sums. However, if you manipulate the limit on both sides just right (e.g. taking -epsilon on the left, and 2*epsilon on the right), it might be possible to construct even more "correct" values. That is the problem with divergent sums: you can manipulate them into being anything. Principal values (in this case being 0) have their uses, but they are more about fixing your intuition then about correctness.
I think this is a very good example why Lebesgue integral and measure theory is needed: the "area under the curve" is not always well defined without it.
If there’s a way to define 1+2+3+...to be -1/12, there must be a way to define this to be zero, too
Yes. It’s called the Cauchy principal value.
@@blackpenredpen Ahh, exactly. I thought someone would bring this up. If Ramanujan can tell such insane stuffs which are not even true in the classical sense, we can easily say this is zero as well.
I can also say π = 3
sin x = x
Nope
Indeterminate does not imply divergent! They are two different concepts. I disagree with you when you say that you need only show the left hand integral goes to minus infinity in order to conclude the overall integral is divergent. When you do both parts you get infinity minus infinity which is INDETERMINATE. Indeterminate does not mean infinite. It just means you have to go back and try to solve the problem another way. And when you do, you get zero.
: )
You can't just say when you do, you get zero! Show your work!
This is exactly the distinction needed to make sense of this problem.
Veexliat takes basically no work
This. Also, you are doing calculus on a non-continuous region of a function, so don't expect a reasonable/meaningful answer.
By the theorem of feelings and irrationality my answer is Zero.
you can't subtract infinities while they have different or unknown 'speed of increasing' relative to each other, but these ones behave theirselves the same way
#teamzero
This feels a lot like quantum mechanics. Something tells me the correct answer is -1/12.
We can (superficially) make the integral approach -1/12 by letting the left integral's upper limit approach 0 exp(-1/12) times faster than the right integral's lower limit
By Riemann's definition, you take the limit of the sum of rectangles deltaX*f(c), where c lies in its corresponding interval [a,b]. If we choose c=a or c=b there will be a point (x=0), where the function is not defined, so the integral is divergent. But if we choose c such that a
looking for this! taking a calc one course now talking about reimanns definition and after solving it myself, it pretty clearly was 0!
That is not correct. The Riemann integral also includes in its definition the necessity that that it converges for *every* partition, not just some specific arbitrarily chosen partition, like in your case. You acknowledged that it is divergent for at least one partition, so the definition tells you the integral does not exist.
Your question is wrong, since it is undefined in zero you can define the integral in multiple ways, the way you broke it into to limits is undefined (you cant add/subtract divergent limits we "do" it just to get a general "feel" of the limit. A proper definition would be lim b -> -0 [integral from -1 to b of (dx/x) + integral from -b to 1 of (dx/x)] and it is clear that it is zero. I repeat, you can't break it into to limits because each of them is divergent and if you define it as such you can't add them, you can't add or subtract infinities, that's just scratchwork before we actually calculate the limits.
if we replace the first b with 2b we get ln 2 instead of 0, we can actually get ln of ANYTHING, that is why a proper limit definition is required.
I think a geometric proof would be needed to solve this completely. Something along the lines of proof by contradiction using geometry. We know the graph is anti-symmetric about zero. An integral like this is the area under a curve. Anything other than zero would contradict the fact that integrals are the area under a curve.
Actually both are wrong.
OP is wrong for the reason you explained very well. But you are also wrong because you can't just pick symmetric bounaries when you feel like it just to make it converge.
If you have any doubt when using these technical methods, go down a level and look at the the Riemann sum directly or use one of the criteria for convergence (e.g. Cauchy criterion, Darboux criterion..)
It would be a contradiction to assume it was non-zero due to symmetry. It could be an infinitely large square, but you can always divide the square in half.
Your limit is actually two limits a and b, tending to zero from + and -. Ony when a equals b do the two areas cancel, therefore for any particular a, a decreasing b can give you a divergent area - the integral is divergent.
This MUST be zero, right?? If the integral is interpreted as an area in the algebraic manner, these two pieces are EXACTLY THE SAME. It doesn't matter that they diverge, as they diverge in the same way. I mean, a rotation of one set, gives you the other set, so they have the SAME cardinality. If you add the fact that you associate a sign to one of the pieces, then they cancellate each other. I see this as a notation/procedure limitation rather than a actual conclusion. Please, BPRP, bring some abstract algebra to solve this problema. Where is Dr. Peyam? This has to have a solution! #teamzero
Carlos Toscano Ochoa ∞ - ∞ is not necessary 0 (in fact, you can find examples where ∞ - ∞ is π)
Amahury Diaz but in this example, both infinities are mirror images of each other. ∞ - ∞ should definite equal zero
Amahury Diaz I know that, it's just that these sets are essentially the same but with different sign, so it feels that the divergent solution is due to a limitation of the approach. I have that feeling that an abstract algebra approach could solve this to be zero.
Amahury Diaz but that’s because when you evaluate those limits they converge to different values. in this case as we evaluate the limit ∞ - ∞ goes to zero for all values
Amahury Diaz that’s kind of like saying the lim x-> ∞ of 1^x =/= 1. of course we know that 1^∞ is an indeterminate case, but if we evaluate the limit for this specific function 1^∞ does equal 1 for that case
Zero...or...divergent...
Zergent
Skippy the Magnificent divero is better
Yoav Shati im a year late but
_J A S O N D I V E R O_
😂😂
I remember getting into an argument with my calculus teacher over a cancellation I did in one integral problem that was considerably less trivial than this, but still had the basic idea of a fraction with a polynomial denomination resulting in a vertical asymptote, and I flipped the result on one side of the asymptote onto the other side and added the two to make the infinities cancel each other out and leave a finite result. I believe they told me that they understood what I did, but this type of cancellation was not allowed because it could lead to nonsense answers if applied to less well-behaved functions, so the correct answer was that the integral diverged, and I wasn’t given credit.
