 Hi, I'm Zor. Welcome to Unisor Education. I would like to continue talking about improper integrals, basically a couple of examples for this lecture. This lecture is part of the course of advanced mathematics for teenagers and high school students. I suggest you to watch this lecture from Unisor.com website, although I'm sure you can find it on YouTube etc. but the website gives you detailed notes for each lecture as well as the ability to take exams, for instance, for any topic. So, I have four different examples of definite integrals which can be called improper because of certain infinities either at the limits or at the values of the function. So, example number one, integral from 0 to 1 to x1 over x2 dx. Now, what's improper for this particular integral? Well, look, we have this denominator which as x approaching 1 it goes to 0. So, if denominator goes to 0, 2x is basically something around number 2, obviously, if x goes to 1. So, we have an infinity. Now, how can we deal with this particular integral considering we have this infinity? So, the function goes to infinity right near the margin. So, it basically asymptotically goes to infinity. Well, obviously, we have to convert it into a limit because this is a definition of improper integral. We go to b where b goes to 1 of this function. So, that's the definition in this particular case because we cannot define this particular function on this segment using dividing the interval of integration into smaller intervals and when the number of intervals goes to infinity and the size of intervals goes to 0, then we can talk about the Riemann sums, so to speak. Now, there are no Riemann sums if this is going to infinity. The function goes to infinity on this segment. So, we have converted into this because on this particular segment, now when b is approaching 1, I'm obviously talking about b approaching to 1 from the left. Sometimes, you can put 1 minus 0, which basically means the same thing, approaching from the left. So, in any case, let's try to find this particular integral. Now, this is a proper integral which we can use formula of Newton-Leibniz to integrate it. So, that's equal to limit b goes to 1. Now, how can we integrate this? Well, let's just, again, make a little intelligent guess. See, 2x is x square derivative, right? Which means I can combine 2x and dx and replace it with d of x square divided by 1 minus x square, right? Now, why did I do this? Well, because now I have a much simpler expression based on x square. So, now, how can I proceed further? Well, I'll just make a substitution. Now, if x is equal to 0, t is equal to 0, if x is equal to b, t is equal to b square, right? So, I have replaced, I made this substitution. So, differential is this, 1 minus x square is this, but my limits are different because limits of the t, if x from 0 to b, for the t, it will be from 0 to b square, right? All right. Now, this is something relatively familiar because it's equal to limit b goes to 1. Now, what is this? Now, you remember that derivative of 1 over t is natural logarithm. So, derivative of 1 minus, 1 divided by 1 minus t would be natural logarithm of 1 minus t with a sine minus, right? Because it's a minus. So, derivative of logarithm of this would be 1 over 1 minus t times derivative of minus t, which is minus. And we have to substitute various from 0 to b square, right? So, this is my indefinite integral from d t over 1 minus t minus logarithm of this. So, now, all I have to do is substitute and the substitution will be, limit b is equal, goes to 1. Now, if I will substitute b square, I will have minus logarithm of 1 minus b square, right? Minus, which means plus because this is a minus, logarithm of 1, right? When t is equal to 0. So, logarithm of 1 is 1 second. Did I make a mistake? Because that goes to infinity, actually. Okay. So, this is 0, right? And when b goes to 1, 1 minus b square goes to 0, which means logarithm goes to... So, this would be infinitesimal value, right? And the logarithm, you remember the function logarithm? When my argument goes to 0, logarithm goes to minus infinity. So, this would be not convergent number. So, it will be infinity, which means that the whole integral is non-convergent. So, this is an example of a non-convergent integral. And the answer is, there is no answer. Well, sometimes people can say this is infinity because this is minus infinity and this is minus, so it's a plus infinity. So, you can say that. However, I probably would prefer that instead of saying that integral is equal to infinity. Quite frankly, it's much better to say integral is non-convergent. So, basically, there is no limit and there is no value. Next. Next is integral from 1 to infinity sine of 1 over x divided by x square dx. Now, here also I'm kind of making this thing a little simpler because you remember that 1 over x is equal to minus 1 over x square, right? So, derivative from 1 over x is, which means that this is equal to minus integral from 1 to infinity sine of 1x d of 1x, right? Differential of 1x is the derivative, which is this, times dx. So, that's why we have it in the denominator. And since it's minus, I have to neutralize this minus, that's why I put minus here. Now, in turn, what is sine t times dt? Well, this is a differential of cosine t with a minus sign, right? Because what is the derivative from a cosine? Minus sine and this minus gives you plus sine and dt. So, now, instead of basically t, I have 1 over x, which doesn't really matter. So, I can say this is integral from 1 to infinity, this minus. And if I will put d cosine of 