收敛性、反常积分与黎曼和

And we figured out that, the monotonic behavior is going to determine the comparison between the right Roman sum, left Roman sum, midpoint Roman sum, and trapezoid Roman sum. Except for whenever we do know that the intervals are all equal, meaning we're dealing with equal interval, then the comparisons become so much easier because you can skip all their shared, and the the middle function value and just boils down to the comparison.Between the lower boundary and the upper boundary. And if we're all good here, are we all good? Now I'm gonna actually ask. Go ahead. So just about like the previous class was that that seemed related to what we did last time. Oh, you're saying there's something I don't remember. I misremembered where we ended up. No, I mean like the or the previous.Oh, yeah, it's something I I just didn't get to teach them. I meant to teach you both, except for I meant to give them a simpler example. But it turns out that because of our okay, my previous session last Monday with them got a little late, so apparently I just didn't get to cover that part of the material. So today I was making it up for them, making it up to them. Alright, so if you recognize it, you're.Supposed to, yes, absolutely, and that's the idea. The derivative doesn't necessarily equal to it could be well defined, but it doesn't equal to the limit of the derivatives nearby. Meaning, derivative function, there's no guarantee for that to be actually continuous function. So that's all there is to it. But our last class here went down to cover remnants, and this is where we ended up. And I'm not understanding the picture correctly.Okay. Now, and any questions? Still no. Okay. Now today we're gonna look at the other side of the coin. One second here. Yeah, I am seeing your homework. Wow. Every time when I look at your homework, I'm rather impressed. Johnson, why don't you send homework to the group? It's a show off.就說的，所以大家可以 bring up the standard，你 know nobody in this group. I thought I posted the link to the group. Yeah, yeah, you did. Okay, sorry, I thought that was personally to me. You know, Elaine, why don't you actually follow Jonathan and see what a beautiful job he does? It's like I said, good organization on the note shows good organization of the mind. So that's my admiration for you. Alright, after we know the Roman sum.And Riemann sums are numerical method, meaning if we don't know all the information about the graph, or sometimes we do, and sometimes we use Riemann sum because you're only given discrete points. Now there's nothing you can do; you have to really just connect them and do a approximation. But sometimes you are given an analytic equation for the graph, but we don't want to do sometimes even this function. This is a famous Bow curve, and there's over two just to get the standard deviation right. But this curve itself.It's actually the famous probability distribution of a huge category of the human phenomena. They all fall into this bell curve. And however, if you find the antiderivative, we can't. It's not because we human beings are not smart enough. We can strictly prove it. Prove here, and it doesn't have an analytic form of the antiderivative. Period. What does that actually mean? Let me you can only do the antiderivative numerically. Meaning. And this is the best we can do. But basically, you just integrate it by using Riemann sums. That's exactly what your calculus doing. Whenever you tell it to do an antiderivative, well, not not to do an antiderivative, but rather to find the definite integral for you. It automatically slices up into an an an chunks and then starts adding the function value. Now I'm actually comparing the two different applications, and I really want to emphasize.There are certain well-behaving integrants, and yet you can never write your antiderivative as a analytic form—meaning as a formula. There's only the choice of numerical integration. Now I want to mention a few AP perennial favorites. Now they could give you these other function values. Actually, let me—I should do that. They give you something like that. And we're saying we're using equal notice. This is.Oh, I give you. Okay, those are all the function values. They could be given as a table, or they could be given as this scatter plot. And that's my x, and that's my y. And I do assure you that I were using equal length of intervals to do Riemann sum. Then could you give me any comparison between the left Riemann sum, the right Riemann sum, the trapezoidal sum, the midpoint sum, and the integral?All you know are these isolated points. Oh, of course, we do know the function is everywhere continuous. Although we do know that's not even necessary for integrability, but for high school students now, usually they want to assure you, "function is continuous, and it's integrable." Continuous necessarily makes it integrable.So doing a the comparison among these. You can take the mean