BPRP, initially I thought this was a silly video but after reviewing Cauchy Principal value, I'm beginning to realize just how well thought-out your videos are. This integral has to do with the strength of the conditions for the theorem. According to my complex analysis textbook (John M. Howie, Springer SUMS, pg. 14 section 1.7 Infinite integrals), The weaker theorem is exactly what you just showed in the video, the limit of the sum of the two integrals, but the stronger version of the theorem says that the integral exists ONLY if the two separate finite limits lim(e->0+) integral(a to (c - e)) f(x)dx AND lim(d->0+)integral( (c+d) to b) f(x)dx exist (sorry for shitty notation, also a = -1, b = 1, c is some value in between, f(x) = 1/x) on an interval a
Umm, Who?
Wow, what a compliment!!! Thank you so much!
I did show both in this video and the rest is left to the comment section, which I AM having tons of enjoyment of reading them.
The function f(x) = 1/x is the textbook example given to introduce divergent improper integrals and Cauchy principal value. It's a nice presentation to be sure.
Sorry yeah you did, now I realize that's why you put the "gap"! My bad ahaha! See, just another reason why your videos are brilliant, you incorporated the "epsilon"-proximity (arbitrary gap "near" infinity) without scaring people by using the actual hardcore rigorous definition for the argument. Please keep doing videos, they are really helpful!
You're the best! ^_^
: )
Please. make another video to rigorously explain that the result is undetermined (or Type 2 Divergent). Please, save your soul !
You aren't dealing with an arbitrary 0- and 0+. The same X applies to both by definition of the initial integral. This absolutely links both sides of that resulting equation. It wouldn't if both sides of the equation were derived from unrelated variables or functions, but that isn't the case here. If there is no X for which evaluating the area to the left of the origin for that X is not equal to the area to the right of the origin for the same X, why does that not allow us to conclude that both infinities, in this special case, are absolutely related and inverse of each other for any evaluation of points along their graph sharing the same X, and therefore accept that the limit of this process conforms? This feels like saying the limit of the function y = 1 at infinity isn't 1 even though the value was 1 at every single point along the graph. I know you've shown why it's dangerous to rely on intuition when dealing with infinities before, but this case just defies intuition so heavily and without an actual application or example to contradict that I can't accept the conclusion.
If we were to assume it is 0 like all intuition says we should, what breaks down? What proof by contradiction arises from such an assumption? If you could show me where assuming this is 0 begins to lead to conclusions that violate other proven mathematical structure, then I would be convinced.
Let J = ∫[-1 to 1] 1/x dx
Take n ∈ ℕ fixed ( n ∈ {1,2,3,…} )
Then we can express J as:
J = ∫[-1 to 0-] 1/x dx
+ ∫[0+ to 1] 1/x dx
= lim k->0+ { ∫[-1 to k-] 1/x dx
+ ∫[nk to 1] 1/x dx }
= lim k->0+ [ln(k)-ln(1)+ln(1)-ln(nk)]
= lim k->0+ [ln(k/nk)]
==> J = ln(1/n) for any n ∈ ℕ
Conclusion: u said the integral is 0, and I’ve said the integral can be made to be whatever value u desire. Contradiction - the integral is well defined and should only output 1 value, not infinitely many
And the limits need not be tied together as at the limit, my construction above still covers all the area under the graph over [-1,1] and is therefore valid
#TeamZero 😶
It is divergent, but the Cauchy Principal Value is 0
Does this work?
Metheth Propbut The only correct answer.
It's divergent of type 2. Meaning it's not even sure it goes to infinity.
Yes--I checked it on Wolfram Alpha
All calculus rules aside, we *know* that the positive area and the negative area are exactly equal but opposite, thus their sum is 0. This simply indicates that there are some problems that the tools of calculus aren't sufficient for.
If you think about it, the infinity on the left side of the integral is the same as the infinity in the right side of the integral even though it is infinitely many digits, but it is still the same thing, therefore it cancels out.
#TeamZero
Usually, when this appears in physics problems, it is zero by symmetry -- for every point on the +x axis, there is a point on the -x axis with the opposite value.
However, as demonstrated, there are ways to take the limit in which it becomes +/- infinity, but in physics problems that would correspond to an obviously distinct physical situation if it happened. When doing physics, one has to think carefully about the physical situation in a problem when potentially ambiguous limits are encountered, in order to know what the true answer should be. Stuff like this happens a lot in Quantum Field Theory, for instance.
Got asked this one on my PhD oral final. I simply discussed the issue of what happens at zero. Since I had used the Cauchy Principle Value in several parts of the work, which involved both analytic and computer solutions to the time dependent transport equations, I passed with a few chuckles from my committee. By the way, I returned the favor on several other students later.
Why can't you use algebra to find that it's 0?
You'll get -ln|0|+ln|0|=ln|0|-ln|0|
Which obviously equals 0
The only thing that stops this is to make a difference between 0plus and 0minus but in regards to ln|x| they are symmetrical and should be regarded as equals
It diverges. You can prove this with e.g. showing it doesn't meet Cauchy criterion or by comparing to a suitable series.
In fact, if I remember correctly: The integral of 1/x^a converges in (0,1] for a1, which is why it never converges on the entire (0,inf) for any value.
I may have got it mixed up though.. :-P
I'm an year late but the controversy seems to be on whether it is indeterminant or divergent. Well the simple explanation is that it's indeterminant only if we are dealing with the same variable in the limit on both sides but it depends on the problem solver to choose different variables approach 0 from either side(i.e. a, b) and you don't have any relation between the two. And therefore he claimed that the integral diverges. After all we are including a point that is not in domain of an unbounded function.
The limit is undefined. When you end up with "infinity-infinity", you are trying to combine two separate limits of two variables into one. This cannot be done as the rate at which each limit approaches the singularity may be different. It is only indeterminate if that result happens within the same limit. Because it's with two different limits, the answer is instead undefined.
