 Hi, I'm Zor. Welcome to a new Zor education. Today, I would like to talk about so-called improper definite integrals. Well, it's those integrals which don't behave properly. All right, so let me just make a little reminder to you. How did we define our integral, our definite integral? Well, we defined it this way. Let's consider we have a function f at x, which is defined on some interval a, b. Now, we assume that a, b is a segment, a closed segment, of real numbers. And function f at x is relatively smooth. Let's put it this way. Well, smooth in terms of, at least it's continuous. Well, actually, sometimes we might need differentiability, but continuous for sure. Now, it means basically that for some other segments, which means, for instance, segments, well, intervals rather, which include infinity, it's not defined. Think about this way. How actually we proceed to define this integral? We divided this into n pieces. Then we took the value of the function. So, basically it was something like this. This is xi. This is so-called Riemann sum. And then we go to a limit as the number of these intervals, which we have divided our segment goes to infinity, and each one of them goes to zero. Now, if, for instance, a or b are infinity, which means just there is no margin on the left or on the right, we can't do anything like this. Because the number of, if you have a fixed length interval, finite length interval, and we have an infinite interval to cover, then there will be infinite number of these small intervals already. I mean, the whole process of dividing and the whole theory actually would not work, because we can't really go into a limit, because it's already an infinite number of intervals. So, that's not really good. So, basically, our definition is not good for cases like this. Another type of case, which our definition is not good is, if the function is not this type, but, for instance, this type, it asymptotically goes to this vertical line to infinity, plus infinity or minus infinity, etc. Why? Because this value, f of xi, if xi corresponds to this point where the asymptot actually is, then it's infinity. So, again, we can't really do this. So, we have to cover certain cases of the function, which seems to be a reasonable function, which means we probably should know, for instance, let me just give you this example. For instance, our function is like this. Is there a concept of area under this curve? Well, maybe yes. I mean, if the function is very, very fast going to zero, then even with an infinite tail, the area might still be finite, and it is, by the way. So, somehow we have to make different definitions for cases like this. And similarly, if, for instance, we have a function which goes to infinity, let's say like this and like this, maybe they're so close to this vertical line that the area of this seemingly infinite height doesn't really present an infinite area. Maybe it's a finite area, and so the whole thing would be finite. So, anyway, I would like to make just a point that our definition of definite integral, which is called Riemann integral, these are Riemann sums, is not good for cases like this. And this lecture is devoted to expansion of our definition towards covering of situations like this. So, that's what we are talking about. This is improper integral. So, we are going to define improper integrals, which are not something which we can call the classical Riemann integral from the functions, all right? So, this lecture about improper integrals is part of the advanced math course for students of high school. It's presented on Unisor.com. I recommend you to watch this lecture from this site. The site has notes for each lecture, quite detailed notes, which you can use as a textbook. And some of the topics which are presented on this site have exams. Well, not all of them yet, but it's coming, it's coming. All right, so, let me start, and let me start from defining different types of improper integrals. Okay, my first example of improper integral is when one of the margins of the segment where the function defined is infinity. Okay, let's consider, let's say, A is equal to infinity. How can we define integral minus infinity, I should say, because it's a left boundary, right? How can we define integral from minus infinity to be f-atex dx? And let's consider that there is no points where f-atex itself goes to infinity. So, we are talking about only one particular distortion from our original definition, distortion only on the left boundary, which is minus infinity. This is a finite point, and the function is also finite and continuous, all right? So, in this case, I define it, this is a definition, as a limit from A to B of function f-atex dx, as A goes to minus infinity. I mean, it's really very, very logical kind of definition. If you have to, this is minus infinity, this is your function and this is B. So, if you would like to define the area here, well, let's cut it here, define this area using the regular integration, and then start moving this. If this limit exists, and this is a very, very important if, if this limit exists, then we can say that this integral exists and it's actually this limit. And it does make sense. Now, sometimes the limit might not exist, in which case this integral does not exist either, obviously, right? So, this is my first definition. Now, my second definition, obviously, is very similar, and it's related to the right margin of this segment. What if B is equal to infinity? So, we need to calculate this. Now, obviously you understand this is limit from A to B as B goes to infinity. Same thing. If this integral exists, well, integral exists, if the limit exists, then we are saying that this is a definition of this integral. So, now we have covered the definition in case my margin are either one