sum of these points, I believe. What would be your interval length? Well, given the data, you can only do that, right? Right, you can only do that. There's nothing else you can do, but there are.To equal intervals. Oh, by the way, we're looking for these are the only data points given to you. We're talking about the the left-remain sum of the function, not necessarily this left-remain sum. I'm basically saying they are all we know about the function f of x. I'm asking if we do the left-remain sum with equal length integral interval on the f of x, and we have the same center of the x, these x.Basically, we cut up our from a to b in a certain fixed manner, and I can do the left region under the da da da. But I'm not talking about this one because it's not equal interval. Not to mention, these function values are too apart too far apart.We're asking how we would do a remuneration. Well, we just we just divided up into an equal intervals and do that remuneration. Guys, let me save you the suspense. I suspect. That you don't know what I'm talking, but meaning it's not a test of our mathematical logic. It's my communicative skills now. I will simply tell you the answer. There's nothing we can know. My only point is, given the individual function values here, there's no way for you to tell why the function cannot behave in this extreme fortuitous or volatile way. You know, it could be behaving like that. So that whatever the point you're seeing, what looks like the concave down and.None of that is reliable. Your function value could be very volatile, as wild as you want it to be in between. So keep in mind that if you are only given discrete function values, although you are promised it's smooth, you know it is internally differentiable, guaranteed smooth, until you don't know because even if I just go that that that volatile, it is still smooth. The derivative would still be really huge. There is not.Nothing you can say about the function. Am I being crystal clear? If it has the sort of loves to give you basically some information on a discrete point, however, on the other hand, there's another category problems they guarantee you the function is monotonic, and then immediately that changes the entire story, and then if it is promised to be monotonic, and they give you a bunch of points now, something like that.And then indeed, you can do something. Although I still don't know between the two points whether it can keep up or can keep down. I still don't know that because it could be actually something like that, right? Nevertheless, you do know it's guaranteed increasing of the function value. So at least you would know if we do know the function is monotonic, and then you would immediately tell the left sum compared to the right remainder sum. But because we don't know can.So we don't know the comparison between the mid Rimansaw. Okay, so that's actually several things to keep in mind when you do definite integrals. Now I'm going to move on to the next topic, which is a big one. We're going to actually stay here for at least two full sessions, possibly two and a half, because today it's it's it's less than two sessions now. So maybe meaning we're going to actually have a next session, next session. These are all about.这叫什么的 divergent or improper integrals？ Because there are two different kinds of improperness, and each one is a different chapter. So these are improper integrals. Let me define improper. The first type will be explicit. It just means the boundary is improper. You're integrating from like infinity to five, or integrating from zero to part of infinity, or.From like infinity, about infinity, or you're into well, that's pretty much it. They involve infinity. That means you can't really because infinity is not a number. That means after you find out the antiderivative, you can't plug in a number. Makes sense. So this is called improper. Now to begin with, I want to give you an intuition. A integral here, maybe it's improper integral. I'm integrating from zero all the way to infinity.But it doesn't mean the total area must be infinite. There are plenty of chances the total area is finite, but it doesn't mean either way because it could be infinite. But is it sufficient to if I tell you that f of x, it's monotonically decreasing to zero, f of x is approaching zero. Does that guarantee? Oh, by the way, it's integrable. Does that guarantee when I do from zero to infinity, f of x will come out as a finite number? We know for that to be finite to be well defined, which is an improper integral yet, and would converge to a finite number. Meaning it's well defined enough. Necessarily requires the f of x to approach zero. Otherwise, if you think about the area covered under the curve, if there's going to be any limit at all, then surely the tail end here, the total accumulated contribution from the f of x simply would amount to zero. They can't contribute anything.Because the total