 So, this says that, what is the advantage of this kind of a theorem? See, this advantage of interpreting Riemann integral as limit is that, now, if you want to look at the algebra of integrable functions, f is integrable, g is integrable, then you want to look at whether f plus g is integrable or not. If you want to go by upper and lower sums, then you have to relate the upper sums of f plus g with the upper sum of f and with that of g. That becomes a bit difficult, one can do that, but now it is being the limit. If I take f plus g, so, what will be integral of f plus g? That will be limit of integral, limit s p f plus g, but the limit split, limit of f plus g is equal to limit of f plus limit of g. So, that is the advantage that sometimes this way of proving that is integrable is useful to get the result. So, one proves the theorem like this. If f and g are integrable, then f plus g is integrable and integral of f plus g is integral f plus integral g. So, this is where the limit operations, because the left hand side will be a limit of norm p going to 0 and that splits into limit of s p f with respect to f, because what is the Riemann sum? f plus g at some point t i into the length. So, that splits into two parts. So, limit of the sum is equal to sum of the limits. So, using that, one proves all these results as a consequence of that limiting operation. If f and g are integrable, then alpha times f is also integrable and alpha comes out, because in the limit of alpha times something is alpha times the limit. Similarly, f and g. So, if you look at s p f with respect to f will be less than or equal to the Riemann sum with respect to g, because f is less than g. The value at a point of t i of f will be less than. So, using that criteria of integrability, one proves these things. This is also not difficult at all, saying that if f is integrable in a to b, then it is also integrable between a and c in between c is a point in between plus the integral between c to b, because the partition of the whole interval can be put it as a partition of a to c and c to b introducing the point c in between. This is because for n f, which is integrable, integral may be negative. So, absolute value of the integral is less than or equal to integral of the absolute value. Again, when you look at the limits, s p f, you look at the absolute value of the Riemann sums. So, that will be less than or equal to mod f into the length. So, this becomes obvious in that case. So, using that, that is a very useful result in proving integrability. Not only that, it frees the notion of integrability from function being bounded. Historically, it is of great importance, because that gave a lot of interest in looking at what is called Fourier series problem or in probability and statistics, you will find what is the characteristic function of a distribution coming and looking at their Fourier series problems. So, this is integration. So, what we have done is, we have given a function f on an interval a, b to r. We defined the notion of, geometrically, the notion of area below the graph of the function. And that we interpreted as, where lower sums, upper sums or where the Riemann sums. There are some situations where you can extend this notion of integral. See, f on a function on a, b to r, f is bounded, the domain of the function is a bounded interval. You can extend this notion when either the domain is not a bounded interval or the function is not bounded on the bounded interval. So, one can define the notion of integral. So, that is called one extension of Riemann integral and that is called the improper integration. So, one looks at the function defined either on an interval which is not bounded, but the function is bounded or the function is defined on the interval a, b which is bounded, but the function is not bounded itself on that interval. So, these two situations can be handled in some cases. So, let us look at some example. For example, look at the function defined on the bounded interval 0 to 1, open at 0, close at 1, f of 1 over square root x. So, at every point, bigger than 0, this function is defined. What is the integral? Integral of 1 over square root x from any point c, if you take a point c bigger than 0. So, that is equal to 1 over x raise to power minus 1 by 2. So, that is the integral. Now, in this, the interesting happens if I let c go to 0, c is bigger than 0. So, if I let c closer to 0, this integral has a limit. When c goes to 0, this goes to 2. So, what we are saying is, even though the function is becoming larger and larger as you come closer to 0, still I can think of saying what is the area of this 1 over square root x on the interval 0 to 1, area below the graph of the function. Now, it is becoming very large near 0. So, this is the situation where we can define. So, this limit will be called as the integral of 1 over square root x in the interval 0 to 1. So, here this is a situation where the function is becoming bigger and bigger in a bounded interval. Let us look at another example. Look at this function. f of x is minus 1 to the power n divided by n and the function is defined on an infinite interval 0 to infinity. x is positive. If x lies between n minus 1 to n, then the function is defined