1 over x, it will negate this minus, right? So, it's the derivative from a cosine is minus sine and this minus will be plus, that's why I have plus, which is equal to, okay? Now, let's do this limit thing. Limit b goes to infinity. And since I already have a differential, I know what's my derivative in definite integral. So, it's cosine 1 over x. Yeah, from b to 1. All right. Now, if I substitute b, it will be cosine of limit b goes to infinity, cosine 1 over b minus and if it's 1, it would be cosine of 1. Now, what is this? Now, this is the constant, that's why it's outside of the limit. So, when b goes to infinity, my 1 over b goes to 0. So, cosine goes to cosine of 0, which is 1, right? So, the whole thing would be, therefore, 1 minus cosine of 1. That's the answer. Okay, next. Okay, next is a couple of problems which are actually related. Here is what I wanted to do. I wanted to investigate how the function y is equal to x to the power of a integrated. Well, you see, we have this is infinity and this is infinity, right? Now, I'm obviously talking for a negative a in this particular case because for positive a, we don't have obviously convergent results, right? With a positive a, it goes like this. So, obviously, integral to infinity would not converge to any number and it's not really interesting, right? So, we're talking about negative a. All right. So, for instance, we have 1 over x, right? One of the examples. Another example is 1 over x square. 1 over x square goes like this. It's higher there and lower here, which means this piece should be smaller as far as compared with 1 over x. But this piece from 0 to 1, the area under the curve should be bigger. Or we can have something like this. Let's say it's 1 over square root of x. This is 1 over x square and this is 1 over square. So, it's higher than 1 over x in infinity, but lower than 1 over x when it's close to 0. So, my question is, you see, considering we have infinity here and infinity there, I don't want to deal with two infinities. I would like to deal with only one. So, I will divide this problem into two. So, my first problem is this one. So, I'm investigating this function, integrability of this function from 1 to infinity. And then my second problem would be from 0 to 1. All right. So, that's my preamble to this problem, to these two problems. So, right now, we're talking about integral from 1 to infinity of x to the power of a dx. Okay. Obviously, let's do what we know how to approach it, right? So, it's limit from 1 to b as b goes to infinity x to the power of a dx. Okay. So, now it's a definite integral with concrete margins from 1 to b and we can very easily basically do the whole thing, right? So, what is the b? What is the indefinite integral? Well, obviously, it's x to the power a plus 1 divided by a plus 1, right? Because derivative of x to the power of a plus 1 is equal to a plus 1 times x to the power of a and it will cancel this a plus 1. So, I will have only x to the power of a. So, this is the correct formula. But is it always correct? Well, first of all, we do know that we are talking about a less than 0. However, there is one point which is a is equal to minus 1, which is not exactly the good one in this case, right? So, it's not really working always. Okay. So, let's put a not equal to minus 1. So, we are talking about function at the power of a for all negative a except 1 over x, right? When a is equal to minus 1, it's 1 over x. And you remember that in this case, the integral is equal to logarithm x. But let's just consider it later on. So, now, my question is, and we have to substitute b and 1, right? Alright. So, what is this? It's limit b goes to infinity b to the power a plus 1 divided by a plus 1 minus 1 to the power of a plus 1, which is 1. Or if you wish, I can take 1 over a plus 1 outside of the limit. And in the limit, I can put b to the power b to the power a plus 1 minus 1, right? 1 over a plus 1 is outside. Now, what is this? b goes to infinity. So, it's infinitely growing variable. If this power is positive, then infinitely growing number, the positive power will be infinitely growing, right? We need this power to be negative to bring down this, this growth, right? Because if this is negative, it's actually like 1 over something in a positive degree. So, it will be infinitesimal variable. So, 1 over infinitely growing would be infinitesimal. You understand that, right? So, basically, our condition is if a plus 1 is less than 0, if it's negative, then and only then this integral has a limit because this thing would go to 0. It would be infinitesimal b to the power a plus 1, where a plus 1 is negative and b is infinitely growing would be infinitely growing denominator, right? So, that would be 0. So, the result would be minus 1 over a plus 1. So, that's my result. If a plus 1 is negative, then this particular integral is equal to this value. However, we have forgotten. Now, let me just write down it a less than minus 1, okay? That's the same thing. Now, b did not really cover a is equal to minus 1. So, for a less than minus 1, this is the correct result. How about for a equals to minus 1? What do I have in this case? Well, in this case, I have 1 over x, right? For a is equal to minus 1. And in this case, my indefinite integral would be logarithm x because derivative of logarithm is 1 over x. Now, if I will substitute b, I will have and if I will substitute 1, I will have logarithm of 1, which is 0. And what is this? Well, logarithm b if b goes to infinity, the function of logarithm, if you remember, logarithm goes to infinity. So, my