or another equal to infinity. What if both of them are equal to infinity? How can I define this? Well, here's how I propose to define it. Now, this is one of the possible ways to define this. So, I just choose a point zero. I can choose point x is equal to one, or 25 or whatever. Basically, I divide my whole straight line from minus infinity to plus infinity into two different intervals. One of them has the right boundary and no left boundary, which is this one, and another has the left boundary but no right boundary. Now, these two I have already defined in the previous couple of cases, right? Now, so this is, you can just substitute B exactly actually equal to zero and this substitute to A take this limit from A to zero and have A limit by A going to minus infinity. And this I can substitute, well, A is equal to zero, so B is equal to plus infinity, which means I have to have it from zero to B and have a limit as B goes to plus infinity, all right? I have defined this integral through these two. Now, again, only if these two exist, because it might not. If either of them doesn't exist, then this doesn't exist. Now, is it possible? Here is a very important question. If I will take another point instead of zero, if I will take another point, P, can I do it this way? Any other point P? Well, actually, yes. And it's the same thing, because if my integral from minus infinity to zero exists, then my integral from minus infinity to P also exists because the difference is I have to just add integral from minus infinity to zero. I have to plus integral from zero to P and then I will get from minus infinity to P, right? And this always exists, so I don't have any problem with this because it's a finite integral, right? And same thing here. In this particular case, I have from zero to P and then from P to plus infinity, all right? So, in both cases, I mean, in this particular case, I can have it this way. I just made a mistake. It's zero, so zero P integral, okay? That's how it is. So, this is equal to sum of these, from P to zero and then from zero to P. This is equal to sum of these, from minus infinity to zero then from zero to P. And this is equal to zero, right? Because these are opposite limits of integration and we know that the sign is changed to opposite if I change the direction of integration. So, I can represent this in this way and this part is also equal to zero. That's why the result will be, if result exists, then it will be the same whether I put any number P here or number zero or anything like that, all right? Okay, so, I think we have covered all the different possibilities of indefinite interval of integration. Left boundary can be infinity, right boundary can be infinity or both can be infinity. Now, let's cover situation when the function itself goes to infinity. And we will do very similarly. We will use limits. So, let's assume that so far we have a finite segment but the function somewhere here goes to infinity, something like this. Well, let me do it simpler than that. Let me just have only one end. So, only at the A it goes to infinity. How can I define area under this curve? Well, very, very simply and more or less analogously to my previous couple of definitions. Let me cut here. Let's say this is A plus D. Now, A plus D is a point from which on my function is finite. So, I can always have integral from A plus D to B of function f of x dx, right? Now, this is fine. That's no problem because in this interval this integral exists so I can calculate it. And now I will take the limit as D goes to zero which means my point goes closer and closer to this one. If this limit exists, then this is the definition of integral from A to B of our function f of x dx. Now, analogously, if my function has infinite, goes to infinity at B and finite on this particular case. So, instead of doing this, I will do this minus G. So, I will step to the left. And again, my integral from A to B minus D always exists and defined, right? So, I can always take this integral and take the limit as D converges to zero. And if the limit exists, then I can just call this is an integral from A to B. Well, finally, if you have this particular point of infinity somewhere in the middle, let's say this is B and this is C. And I have to define integral from A to C. Well, I will define it as integral from A to B plus integral from B to C. Now, this is already defined because in this case, my function takes an infinite value at the right boundary of this particular segment and they know how to do it using this type of things with B minus D. Now, in this case, from B to C, it's on the left where a function goes to infinity. And I do it as I did before from A plus D, right? So, these two are already defined using my previous logic, which means I define this as a sum of these if and only if these two exist in terms of their limits, corresponding limits. So, basically, what I would like to say that I have defined all the different peculiar cases and my definite integral is defined in a broader range of segments, which can be infinite, or function which can have certain points where it's going to infinity, plus infinity or minus infinity, whatever, doesn't really matter, all right? While at the rest of the points, it's fine and finite and continuous, et cetera. Now, if there are more than one point of peculiar behavior of our function, then, again, we will just define as long as there are finite number of these peculiar points where the function goes to infinity, I can always divide my big segment into small ones and calculate integral in each one of them. Considering it exists, then I will just summarize them together. And now it's about time we will just go to a couple of examples if you don't mind, all right? So I will just show how to do this. So I have only two examples. One for infinite segment, rather interval, I should really say. The word segment is not correct. So one is interval. So my function is 1 over x square, which is this, and I will integrate it from 0 to infinity. Okay, so integral from a, sorry, from 0 to b of this particular function equals. Now, what is indefinite integral of 1 over x square? Well, it's minus 1 over x, right? 