Finally the right answer!
@@KokomiClan Just because the rates *can* be different doesn't mean they have to be. In this case the solver gets to choose their variables, and makes a mistake in choosing two distinct arbitrary variables with no relation to one another rather than one variable that forces the integration to be symmetric (or at least an assertion along the lines of let b = -a or something similar). Let a -> 0+ and then integrate over -1 to -a and a to 1. The value of a is still arbitrary, but by now we get to conclude that the resulting infinities are indeed symmetric and therefore cancellable.
It's not two limits. It's one limit, and the same limit with a -.
A thing - the same thing is zero.
If the definition does not work, the definition is wrong.
It's not two different apples. It's an apple, remove it, no apples.
It's literally a mirrored image, point by point.
It's not subtracting two infinites, it's subtracting an infinite from itself.
From some definition, lately, women may have a penis.
Definitions may change language, not reality.
What is exactly the definition of 0+ and 0-?
I say it would make sense to define it as:
lim d-->0 of d and lim d-->0 of -d.
And then you have: area = lim d-->0 ln(d)-ln(1))+ln(1)-ln(d) = lim d-->0 0 = 0
Really late to post here but I just want to add that you don't know if 0+ or 0- is bigger because you defined it that way. If you did it such that one integral goes to (0 - b) while the other goes to (0 + b) and then take the limit as b approaches 0 you preserve the symmetry of the integrals. It still means that b can get arbitrarily small and both halves will cancel each other out
While that's true it doesn't really help, because it's not true that lim x->a (f(x)+g(x)) = lim x->a f(x) + lim x->a g(x) in general, and the existence of an improper integral requires the latter.
For me the divergent explanation doesn't make sense. You can operate the ln before subtitute it, obtaining that ln|0-|-ln|0+| its in fact 0. Its like saying that the lim x->inf of x-x diverges because you substitute before operate the "x" and you obtain inf-inf that is inditerminated form. No sense for me.
#TeamDivergent
The theorem states: the integral of any odd function on an interval, which is symmetric with respect to the origin, equals zero, if the function is continuous on that interval.
The function 1/x is not continuous on [-1, 1], thus the theorem cannot be applied.
Exactly!
Agreed. This theorem will hold true for functions such as the cubic function and sine function.
Who cares about the theorem when both of the functions behave the same whether x is positive or negative. The infinites are symmetric and are the same value. This integral must be 0.
If we make a function F(t) = ∫₋₁⁻ᵗ(1/x)dx + ∫ₜ¹(1/x)dx, it is 0 everywhere for 0 < t < 1. That would make it seem that lim(t->0) F(t) = 0.
Forget Yanny vs Laurel, Zero or Divergent? #YAY!
15schaa 15schaa I like this!! I am changing the title if you don't mind!
Team Divergent!
You're all wrong. The area under the curve is complex.
EDIT: The explanation:
The integral of 1/x is lnx, so we have ln(1)-(ln(-1)) = 0 - ln(e^iπ) = -iπ. Hence proved.
EDIT 2: This essentially violates the fundamental theorom of calculus, since the function isn't continuos. But it Is always fun to venture into the complex world.
Can't you just say
Lim b->inf : ln(b)-ln(b)
= 0
Instead of having an indeterminate form ?
His graphical explanation of the 0+ and 0- corresponds to using two separate variables for the limits, such as ln|b|-ln|a| (rather than ln|b|-ln|b|). While intuitively it would be nice to call both limits the same variable as you do, I'm not sure if that can be rigorously justified.
This is the Cauchy principal value
If you want to define the integral as the area under the curve, then I think it's not rigorous, why you should integrate from -1 to -b and from b to 1 instead of from -1 to -b/2 and from b to 1? At the limit b->0 the two options define the same area, so the result must be the same for both, but it's not.
math.stackexchange.com/questions/2665679/cauchy-principal-value-and-divergent-integrals/2665708#2665708 this action(of setting both sides to have the same limit) is called Cauchy PV
Roger Balsach why not b/2? Because that's inconsistent. The goal behind the limit is to capture everything except the point zero. If you're already as close as you can possibly get, or taking a limit, it doesn't matter if you divide by 2.
It is very simple. It is a matter of definition. Unfortunately, the integral of a discontinuous function with a single discontinuous point in the middle of the integration interval is defined to be the sum of the of the limit of the 2 integrals to and from the discontinuity point provided that each of the 2 limits converges. If anyone of the limit diverges, the whole thing diverges.
Of course, one can change the definition from the "sum of the limits" to "limit of the sum", in that case, we can show that this integral is 0.
But this changed definition is not the standard textbook definition. If we really like 0 to be the answer. We will need to use a non-textbook definition.
{-(Infinity) - 0} + {0 + (Infinity)} or 0? This is what led to the Infinity War.
This is a great example of why Riemann integration on open intervals doesn't work. However, in the video the problem is represented as a sum of two integrals: from -1 to 0 and from 0 to 1. Riemann integration works with closed intervals, so that corresponds to [-1, 0] and [0, 1], and the intervals actually overlap. Another mistake is using a limit to calculate an integral at the point where the original function is not actually defined.
We're actually trying to calculate a sum of integrals over 3 intervals: [-1, 0), [0], (0, 1]. We know that the function 1/x we're trying to integrate is smooth and finite on the intervals (-∞, 0) and (0, ∞), therefore the integral of the function must also be smooth and finite on any subset of those intervals, so for any real b > b' > 0, integral over [-b, -b') has to be finite, and equal to - integral over (b', b]. This can be applied recursively - no matter how small b is, the integral over (0, b] can be written as a sum of an integral over (b', b] plus an integral over (0, b']. This gives us two infinite sums a = a_1 + a_2 + a_3 + ... and a' = a'_1 + a'_2 + a'_3 + ... such that a_1 = -a'_1, a_2 = -a'_2, a_3 = -a'_3, ... This means a + a' = 0, therefore the integral over [-1, 1] equals 0 and does not diverge.