area has to be finite, so we know the other way definitely has got to work. But my question would be: Can I infer forward? Do I know for sure if the integral is approaching zero, then the entire integral is finite? You could try to give me an example or counter example.Ah, I do not believe so.Very good. Well, if your conclusion is negative, you could simply give me a counter example. So, um, maybe like a rational function. With one end that goes to infinity and one end that goes to zero. Okay, I. Oh, oh, okay. Well, then that the trouble happens at the zero. That's the second kind of improper. That means at.That point is undefined. The zero is a finite number, it's concrete, but your function value is undefined. Let's call it the second type. But Johnson, you're totally right. You're totally right. Johnson saying he's really showing me the kind of integral where you have a finite boundary, but your integral is already going to infinity. So surely, by the way, that is called the second kind. The first kind only has to do with a boundary is equal to infinity part. So when I said that, maybe I'll use.One, you can still make x minus one and do this now. And nevertheless, let's avoid meaning at any finite point. The function is well defined. In other words, we call that the function does not have any singularity in this domain. A singularity is where the function value is divergent. Divergence simply means it's approaching infinity. But if we just take this.例如呢？张德森，suggesting a rational function。那 why don't I try one to infinity？ Sorry， more over the x， the x。 Is that finite area or infinite area？``` Yeah, infinite. Yes, because the antiderivative solves natural log. Right. So meaning you can plug in the lower boundary, but when you plug in the upper boundary, it's going to diverge. Eli, are you comfortable? Yeah. Well then, I want to make a comment though. Ah, don't just say in.Narrow down to the formula, and in fact, we do know function that's decreasing. Whether the total area is finite or infinite, actually let me not give us a singularity. This is a whole function, a function decreasing, but the whether this entire area is even the finite actually depends on whether it's decreasing fast enough. Especially at the tail end, for example, one over the x here doesn't descend or decrease fast enough. For comparison, though.If you want to do one over x one point five power, and I'm still integrating from one to infinity, try this one: see if the total area is finite or infinite.中文`````Ì have it. Go ahead. So I got it equals one. Yes, indeed, that's fast. You find the anti actually, and were you careful enough with the coefficient though. I just give us anti derivative first.I have one over the square root of x. Oh, and then. I that's where you have to be careful. The power is right, but if you start from here, I love your procedure. We want to get the box figure, get the power right, but I always take the derivative to double check the coefficient thing. Times and.Negative one half. Yes, so this is actually times negative a a a a. I mean, you are right. If you're telling me, when you take the derivative, there will be times negative one half. But because I want the coefficient of one, times negative two. Yes. Let me write it here. Correct. So that's our antiderivative. Well done. Now when you plug the boundary, and we notice here, after you find the antiderivative, we still end up with positive power on the denominator. So it still converges. The answer is simply going to be positive two. Wait, why is the one's coefficient a negative two? Try here, Elaine. Why don't you take the derivative?Of this guy, exit ratio to ninety-one horsepower. Double check whether it's gonna give you the desired antiderivative.``Òh. We got confused with the notation for a second. Okay. Are we good? Yeah, well done. So done. We know that's actually convergent, as opposed to one of dice. It's divergent. What? How do they differ? For the simple fact, this one the tail and this thinner, so because of the higher power on the denominator, it just decreases faster, as opposed to the one.Over the x, then can we put them into a family? If I go for one over the x to the p. By the way, this is called a p-test. In the future, there's another version when we do the power expansion. So basically, it's the power sequence, not power expansion. It's called power sequence. So anyway, now for this one, and if I integrate from one to infinity, can you tell me what is the if and only if condition for the p so that the function could be convergent? That's the dx.I think you pretty much have the conclusion already. You want that to decrease fast. We have already checked. Well, I'll let you finish it. Look at the p here as a undetermined parameter.