1 over x derivative would be 1 over x square with a minus sign and minus this minus will neutralize that minus and I have to substitute 0 and b, right? Actually, it would be probably better if I would do it from 1. I mean, this is 1, this is 1 because I don't want to have 0 in my denominator. So let's start from 1. From 1 to infinity, this. So from 1 to b, from 1 to b, so if I will substitute b, it would be minus 1 over b. And if I substitute 1, the 1 minus 1 over 1, right? That's what it will be. If I substitute b, it would be 1 over b with a minus sign. So this is the Newton-Leibniz formula, right? So I took the indefinite integral under anti-derivative and have to substitute first the upper limit and minus substitute the lower limit. So which is equal to, this is 1 minus in the minus, so it's 1, 1 minus 1 over b. Now what happens if b goes to infinity? Well, this disappears, right? And I have only 1. So this integral is equal to 1. Well, this integral is infinity here. So this is an example of how I apply my new definition, the integral which involves infinite right limit. I have basically replaced with integral with a finite right limit, calculated it, and then went to a limit as my right boundary goes to infinity. So that's my first example. Now, my second example is related to a function which goes to infinite value on its segment where it's defined. Well, again, it's not even a segment. So consider the function logarithm x. This is 1. That's how logarithm x behaves, right? This is a natural logarithm. If function is equal to 1, I mean if argument is equal to 1, the function is equal to 0, and that's how it behaves. So I would like to know if this area exists. But you see this is where the function goes to infinity. So my question is, I mean obviously the area can be large and negative by the way, but maybe it's large which goes basically to infinity, or it's still kind of a finite number. So how can I find out? Well, what I would like to know is integral from 0 to 1, from 0 to 1, function logarithm x dx. Now, again, 0 is this peculiar point where the function is not defined basically, right? So I have to define it as a limit from a to 1, logarithm x dx, as a goes to 0, right? This is a. So this is definitely a finite piece, so I can calculate it. So let's calculate it. Now, what's the indefinite integral of logarithm x? Well, I happened to prepare for this lecture and I remember, but it was actually a subject of one of the previous lectures where we were talking about derivatives. So we did actually talk about this. So it's x logarithm x minus x. That's what my indefinite integral is. Just for a check, let's do derivative of this. Derivative of this, it's a product, so it's derivative of this, which is 1 times another plus x times derivative of logarithm x, which is 1x minus derivative of x minus 1. Now, this is x and x, this is 1 and minus 1, so this is cancelling and I have only logarithm x, you see? So this is the correct indefinite integral. Okay, so now I have to substitute upper and lower limits of the integration. Now, if I put upper 1, so logarithm of 1 is 0, so this is minus 1. Minus, if I will put a, it would be a logarithm a plus a, right? Minus and minus in front would be plus. Now, as I goes to 0, what happens with these members? Well, this obviously goes to 0. Now, I'm also stating that a times logarithm a also goes to 0 as a goes to 0. How do I know this? Well, first of all, I know that logarithm is a slower changing function than the power function. So that makes me to believe that, but that's an infinity. Now, whenever we are going to 0, let me just do something similar. I will do this, logarithm a divided by 1 eighth, right? This is the same thing. Now I can use the L'Apitales rule. So, instead of finding the limit of this ratio, I can say that this is the same as limit of their derivatives and what is their derivatives? Derivative of logarithm a is 1 over a. Derivative of this is 1 over a squared with a minus. So this is limit of minus a, right? As a goes to 0, which is 0. So this is also 0 and what's remaining? Minus 1. So that's the answer. That's my integral. Now, by the way, why is it minus? Because the function is negative, right? So whenever the function is negative, obviously, all the area so-called under the function, it's actually above the function. That's why it's negative. But anyway, this is how we are taking integrals in case of either one of the or both margins are infinite or the function is undefined actually in one of the points and goes to infinity in this particular point. So this is what's called improper integrals. They are improper because we cannot really use a classic definition which is using these Riemann sums of f of xi times delta xi. We have to really use some other mechanism to define these so-called improper integrals. And once we have defined them, we basically can calculate it using the same formula, Newton-Leibniz formula, for instance, or in any other way, whatever. Okay. I do suggest you to read the notes for this lecture on unison.com and try to take these two integrals just yourself, whatever I was just doing. They are presented as part of the notes with solution, but don't look at the solution. Try to do it yourself. It's always helpful. And I might actually think about some maybe exam questions which I will also later on put onto the website. Okay. That's it for today. Thank you very much and good luck.