this way. So, we would like to know, can we say something like the function has some integral from 0 to infinity. So, unbounded interval. So, what we do? We have only defined the notion of function integral when the function is bounded over a bounded domain. So, let us look at the integral of this from 0 to some point. Then it becomes a bounded interval. So, let us choose a point say i m, which is 0 to m and look at the integral of this function in this interval. So, it is a continuous function, a piecewise continuous function in that interval. So, what will be the integral? This is defined as constant function in this interval. So, what will be the integral? The value of the function into the length of the interval summation. Now, this as m, m is the interval i m 0 to m. Now, let us, let m go to infinity. Then what happens to this series minus 1 to the power n divided by n? Have you come across series in your courses? So, this is a convergent series. This is a alternating series actually. So, this is a convergent series. We will do it again also later on and its sum is equal to minus ln of 2. ln is a log function. So, now, the function is defined over the whole interval 0 to infinity, which is unbounded. But in every bounded part, integral is defined and as we stretch that interval to infinity, the limit exists. In the earlier case, it was 0 to 1 and c was being pulled to 0. So, in all these cases, there the function was unbounded. Here, the function is bounded, but the domain is unbounded interval. So, such things are called improper integrals. So, let us make a definition. Let us say i is an interval a to infinity and f is a bounded function on that. The interval is 1 to infinity, but the function is bounded. So, if I take part of a to something, it will be a bounded function on a bounded interval and suppose that integral exists and as we take the limit of that point going to infinity, that also exists. Then we can say f is integrable on the interval a to infinity. So, let us define that. So, take a sequence bn of numbers, which is increasing to infinity and i n is a to bn and suppose f is integrable, r is the symbol Riemann integrable on the interval i n, that is a to bn. If that exists, take the limit of this and suppose that limit is equal to some number is convergent to alpha and that number alpha is independent of the way you go to infinity. It should not depend upon that. Then you say that the function has an integral and that integral is called improper integral of f over the interval i. These are called here the domain of the function is unbounded. These are called improper integrals of type 1, where the domain is unbounded, but the function is bounded. The other one, which we saw earlier, where the domain is bounded, but the function became unbounded and still the integral existed. So, that was this situation a to b and from c to b, the function is integrable and limit c going to a exists. Then we say this is improper integrable function of type 2. So, domain is bounded, function is unbounded. In the other one, function is bounded, but the domain is unbounded. These kind of situations arise for functions, which are both important in mathematics, probability and statistics. The improper integrals. So, we will give some more examples of this. One way of checking is called comparison test, which says if how to check whether some integral will exist or not. So, if f is less than or equal to g and if, so saying that the integral a to b is finite, that is saying that the improper integral exists. Then, so if integral of the dominated function improper integral exists, then of the function, which is being dominated, that also exists. And if this is infinity, then g being bigger, that will be also. That means when the improper integral exists, you say it is less than infinity. When it does not exist, you say it is equal to infinity or one uses the word convergent and divergent. Improper integral is convergent. That meaning that integral limit exists. Divergent is another way of saying. Here, because we have 0, so that is another way of writing. For example, look at 0 to infinity e raise power minus x square. Here, the function e raise power minus x square, the interval 0 to infinity, that is unbounded. So, let us try to analyze this in two parts. We will see e raise power minus x square behaves differently between 0 and infinity. So, let us look at, if x is bigger than 1 in this, then what is the inequality? e raise power x square is bigger than e raise power x. So, e raise power minus x square because the function is e raise power minus x square is less than e raise power minus x. So, this is less than. And look at the integral 1 to infinity of e raise power minus x. What is that integral? 1 to infinity. That means, look at integral 1 to some finite quantity of e raise power minus x and limit of that as that point goes to infinity. So, what will be that integral? So, that goes to infinity. That other part will go to 0. So, this limit is equal to 1 over e. Is that okay? Integral of e raise power minus x from 1 to some point c. Exponential derivative itself, integral itself is negative sign. So, negative. So, this integral exists. So, that means, e raise power minus x square from 1 to infinity will also exist by comparison theorem because this integral is finite. So, by comparison test the integral e raise power minus x square will also exist. And 0 to 1 e raise power minus x square does that exist? Domain is bounded. 