I'm going to give you the classic response from a mathematician:
It depends on your definition of an integral.
The most useful definition of an integral is probably the Lebesque integral. By this definition, the integral diverges.
8:06 notice how you’re “subtracting” infinity from infinity in one step, but the step before, you’re subtracting ln|0-| - ln(0+). but that’s just a shorthand for the limit as x->0+ of (ln|-x| - ln(x)). but we know that |-x|=x whenever x is positive so you get the limit as x->0+ of (ln x - ln x). ln x subtracted from itself is 0 for all positive x. therefore the integral is 0 and i strongly disagree with anyone who says it shouldn’t be 0.
Your arguement is not convincing when you chose to arbitrarily give different values to "0" on the negative and positive sides of x-axis. You don't have the luxury of treating the two parts of Y=1/X equation differently!
The two parts of this equation approach Y-axis asymptotically (divergently.)
That is an odd function from -a to a therefore it’s zero right?
These comments are great btw and also I realised that the theorem above doesn’t hold in this case because the function isn’t continuous on [-1,1]
The integral diverges but it’s Cauchy principal value is 0
Ye it’s a really interesting debate
Integral of 1/x from 0 to 0?
A black hole.
Still infinite
It's gonna be 0. Area is 2 dimensional so infinity*0 = 0
0
@@Exachad Dude ur pic are everywhere on yt comments, whats up with that?
There is a continuous one-to-one correspondence between all points left of zero and right of zero, so everything must cancel. Imagine an excluded band inside of -1 to +1 centered around zero that is shrinking. If we add the left-side integral from -1 to -f to the right-side integral from +f to +1 (where 0
The issue is, the way improper integrals are classically defined, the "f" used as the limit for the right-side integral doesn't relate to the "f" used as the limit for the left-side integral. It is not defined to be a symmetric band. There is a way to make it symmetric though, (en.wikipedia.org/wiki/Cauchy_principal_value), but again, that differs from how improper integrals are usually defined.
#TeamZero This is where I'd favor the sensible answer (0) and say that improper integral as a technique just breaks down. Diverges is just so unsatisfactory of an answer.
Define F(c) = lim b -> c Integral from -1 of -b of 1/x dx + Integral from b to 1 of 1/x dx, c in (0, 1). Then F(c) is zero everywhere. Notably, as c -> 1, F(c) = 0, so F(0) is a removable discontinuity whose limiting value is 0.
"If one is piece is divergent then the whole thing diverges." Let's try that, ln|x| diverges as x approaches infinity. So I do the limit of ln|x| - ln|x| as x approaches infinity. Both pieces diverge but the whole thing converges, so no, your reasoning doesn't quite work out. But, yes I agree it does diverge.
And what's ur reason why it diverges?
Btw, when I said that, that was meant to be a quick way to draw the conclusion. I did provide another reason why it diverges at the end with the graph.
I have heard the infinity minus infinity argument and the graphical argument, but neither helped me understand it, because the symbolic answer does not make sense intuitively. My reasoning therefore is that if you were to numerically integrate this on a computer with infinite memory, a Turing Machine, then it would return different answers each time you ran the calculation, i.e. one time you get 0, another time you get 243534574, and another time you get 12. Its's a little wishy washy since we don't have true turing machines, but when I learned this, I needed the numerical argument for me to get it.
There's a big difference between taking
lim(x→∞) (ln|x|−ln|x|)
and taking
lim(x→∞) ln|x| − lim(x→∞) ln|x|
MuffinsAPlenty That seems correct to me, but it's not really intuitive, and you certainly wouldn't have convinced me with that a year ago. BPRP was trying to answer this in terms that would make sense to students, but this is something that will never make sense. So I would say go for a rigorous proof instead, considering that will make just as much sense, and it's not like BPRP hasn't done rigorous proofs and such before.
What if you take the following limit of integrals:
Limit as c goes to 0 from the positive side.
First integral: -1 to -c.
Second integral: c^2 to 1.
You'll get 0 - 2 ln (c) + ln (c) - 0
Simplifies to - ln (c) for any small positive c.
In the limit as c goes to 0, this is infinity.
I think you should've evaluated the integral on "b" and left thrlimit for the end, so you couldve saved the indetermination.
Limit b->0 [Ln (b) - Ln (b)] = Ln (b/b) = Ln 1 = 0
You can do this in this way because the function is symmetrical by turning the axis 180°
Approaching zero from both sizes (positive and negative) in an absolute value is equal, therefore we can simplify to one limit, either positive or negative, and therefore simplify it to 0.
#TeamDivergent. I did the limits on both, but on one I used epsilon, and on the other k*epsilon. So, if the answer is finite, it must not depend on k. Then, I ended with lim epsilon -> 0 ln|k*epsilon|-ln|epsilon|, which, by the log properties, throws ln|k|, which contradicts the suposotion to the integral being finite. So, it does diverge.
usuario0002hotmail I am not at high math level,,, only took calc 1 ,,, but seems divergent because both are sides sum to same, (one side neg Infinit , other side pos Infinit,),, , yet never zero. Because for he fact that they never actually touch. The true origin / 0 , they diverge. ( I'd argue in a real world In real world I'd be 0, both are infinite and cancel out) But !,,, Since They're is a gap between both 0- and 0 and also gap 0 to 0+. #DiVERgEnT 🔥💯🙏🙏🙏🙏
why? ln(k) is finite unless k is 0, which is not allowed
ah, nevermind, I made a thinking error
Finite or undefined ?That's a difference.
You did not show, that it cannot be finite. Just that the value can be made into any real number ln (k) by choosing any positive k you like.