``Ì'm sorry, could you repeat what P was again? P is the exponent. Because remember, we compared the several scenario, we decided that this one when P equal to one point five. And that was convergent. This is finite. It's convergent. When p equal to one, it's divergent. But we surely need p to be greater than zero, otherwise the function doesn't even approach zero at infinity, meaning the function doesn't even decrease. But even if we guarantee the p is greater than zero, we have something on the denominator. It doesn't always guarantee convergence. So I want you to just deal with the whole family.List negative power functions categorically and find me what is a cutoff line for convergence.It's greater than one. Exactly, it's strictly greater than one. You said it perfect. It doesn't include the equal to one, and the reason is that if you do have a, let's say x point x to the one point zero one power, no matter what, by using the power rule in reverse, you end up the power of power on the denominator. There's a life over beautiful, beautiful. You got it. That's why I figured it out because the power rule subtracts by like one. Yeah, that's exactly right. That's really great.Oh well, I shouldn't bundle them up together because we're talking about the type one now. But since Johnson mentioned it before, now let me actually give you this: What if I should give you one over the x squared going from zero all the way to infinity? Does that converge? What if you go from the zero to infinity one over the square root of x? Does that converge? Can you prove to yourself, no matter what p you put there, if you're integrating from zero to infinity one over the x to the p dx, always.Diverges. You don't have to do redundant work. We have already figured out to guarantee convergence at infinity, meaning type one convergence, which is caused by explicit infinity on the boundary. We need p to be greater strictly greater than one. But now let's look at what's going to happen when you're looking at what's going to happen in the vicinity of zero.The P is zero. Oh, okay. Now P is zero. You could also, they are. That's a legit case. We can't exclude that. But obviously, that's divergent. Because you're just integrating a constant one all the way to infinity.`````` It is because the denominator is almost always zero. You are saying there is a singularity, meaning the integral is approaching infinity. In the vicinity of zero, does that automatically mean the total area is zero? No, it doesn't. Why don't you find the antiderivative? Have you found.Under the rule for this one.``Ìt is Square Root X. Correct. So of course, oh, I think we know. Over two. Hey. Two, I got two X, two Root X. Yes, yes, the other way around. Don't fall in the same.In place twice, so this is a yeah. That's the antiderivative. Of course, the infinity and its it's hopeless. Definitely approaching infinity, but at the eight zero end, though, it is actually finite, isn't it? Fundamentally, it comes from if you graph the antiderivative, if you graph the derivative, now you do end up with infinite tangent line. Remember earlier when we analyzed that given the graph of the f derivative, let's find what's the behavior of the function. We encountered the fact that.Your derivative could have a vertical asymptote. It could be approaching infinity in the vicinity of a finite number, and yet we were undbounded. We say we could still reason out what's the behavior of the antiderivative of the function itself. So, so this is well behaving. So basically, the lower end is safe. It is finite, or be it divergent. I mean, or be it encounter a singularity for the integral. This integral is very infinity, but the total area is finite.But combining the two sides together, though, you I think could be ready to answer my question.``` `````Ìf x is greater than one, it approaches positive infinity, and if x is less than or equal to one, it approaches negative infinity. Ah, you're saying that once you plug in the endpoint, a. In other words, when the x is, you're right. When the p is less.On equal to one, what's really divergent is at the zero end. So meaning it's a lower boundary. When x is approaching zero, then we have trouble because that's going to be. Wait a second. Let me say again: when p is less than equal to one, what's having trouble is x approaching infinity. And there it becomes divergent. Johnson said, "The whole anti derivative when you plug in the boundary is going to be approaching positive infinity." I totally agree. Now when p is greater than one, what's really having trouble is actually.At zero end, that's absolutely right. It is because when you actually plug in the end meaning, you end up with some kind of a negative power on the denominator. But when you when you minus was going to show up on the denominator, you're definitely getting a divergent. So the issue with the bottom, however, Jonathan, that's not approaching like infinity though, because when the the p is greater than one. Yeah, it's one over zero. Yes.It is one with zero, but it's a when you're minusing a negative number, so whatever the antiderivative is still positive though. I mean, okay, okay, meaning if you plug in the infinite minus zero side, all right. So these are the two types of improper. I just happen to bundle them and ended up bundling them up in one now. So fundamentally, when the boundary is going to infinity, and then you go for the P test, but on top of it, not every.