integral is not converging. So, there is no value. So, this particular value a is equal to minus 1. That's when it's not integrable. So, only for a is less than minus 1. This integral actually exists. And let me just as an example, as an example, give you a concrete value. If a is equal to minus 2, for instance, which is less than minus 1, the integral is equal to now a is equal to minus 2. So, it's minus 2 plus 1 minus 1 minus is 1. So, integral, so this is 1 over x square from 1 to infinity, this area is equal to 1. If a is equal to minus 3, for instance, I will have one half. It would be even closer to x axis in this case. So, it would be half of this area. All right. Okay, fine. Let's do the next problem. And the next problem is, instead of limits from 1 to infinity, I will have from 0 to 1. Now, again, in this case, it's not interesting actually to consider positive a, because in this case, oh, sorry, because in this particular case, obviously, the integral would be definite and proper, right? So, we are not talking about positive a. We are talking about negative to get the infinity here. And we are talking about this area from 0 to 1, but this is infinity. So, we have a problem. We have improper integral. So, in this particular case, we can say that this is limit from b to 1 x to the power of a dx. Now, again, they are considering these and again, let's consider the case when a is not equal to minus 1, in which case I can say exactly the same thing as in the previous case that this is, I forgot the word limit here, as b goes to 0, limit b goes to 0. And what's the indefinite integral? x to the power a plus 1 over a plus 1 from b to 1, okay? So, from 1 over a plus 1, right? If we want, if I substitute 1, minus b to the power of a plus 1 divided by a plus 1. Again, let me factor out 1 over a plus 1 and I will have 1 minus limit b to the power a plus 1 divided by, not divided, b goes to 0. Now, b goes to 0. So, b is infinitesimal, variable. If a plus 1 is greater than 0, or even equal to 0 actually, then infinitesimal, no, let me just forget about equal to 0. If it's greater than 0, so this is the positive power and this is infinitesimal variable because b goes to 0. So, whenever I'm raising my infinitesimal variable into the positive power, it's still infinitesimal. So, under this condition, this thing is infinitesimal and limit is equal to 0, so my result would be 1 over a plus 1. Now, I specifically did not put 0 here because that means a is equal to minus 1 and we have agreed that this is not the case, right? So, a is greater than minus 1, or if you wish from minus 1 to 0, since the positive a we don't consider at all because it's a trivial case, we are considering always over a negative a, but in this case it's greater than minus 0, so from minus 0 to 0. Now, let's consider minus 0 separately. Now, if we consider minus 0 separately, then we will have here logarithm x, right? If a is equal to minus 1. Now, logarithm x, if we will put 1, it would be logarithm of 1 would be 0, but if we will put b and b is infinitesimal, so what happens is again, this is logarithm. If b goes to 0, logarithm goes to infinity, which means there is no limit, which means integral is not convergent to any real value. So, basically this is excluded case and the correct is only this one. So, the whole thing, the whole integral is equal to 1 over a plus 1, but only if a greater than minus 1. For instance, greater than minus 1, for instance, minus 1 half, just as an example, what happens if a is equal to 1 half, minus 1 half, I mean. So, integral from 1 to 0 to 1, x to the power minus 1 half g x, it's integral from 0 to 1, 1 over square root of x g x. Now, if a is equal to minus 1 half, that would be 1 half, it would be 2. So, that's the answer. And again, remember this graph, this is 1 over square root of x. It's higher, this is 1 over x. So, it's higher than 1 over x from 1 to infinity, but it's lower than 1 over x from 0 to 1. So, what's interesting is, 1 over x is divergent on both plus infinity and 0. So, both integrals from 0 to 1 and from 1 to 0 are not convergent for 1 over x. But for those curves, which are slightly turned over 0.1, 0.1, in this way, which means it's higher than 1 over x on the infinity, but lower than 1 over x close to 0, these are convergent here and divergent there. The opposite thing, like this one, like 1 over x square, which is lower than 1 over x on infinity, but higher when it goes to 0, then it's convergent in this case and divergent from 0 to 1. So, this is something which you can feel about all these integrals when they are convergent and under what circumstances. So, these are very typical examples. And you can always think about if function behaves like 1 over A x, A x plus B, and you can even put a constant C here, whatever, doesn't matter. If it's something like this, it will be divergent everywhere. Now, if you have some degree here, some kind of a polynomial, let's say, of degree greater than 1, then it will be convergent on the infinity, but it will divergent from 0, around 0. And vice versa, if you have something like lower than 1 degree here, again, plus constant, whatever it is, then it's vice versa. It will be divergent here, but convergent around 0. All right, so these are just examples of improper integrals and how to approach them. Basically, again, all about the limits here. It's not a big deal, really. I would suggest you to go through these examples just by yourself. They are obviously on Unizor.com. Check if you have the same answers. And well, that's it. Thank you very much and good luck.