"Real" divergence is also possible, eg by using nonlinear limit variables (b, b^2). I don't really like the word divergent for indeterminant expressions ^^ sounds too much like "approaching infinity". I know that's stupid :D
Edit: i think the original commentator also took "divergent" as "going to infinity" and tried to show that ;p
You can make the left hand integral ln(0+), which is the same as the right hand integral. You can then say it is the limit as t-> 0+ of ln(t)-ln(t). This is simplified as limit as t->0+ ln(t/t) which is ln(1), so the integral does equal 0
Man, you're incredible, you explain so good... With you I'm learning all the calculus because in school I haven't learnt it. Thanks, man.
#TeamDivergent
#YAY
Salute from Spain.
Use the Cauchy principal value and you have the ans to be 0 and remember this integral is divergent
I got a similar question on a test and said 0. It was the only question I missed and I got into a debate with my teacher about it and got the points back lol
If you approached this problem in the manner of Riemann sums, for every x value on one side of the y-axis, the opposite x value will return the same "area" of its corresponding rectangle, just negative. This is true for any x value on either side of the curve, no matter how small the rectangles get, so, logically, it'd make sense for the integral from - 1 to 0- and 0+ to 1 to be 0, as whatever 0- or 0+ is, its counterpart can be just as small, just opposite.
you can't do Riemann sums because for a function to be Riemann integrable it has to be defined over an interval (this function is not defined on all [-1,1]) and it must be bounded on this interval (this function is not bounded) so the only way to approach this problem without using more advanced theories of integration is to use the limit, which gives 2 integrals that MUST individually converge, but they do not.
The argument that it's zero can be used to show it has any value.
can you elaborate on that?
Zone Goku Essentially with principal value you're taking the integral from - 1 to - z and from z to 1 then taking the limit as z goes to zero, but there's no reason it couldn't be say taking the integral from - 1 to - b*z and from a*z to 1 then taking the limit as z goes to zero, the should give us a value of ln(a)-ln(b).
That's brilliant - it's made me #TeamDivergent
if you take the limit as b approaches zero of Int (-1,b) + Int(b,0) it is zero, so it is zero for me
But what if you calculate the indefinite integral first and then evaluate from -1 to 1? You always get 0. Only when splitting the integral at the asymptote do you get an indeterminate value. Also, using the limit definition of an integral, even when you do split the integral at the asymptote, you still get a limit of 0. Because infinity minus infinity still means something in limits.
you cannot integrate between -1 and 1 because for a function to be Riemann integrable (the most well know and common integral) it must be defined in the whole interval (this function is not), bounded (again this function is not) and also another condition that i can't really type on here but already the first two aren't satisfied. There exists a natural way of dealing with non bounded functions, and it's splitting up the integral in two distinct integrals and to take a limit. Both of these integrals must converge for the original integral to converge. Maths doesn't really care about intuitive results, but only rigorous definitions and what follows from them, so talking about "symmetry" or "intuitive results" doesn't work very well
@@OnamKingtheKing Nobody asked
I don't know alot about calc but it just kinda makes sense for it to be 0
As Lebesgue integral, it's obviously divergent.
This is satisfying :D
ljfa Shut up with your lesbian integrals, it's most definitely zero.
CAUCHY PRINCIPLE!!!
U can change the integral of 1/x from -1 to 0
Into integral of -1/x from 0 to 1 which makes the answer 0
#TeamZero is, sadly, wrong, thought it feels so right... Here's more proof
Let's apply the same reasoning to the the integral from -1 to 1 of 1/x^2
1
∫ 1/x^2 dx = ?
-1
If we don't pay attention, we just use the anti derivative and get
1
-1/x ] = -1/1 - -1/-1 = 0
-1
We can already see that this is wrong, but let's keep going
Trying to integrate over dx, split up the improper integral into two parts, one from x=-1 to x approaches 0-, another from 0+ to 1
a 1
lim ∫ 1/x^2 dx + ∫ 1/x^2 dx
a-->0- -1 b
b--->0+
The anti-derivative of 1/x^2 is -1/x, so when we plug in 0 like you did at 6:28, we get that the first integral diverges:
0-
-1/x ] = -1/0 - -1 = undefined
-1
Likewise, the second integral diverges, and as a result:
***Using **#TeamDivergent**'s reasoning, the integral from -1 to 1 of 1/x^2 is undefined. Which is correct***
Now, lets approach the integral a different way. Instead, we'll integrate over dy. I can't draw a picture here, but the are bound by x=-1, x=1, y=0, y=1 is within the area of this integral and has an area of 2 square units. So, we only have to integrate above y=1.
y = 1/x^2 -------> x = ±√1/y
∞ ∞ ∞ ∞
∫ xdy = ∫ ±√1/y dy ===> ∫ +√1/y dy + ∫ -√1/y dy
1 1 1 1
The anti-derivative of √1/y is 2√y. So, we get
∞ ∞
2√y ] + -2√y ]
1 1
Which we can already see diverges, because we have √∞
The same logic applies to any 1/x^n.
Therefore, it's important that we always split the integral at any break in differentiability, and #TeamDivergent is right
+Oleksiy Bazhenov
Just a minor correction which does not affect your argument, quite the opposite:
If we don't pay attention, we just use the anti derivative and get
1
-1/x ] = -1/1 - -1/-1 = 0
-1
Or we don't pay attention to the numbers :). The first fraction (-1/1) equals -1. From that, we subtract 1 (-1/-1). -1 - 1 = -2, not 0.
This makes things even weirder, because we are adding positive numbers and get a negative result (Zeta function anyone?).
Since it contradicts your second solution using #TeamZero logic (integrating using dy), it further proves that #TeamDivergent is correct. Unless we might miss something.
Sincerely,
ZfE
The problem with 1/x^2 is that both halves of the integral are positive. Adding +inf to +inf is always going to diverge.