一个函数是power function，you have to really analyze the case by case，find the antiderivative until，however，sometimes it's very complicated to find the antiderivative，for example，let me actually give you this，and I want to judge whether this is convergent or divergent，let me save you the the suspense now，I'm not going to actually give you two ends，both are in contrary issues here，this is a type one improper integral，and I'm giving you the x natural log.Of the X, the X. Right, my purpose is just don't don't try to find the anti. No, no, no, no, you can. Okay, sorry, sorry, madam. Madam, I did not mean that. I actually mean this. Meaning, I don't want you to try to find the anti. Derivative. Although I can't say one hundred percent sure we can, meaning there's not analytic form, but very likely there's not. At least I don't know how to find the real antiderivative of this, but I still want to find its convergence or divergence. Remember though, we can use a lot of leeway by comparing. If you know this is slightly bigger than the divergence integral, but that's easier to tell whether it's convergent or divergent. Okay, it is slightly bigger than the divergent one. This is definitely divergent. If.It is slightly smaller, now the convergent one is definitely convergent, so that's the idea. By comparison, you could actually simplify the form into something that's eyeballable.```We simplify it down into one over x plus one. But this is a lot smaller than that divergent integral. You can try to compare with this, but the issue is that's a lot smaller than that, and this is divergent. So your comparison would be inconclusive, right?Does it make sense? It's divergent, but it's smaller than this one. Wouldn't this one also be divergent because it's larger? That's smaller. I wrote the inequality. By the way, you could have the sense we're dividing by something that's approaching infinity on the denominator.Ah, do you follow what I'm saying? Yeah. Right. So one of our natural logics, but that's also significantly smaller. Elaine, there's something you shouldn't be doing. Meaning the bug figure, what is actually making the fast or smaller. Well, to begin with, we know this is.Really sitting on the boundary, you can't tell for sure because if you don't have the natural log of x, you just have x plus one. The p is only one, it's not big enough. We need to add something that's approaching infinity on the denominator to give it any hope of convergence. If I just give it x plus one to the one point one, that's really good enough to be convergent, right? But unfortunately, we have to know the natural log of x will be at approaching infinity, approaching infinity slowly, rather than any any tiny power function.So if I actually compare that with the x to the one point zero zero zero zero point one, that's still approaching zero, meaning whatever you add it on the denominator makes it possibly convergent, but it's not enough to compare it with any power function. So that's where we really don't know. It's and it's getting very close now. That's why I said earlier, it's not reasonable to compare it with the natural log of x because we have lost the the real convergent force. I mean that that. Power, it's so much increasing so much faster than that. Meaning, if you just compare with this, we haven't even captured the bulk figure of the denominator. Do you follow me? Yeah.中文嗯。``````中文嗯。Oh, I tried multiplying both the numerator and denominator by x minus one to raise the power of x on the denominator. But you're raising it too much, though, and this is smaller than that, and that one is convergent. So the issue is.If you, the point is, you can't make it smaller than the divergent one because that's not going to be conclusive, and you can't compare with a one, meaning you can't make it bigger than the convergent one. But again, both are inconclusive. How do we make it integrable? Try this. I'm not going to change it much. You, you should have the sense that that's really cutting very close. I didn't make much of the change at all. Can we see these are really on par with each?I haven't significantly changed, although this is just a little smaller than the other one. I mean, a little bigger than the other one, right? That's a little bigger than. I mean, this, I made my integrant a little bigger by doing that. Because I just changed the numerator from x plus one into x now. My x is big. That much of the change is nothing. Do we agree? Yes. Okay, then can we just find the antiderivative of this? By the way, that's in the IP level. They would antider, they would expect you to do to know how to integrate that.