Improper integral of 1/x from -1 to 1 (THE DEBATE?)

The debate of the improper integral of 1/x from -1 to 1. Is this zero or divergent?

Let me know what you think!
blackpenredpen

[isambo]

I wish Euler was still around, he'd know what to do

[reetasingh]

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.

[snejpu]

-infinity + infinity = whatever #TeamWhatever

[garrytalaroc]

By the theorem of feelings and irrationality my answer is Zero.

[Idran]

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.

[jianxiang]

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

[twwc]

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.

[UniMaestro]

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

[fullfungo]

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<c<b, for example, c=(a+b)/2, then the limit is 0 at any step. Therefore the limit converges to 0. So we can conclude that Riemann integral of 1/x from -1 to 1 is convergent and equals to 0..

[jonshonjohn]

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.

[brianxx]

This feels a lot like quantum mechanics. Something tells me the correct answer is -1/12.

[KnakuanaRka]

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.

[ummwho]

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<c<b, c is a singularity. In your video you showed the weaker version, so it has a Cauchy Principle Value of 0 BUT the actual integral does not exist because both its parts are divergent.
Lesson learned: always pay CLOSE attention to the theorem's requirements, also infinitesimals is a bad way to argue things rigorously.
You're a genius btw and a seriously gifted teacher. I love your videos, thank you for doing them!

[benmcconnell]

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.

[CarlosToscanoOchoa]

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

[teslapower]

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.

[BTheBlindRef]

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.

[schaa]

Forget Yanny vs Laurel, Zero or Divergent? #YAY!

[shacharh]

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 a<1 and in [1, inf) for a>1, which is why it never converges on the entire (0, inf) for any value.

I may have got it mixed up though.. :-P

[ethiu]

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