In the case of an odd reciprocal power we are adding -inf to +inf which only potentially converges - to show it actually converges we need to show both sides approach the limit at the same rate. One way to do so was shown in my separate post for #TeamZero
Wow, great explanation! :) And I have similar example. I have to integrate -e^(-x^2/2)/(x^2) from -Inf to Inf, but this function is undefined at x=0. And with integration by parts(u=e^(-x^2/2), v=-1/(x^2)), we get sqrt(2Pi). However, if we integrate from 0 to Inf, it should be half of sqrt(2Pi), but in this time it diverge! So I'm really confused about this. I need to calculate this integral to solve my problem, and if this integral diverge, then I have to throw away this integral and find the other way. lol In fact, I have some sequence, and my integral is -1th term of that sequence. Also, 0th, 1th, 2th, ... term are all sqrt(2Pi). But -1th, -2th, ... term are all "undefined integration" like my example. So,, I hope that (integral of -e^(-x^2/2)/(x^2) from -Inf to Inf) is sqrt(2Pi), but after I read your comment, I think it may diverge. ...... What is correct answer?
@@trueriver1950 You cannot actually prove that the limit on both sides converge to their respective values at the same rate. What you are doing is simply calculating the Cauchy principal value.
What if you took the integral of the function (x / (x^2 +a)) with respect to x, for positive constant values of a. Then, taking the limit as a approaches 0+ should give you a consistant value equivalent to the cauchy principle value of 0 for the - 1 to 1 bounds. Of course, since the limit a approaches 0 - does not exist, you can still argue that the integral diverges as well.
#CivilWar #teamzero
#teamzeroforever
Nah, fam, Divergent for life!
Alexander Nenninger In the second case we have S(n)= [-1;1/n] u[exp(-x);1] but 1/n and exp(x) are always >0 so I think that exists an interval (0;ε),ε>0, that it's considered two times; in other words the intersection between [-1;1/n] and [exp(-x);1] ≠0
Alexander Nenninger ah ok ahahah
My problem with this is that the integral from -1 to 0-E plus the integral from 0+E to 1 equals zero, even for arbitrarily small consistent E. The reason that the integral fails to converge is only the point x=0, a fact which I feel like this video glosses over. If you take the limit as E approaches zero but remove that one point, the integral converges, but that’s not an equivalent integral. The indeterminate form comes from the fact that the area of the “rectangle” with zero width centered at x=0 is not necessarily zero, as the rectangle has infinite height (0*infinity is undefined)
(Also, I realize I was vague as to why the altered integral converges. If you take the limit of the Riemann sum that represents the full integral, separated into its two parts, with the terms grouped so as to cancel one another, only one term in the sum is undefined: the infinity minus infinity term. Removing the x=0 point changes these from rectangles of infinite height to rectangles of arbitrarily large finite height, and thus cancelling one another out.
)
Another way to understand (using definition): If we divide [-1, 1] into N pieces equal in length and N is an even number, Ci is the midpoint of each piece. The sum of all f(ξi)·Δxi = Σ(1/Ci)·(2/N) will always be 0. So the same is true when N goes to infinity.
HOWEVER, if N is an odd number, for every piece which doesn't contain 0 we choose their central point. For the central piece (the only piece contains 0) we choose one point bigger than 0 from it. We call this point Q. The sum of all f(ξi)·Δxi will be (1/Q)·(2/N) = 2/(Q·N). One thing we can know is Q can be any value in (0, 1/N]. Thus (Q·N) can be any value in (0, 1]. When N goes to infinity the sum can be any value in [2, +infinity).
Thus by definition the integral of 1/x from -1 to 1 cannot exist. I hope this is more explanatory.
I like this answer a lot! And of course, the definition of an integral doesn't require your sample points to be chosen in the same location for each piece. So, playing around with that, with careful selection, you can probably get _anything_ you want, including in (-infinity,2) as well.
My interpretation is that the integral is certainly divergent, but the area the integral is supposed to represent is 0. Thinking about the way the integral defines the area underneath a function as many tiny rectangles, we see that as the number of rectangles goes to infinity there is only a single rectangle which doesn't cancel with a mirrored one: the one at x=0. Now for the integral this has disastrous consequences but in terms of intuition about area it is an infinitesimally small strip that can be disregarded. Therefore the area is 0, but the integral is divergent.
Alright, as a proud member of #TeamZero, here is my rebuttal. When we evaluate ln(|x|) [-1,0-], we get 0-lim[x->0-] ln(|x|). When we evaluate ln(|x|) [0+,1], we get lim[x->0+] ln(|x|). If we dont evaluate the limits yet and add them we get integral from -1 to 1 of 1/x =
lim[x->0-] ln(|x|) - lim[x->0+] ln(|x|). Since it was established that ln(|0-|)=ln(|0+|), then this can be rewritten as lim[x->0+] ln(|x|)-lim[x->0+] ln(|x|)=lim[x->0+] ln(|x|)-ln(|x|) = lim[x->0+] 0=0. Therefore, the integral from -1 to 1 of 1/x dx is 0
Int(1/x)=ln(x)
And since ln(-1)=ln(1)
It becomes ln(1)-ln(1)=0
Your way ends up with not defined, therefore you can't say if it diverges or not. Just like you are able to create a not-defined almost anywhere. For example before using l"hopital.
ln(-1)≠ln(1), as ln(-1)=i*pi, and ln(1)=0
0:58
"LET ME PUT dx OTHERWISE UR CALCULUS TEACHER WOULD BE MAD"
DAMN TRUE !!! 😂😂😂
I still say it's (2n+1) i pi for all n in Z by how the ln function works for negative numbers.
blackpenredpen said that the two areas (left side area and right side area of the curve) are not equal because ------->
+(0)[very little right side of zero] is not equal to -(0)[very little left side of zero]
can someone tell me why?
i think this is the only reason why the area is not zero
Both sides grow/shrink to ±∞ at the same rate, so personally I wouldn’t have two separate limits for the two integrals but rather:
lim_{b→0⁺}(∫₋₁⁻ᵃ + ∫ₐ¹), where both integrals are of 1/x dx. If we computed it like this, it would indeed turn out to be 0, and since infinity is reach from both sides at the same rate I think this is how we should do it.