``````嗯，嗯。This is remind you chain rule, rather reverse chain rule. Do we notice part of the integrand would happen to be the derivative of some other part?What did you say? The one over x part. Yeah, brilliant. That's good. That's good. I was using the power. I mean, the the most location rule. I'm checking my answer.`````` 嗯。Can you tell me what? You guys did something wrong, so I'm redoing mine. I'm having trouble on just getting the natural log x. Hmm, but we agree though. The one with x happened to be the derivative of inner function, so whenever you can't see through it.Maybe we shall do the u equal to the natural log of x, so that you know the du equal to the dx over the x. So now you can rewrite this antiderivative. And you will you will be able to see.``````So I got the integral u d u. Which is u squared over two. Eh. Wait. U d u. I mean that is the d u, alright. Ah, d d u over u. That's better.So the solution is natural log of natural log of x. That's right. Yeah, I mean the natural log of the natural log of x. Yeah, so that's divergent.At infinity, right? A bit of intuition. Yeah. When you're trying to find the antiderivative of one over x, now this is a decreasing area, and the area grows pretty slowly as a natural log of x, but it's still growing to infinity. But when we actually put a natural log on the x on the denominator, we're making the decrease even faster, faster now, making the area grows even slower. But it's slower by taking. In the natural log, the natural log, so it's flatter, and yet it is still going to infinity. So actually, this is still divergent. However, we are in a little bit of a trouble because I said you can't compare meaning. This integral happened to be smaller than a divergent one. This is smaller, smaller. So when you write, it gives you the intuition out because the between x plus one and x, there is really hardly any difference. But when you write a formal proof, you can't do that.Because our interval is smaller, we can't draw any conclusions. What do we do? I played that trick before with you guys, or not with you, but rather for you. So what do we do? Just so that we could have a disproof. That it's either convergent or divergent. No, about how to make it bigger than something without changing.The nature of the function we have found. Meaning, get rid of the plus one to make it integrable. Well, the original is integrable, but I mean, make it an analytical integrable. Meaning, you can find out what is antiderivative easily. Oh, why can't we compare it to the dx like the integral of the dx over x natural log x? If it's greater than that.Oh, sorry, this is smaller than that. You're integrated smaller, so although we did find that this is divergent, but you can't say anything smaller. Oh, oh, I see. You can't say anything about. You can't say anything about x plus one, which makes the denominator bigger. Oh, that's right.X minus one times natural log X minus one. Eli, natural pretty move. Let me ask him then, and did you catch what he said?X minus one times natural log minus one. As plus one. Knowing that we can't make the x smaller, hence this whole integral bigger. So I can make the x bigger to make the whole integral smaller. So he actually did the move right here, Johnson. I'm saying what you did, and that's really a brilliant move. He's saying, "Eli, whatever our integral will be bigger than this one when I."Added one to it, but that just a rename of the new variable. That's also the antiderivative of that. Whether added a plus one, it's still natural log of the natural log of the one plus x. Jonathan, I begin to have the feeling you gradually ease your way into calculus now. That frees your intellect. Earlier, for a month, I really. By the way, there's nothing unhealthy. It was very healthy. But I did notice here. I think your mind was still tied up in getting to know what this is about.Just knowing what the language means, so that you couldn't demonstrate your flexibility, your creativity, your smart, but gradually now you can. That means your mental energy is not just so preoccupied with simply understanding what is the formal notation here and what is this concept. Once you do know the fundamental concepts, now you can really show your your creativity as a as just a mad savvy person. So that's really smart, Elaine. Do you agree? Yeah, so we're done. So it turned out to be divergence. More or less, in the future, if you're gonna try to judge the convergence or divergence, you have several rules which handles a big category of functions. For example, power functions. That's the P. It's the P test, and to to judge whether the function decreases or or increases fast enough or slow enough. And there will be natural logs, there will be exponential functions, but what if you just straddle.In this awkward kind of form, it's none of the above. For example, the one I gave you, and you can't eyeball what is