But what if you, before you work out those 'ln' things at the end, leave those and simplify it using those. That way you'd get ln|0-| - ln|-1| + ln|1| + ln|0+| = ln|-1| + ln|1| = 0. That way you don't have to deal with an 'infinity' and it is simply 0
Interesting... ln(0) is undefined.. but can you still cancel them?
However, in the explanation that leads to a divergent series, the ending result after solving the logarithms is infinity-infinity. But infinity-infinity is an indeterminate form like 0/0 and can therefore take any value. The trick, as with the indeterminate form 0/0 generated by computing derivatives, is to find which value is the expression really approaching, it's the essence of calculus. And since both infinities seem to grow at the same rate and both infinities seem to get closer to zero at the same rate, one could then argue that the expression actually reaches zero. This is the reason why I chose to believe that the integral is equal to zero.
#TeamZero
taking the integral of this by not equal spaced intervals doesn't work, or at least to me, if they are not equally spaced, then we are always missing a bit "almost zero" here but not there, or an infinitismal. An integral is taking the ENTIRE function, not almost this function, but it's still okay to do equal intervals to get all values except the singularity. 0 is the Cauchy Principal Value of this integral, the main sensible value.
i.e. we are getting away 1d line rather than 2d rectangle/trapezoid/area that is almost as thin as a 1d line
The theorem for odd functions holds whenever the integral exists. Continuity is not necessary but sufficient (continuous images of compact sets are compact, compact sets are bounded, bounded piecewise-continuous functions on compact sets have an integral). The theorem already holds for, e.g., the integral of x^-3.
Thanks math, but I'll stick with logic. It's zero for me. :)
Math is logic. I think u mean: stick with common sense. But common sense is based on experience and u have no experience with infinity in real life.
@@mobilkonto__free__9726 The symmetry is both logical and provable. You just have to reformulate the problem so that the symmetry info is not lost.
Assume deltaX as a distance to 0. Use -deltaX and deltaX respectively as the boundaries of the integrals. Solve for the area as deltaX goes to 0. Wouldn't this yield a proper answer? You still approach 0 but with an equal distance from 0 at both sides.
Here's proof that it diverges , we can first of all devide the integrals into two parts one from -1 to 1/n and the other from 1/n to 1 under these conditions we get ofc 0 , however if we change 1/n by 1/2n if it were convergent we would still get 0 as n goes to infinity but infact we get infinity and so we have shown the existance of two possible values for the integral wich is ofc impossible and therefore it is divergent
Destroctive Blade i like it!!!
Destroctive Blade But if we integrate 1/z over the upper half of the unit sphere in C, since 1/z is Holomorphic in C\{0}:
integral_0^1 exp(-i (π t)) d/(dt) exp(-i (π t)) dt = 0 #teamzero
😁
Under some other comment I have shown that it's (infinity)
If you let b go to 0 *simultaneously* for the two limits, it will actually be 0, for instance A = \lim_{b\to 0} \ln(1) - \ln(b) + \ln(b) - \ln(1) = \lim_{b\to 0} \ln(b/b) = 0.
#TeamZero - I like so much your videos. Actually, there are infinite that diverges faster than others. For example, the function y = x^x and the function y = x. The both functions diverges when x comes to infinity, but y = x^x diverges faster. Also, we know that there are more real numbers than natural numbers, even when there are infinity real and natural numbers. In this case, it makes sense that the both sides comes to infinity in the "same velocity", what lead this integral to 0. Well, that is just my positioning, but I am just a boy in math, I hope learn even more if your videos. Thanks!
The definition of a convergent integral states : int from a to b of f converge if and only if for all [c,d] in [a,b], the integral over [c,d] converges. However, int from 0 to 1 of 1/x diverges. Thus int from -1 to 1 of 1/x diverges. Simple as that.
Does it make a difference if you write one limit as epsilon->0 of the sum of the two integrals (i.e. from [-1,-epsilon] and [epsilon, 1]. In the end you have the difference of two logs which is the log of a ratio which equals 1 hence the answer is 0?
According to the video the limits don't nescessarily approach zero at the same speed, so you can't use epsilon for both limits.
But you can make them approach at the same rate by using one limiting variable. The Cauchy.Principle Value theorem uses this fact.
I also thought of using something along the lines of an infinitesimal like epsilon ε. Also, 0 fits nicely with the properties of odd functions.
Yes I know. But as you say yourself this is something else, namely the cauchy principal value of the integral (or whatever you would call it in english) and not the value of the 'normal' integral.
OK. So this strikes me as something like the alternating harmonic series in that we have a conditionally convergent function as long as you evaluate it in a particular way. If you add the terms of the ahs out of order, you can make it equal anything. If you evaluate the given integral as a left-to-right Riemann sum then you'll soon get to infinity and all bets are off.
How about the next best thing? Define a function f(a) = integral from -1 to -a of 1/x plus integral from a to 1 of 1/x. Then f can be continuously extended as f(0) = 0. That way you can tell people they are wrong about the integral while also satisfying their intuition.
The first integral you wrote does not exist. The reason is that 1/x is not continuous at x=0, so it's not Riemann-integrable at that point. As for the split integral with lateral limits, it diverges if you separate the limits, but converges to 0 if you consider a single limit of (int - int)
#TeamDivergent
I have a suggestion for strengthening your argument at the end: by letting 0^+ be +a and 0^- be - a*b, with arbitrary positive b>0 and letting a>0 approach zero, you can show that the combination of the limits approaches ln(a*b) - ln(a) = ln b. That shows the indeterminacy in its pure form: by being able to choose b>0 arbitrarily, the integral can take any real value. The number b can be interpreted as the relative rate at which 0^+ and 0^- approach 0. Taking b = 1 would give you the Cauchy Principal Value of the integral
Can you talk about how the Cauchy principal value can solve this debate?
Yup! CPV says 0. Otherwise it diverges. Problem solved : )
Here is my take... when working the improper integral, we get (before you simplify ln|0| as -oo) ln|0| - 0 + 0 -ln|0|...
we know that ln|x| converges to -oo at 0 from both sides (absolute value functions are symmetric), so lim x->0- ln|x| = lim x->0+ ln|x|.
Now we get into semantics. I argue calling ln|0-| = -oo is a generalization - that is, we lose information by representing it this way. However, if the above limits are true, then there is no need to generalize them this way. ln|0| - ln|0| = 0 because ln|0| is always = ln|0|. This would mean we could simplify your equation to 0.
Note that oo - oo is still undefined because many expressions “evaluate” to infinity, like 1/0 and 2/0. That said, by my argument, 1/0 = 1/0 ≠ 2/0 because if we had the means of solving this expression we would get the same result using the same numbers.
So, in short, I claim by semantics that calling ln|0| = -oo is a generalization (we lose information), and that ln|0| - ln|0| = 0 because ln|0| would consistently map to the same answer, if we had the means of calculating it, because the input is the same every time.
Thoughts?
I would note that the idea of integration is to accumulate all of the values of a function as an infinite sum, the notion of infinity is not a value rather a symbol that represents the continuing of numbers to no end. In a similar fassion to the problem here, we can, for example, integrate the gaussian from "negative infinity" to "positive infinity" which really just represents the total value the function has as a sum, if we were able to evaluate infinitely many terms. An integral is more so asking, what does this infinite sum APROACH as I use smaller and smaller portions of the domain 'dx' to accumulate a function, in a similar way, we know that by nature of 1/x, all finite inputs cause it to act as an odd function, hence, as we choose smaller and smaller portions of the domain to examine and accumulate, the sum doesn't exist, as 1/x as x approaches 0 leaves a figuratively thin sliver of infinite height in the center of the integral, which even as dx gets smaller and smaller, still gives an infinite accumulation. #teamdiverge
FYI: limx->0+ doesn't mean approaching to a number a little bit more than 0, since for that matter you can also approach that from the left. However it means the x approaches (but never equals) exactly 0 from the positive side. Although I do agree that the integral diverges because the bottom and top limits of integration are independent of each other
#teamzero I mean just look at the first graph he made. Like, it's obvious the area cancels out. Just to be clear I haven't studied integrals thoroughly so what I say might not be mathematically rigorous, but based on my intuition the integral should be 0
That might be true for all the other values either side, but what happens at 0? Both sides are trying to claim that piece of area to cancel the other one out, and they can't both have it, and both cancel out.
Horep I'm not quite sure I understood what you mean there. If you are talking about the slice at x=0, then the areas can't split it as it's not in the function's domain.
This improper integral is also the solution of -ln(-1). This is because the natural logarithm function is the integral from 1 to n of the function 1/x respect to x. The negative natural logarithm is formed when you switch the upper and lower bounds. Using this rule, ∫_(-1, 1) dx/x = -∫_(1, -1) dx/x = -ln(-1). Since e^(pi*i) = -1, ln(-1)=pi*i, -ln(-1)=-pi*i, and the solution to this integral is -pi*i. This is because this integral can be broken into 2 parts: The positive side, and the negative side. ∫_(0, 1) dx/x = infinity, and ∫_(-1, 0) dx/x = -infinity. Therefore, the integral brings us to an infinity-infinity paradox as now we can see that infinity-infinity can create imaginary numbers as well. This is why infinity-infinity is indeterminate without using 0/0, 0*infinity, or infinity/infinity.
if you accept 0 result you could get some paradoxes.Something that Maths hate.One example is Apu´s Paradox
Exact. You can prove the integral( 1/x ) over 0, 0 is 0 ! :D
But Zero is a more comfortable value to accept. The shame is that the mentor ends the video with "take a stand", like if we could choose the result we prefer in Math !
At 0, the function approaches infinity and infinity-infinity is undefined. It gives a feel that due to symmetry, the areas below and above x-axis cancels out each other to give 0 but actually its not. It is undefined.
You can't just separate it and then make zero not equal to zero. So I still think answer is zero
If you instead take a complex integral, using any path not containing the origin, then the integral from a to b of 1/z dz ln|b/a| + (angle(b) - angle(a))*i. This means that the real value is always zero, but the imaginary component is (2*k+1)π, depending on the direction and number of times the curve loops around the origin.
Maybe we should accept that it is unknown rather than be on team zero or team divergent.
My intuitive opinion is but I don't know the mathematical proof is if both sides diverge at the same rate in opposite directions then they would cancel each other out. If one side diverges more or quicker than the other than the whole thing diverges.
It's definitely one or the other.
Better safe than zero I guess
Ayman Abdellatief Believe me We integrate over the upper half of the complex unit sphere: integral_0^1 exp(-i (π t)) d/(dt) exp(-i (π t)) dt = 0 😁
Edit: BS calc. It's already way too late
It depends on how we define integral. There are multiple definitions, and in each definition there is an unambiguous answer.
It does not have an improper Riemann or Lebesgue integral, but it does have a Cauchy principal value of 0. The Cauchy principal value splits the integral into left and right (proper) integrals, then takes the limit as both of them approach the singularity at the same rate. In this integral, the value is 0 for all δ>0, so of course the limit is too.
Look up Cauchy's principal value. Instead of using 0-minus and 0-plus, use -a and +a (same magnitudes). Integrate, combine results, and let a->0. The limit is zero.