antiderivative. Well, you can tweak it. You have to recognize what captures about figure. What kind of change won't really derail your convergence or divergence? Meaning, they're about the same, and you tweak a little, and then so to the point that you could conveniently find your antiderivative. I do admit though, AP test very seldom give you any interesting.Problem at all. You'll be lucky if you encounter such a problem that takes a bit of thinking. They don't give you that. They dare not give you that. It's not their fault that students are too bad. But anyway, so knowing that this is the idea. Are we good so far? Alright, now let's actually introduce the second type. By the way, the second type is more nuanced. The type that where we have I have a lot more to say, and the questions could get a little trickier, a lot trickier. So the second type of a divergence.Or convergence here, or improperness happen at the basically, while behaving boundary integrating from a to b, a and b are finite numbers, and then you're doing the f of x dx, but very slightly, the function value it's approaching infinity at a certain point in between, at the x equals to a particular c in the open interval, including the a and b, including the a and b, it could be on the boundary. But the boundaries are not infinity, and they are of course.The second type, meaning what's giving us trouble is the function value, its divergence, not the boundary. We're good. Now we want to find out: does the integral get well behaved? Although on top of it, it's also improper if you have a point simply undefined. But we covered that previously. No, for finite points here, you could have just moved them around because whenever you could cover the sliver.Of the discontinuity within infinitesimal area, according to the spirit of Romans, it doesn't matter, it doesn't hurt, we're still good. Alright, now let's look at a few examples. I can give you one over the cosine theta minus the one, we're integrating the d theta, theta goes from zero to actually, I can integrate from negative half of pi to positive half of pi.By the way, we lately did this interview. Jonathan did a beautiful job. You may not remember though, because back then I don't think you were fully immersed in what the substitution means. And do you remember how you did this? Not really. You did the substitution using half of the Italian data as the.Oh, yeah. I think I thought about using like a trig rule. Ah, beautiful, beautiful. You could also do half angle. If you do half angle, the denominator is really negative sine half of theta squared. But now we run into the type where you have an even power of the sine. Those are not that easy. You have to use double angle formula. You know that just go right back. In fact, it shows you what you're gonna do in the future if you start from here. And but okay, we went through this before. You can check your homework. You both did a beautiful thing. And by the way, you don't need to memorize such stuff. In what to begin with, AP tests are always very very simple on finding the antiderivative. I think your skills are fine. We are going to do last round of the just slightly trickier problems before the test. But now I want to do not being able to you don't have to find the antiderivative at all. We just want to judge the convergence and divergence.But to begin with, it looks innocent enough. Then the issue must happen somewhere in the middle. I'm telling you in the future when you look at the integral, be sure to double check. You don't encounter infinity somewhere hidden between the lower and upper boundary because they don't tell you. I'm not saying this is improper integral. I want you to integrate and find out what the number is. Okay, can you write it down as a advice to your future self? If you're seeing a definite integral.First instinct to double check whether there are singularities in the middle, because they might encounter improper integral. Improper integral is actually a bigger chapter required by the APBC. People love to test you out, test you on it. Yeah, okay. Then go ahead and think about how.We could decide. If the denominator is zero, then it there's an infinity. Not necessarily. Yeah, yeah. I mean, the integral is infinity. If that's what you're saying, yes. Right. Yeah, yeah, that's correct. It doesn't mean the whole integral is infinity.```````````` 中文嗯。When baking a graph, help. I think isolate the singularity helps. Oh shoot, I'm late. I didn't even notice time. I'm really sorry. And I take it as homework. I need to go. And take it as a singularity or isolate the singularity and look at exactly how fast it approaches zero denominator. That means the whole function approaches infinity. I will do maybe doing a bit of a power expansion. That you guys are very.Good then. Okay. Take care. Have fun. Keep up your good work. Bye. Alright.

收敛性、反常积分与黎曼和

收敛性、反常积分与黎曼和

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