0 to 1 e raise power minus x square. What is happening to the function? It is a continuous function. So, integral exists. We know that integral, that is ordinary integral. So, the integral 0 to infinity will exist. And, this is something similar to what is called normal distribution, normal density function that will come in statistics. So, that is improper integrals. So, that is one of the uses of improper integrals. Here is another application which we will not go much into. I want to look at the integral. I think I pointed out earlier. Look at the integral of 1 over 1 minus t square d t minus 1 to 1. So, let us try to split 1 minus t square factor is that is 1 minus t and 1 plus t. And, the 1 minus t into 1 plus t, t is between minus 1 to 1. So, what happens to 1 plus t? That is always 1 over 1 minus 1 plus t, that is x square root. So, this quantity is less than 1 over 1 minus t. So, if between 0 and 1, look at the integral, integral 0 to x f t, because this is less than this quantity. So, it is less than or equal to this quantity, 0 to x. So, I am trying to look at the integral from 0 to x of 1 over 1 minus. I am not going to 1, because if I try to go to 1, the function becomes unbounded 1 over 1 minus t. 1 over 1 minus t square root, as t goes to 1, it is becoming unbounded function near 1. So, I should avoid 1. So, to look at 0 to x from 0 to 1 only, that integral exists and that always remains less than or equal to 2, because 2 minus something. So, that means what? 0 to x, f is non-negative. So, this is an increasing function bounded above. So, limit of this 0 to x, x goes to 1 will exist. Are you following? Because 0 to x, non-negative quantity, the interval is increasing as a non-negative function, integral will be non-negative. So, these quantities are non-negative, increasing bounded by 2. So, limit of this will exist. So, this limit exists, because this limit exists. So, similarly with minus 1 to 1, also the limit exists. So, you can say that this integral exists. Now, the integral is unbounded, the other function is unbounded near the value 1. Now, because of this, here is one application of this. Look at the integral, this integral just now we said exists, minus 1 to 1 of 1 over 1 minus e square. Look at 1 over 1 minus e square, that is the derivative of what? 1 over 1 minus x square square root. It is the derivative of sin inverse function. So, if I integrate the derivative, if I am able to integrate, I should get back the function. But here, the integration is coming via improper integral. When you are away from minus 1 or 1, you are applying fundamental theorem of calculus and getting value of sin x. And at the end point, it is the limit. So, that is also giving you the continuity of the sin inverse function, because the way it is defined. So, I will not go into this. Just for the sake of exposure, saying that the improper integral can be used. In fact, here we are saying that we define sin inverse as improper integral of its derivative. And once sin inverse is defined, this is defined from minus 1 to 1. So, what will be the value of sin inverse at the point minus 1? What do you think should be the value? That is, right. Now, as such, we are not defined pi by pi or anything. So, we are defining a sin inverse function between minus 1 to 1, taking values in R via integration. One proves it is continuous, differentiable and all those properties. It is a continuous function. It is a one-one function. What will be the derivative of this 1 over 1 minus t square positive? It will be monotonically increasing function. Derivative is nowhere 0. It will be 1 to 1 on 2 function between minus 1 to 1 to R. What will be the range? It is a continuous function on minus 1 to 1, an interval. Range also should be a close bounded interval. So, it should have a left end point. It should have a right end point. That left end point is called minus pi by 2 and the right end point will be plus pi by 2 because you can see from here it is what is an odd function. So, this is the way of defining sin inverse, getting what is minus pi by 2, what is pi by 2 and then you invert that function. You get sin function between minus pi by 2 to pi by 2 to minus 1 to 1. If you look at the graph of that sin function because sin inverse is 1, 1 on 2. That also is a 1, 1 on 2 function between minus pi by 2 to pi by 2 and then you extend it periodically everywhere. So, that is a way of defining trigonometric functions and also in between you define what is pi. We defined pi also via sequences. Look at the area of the circle. So, that was the beginning of our story of saying that a monotonically increasing sequence which is bounded above must converge. So, that the area of the inscribed and gone was monotonically increasing and bounded and similarly outside was decreasing. So, this is a way of defining, a way of extending Riemann integral when either the domain is not bounded or the function is not bounded. There are many other functions which in applied mathematics also it comes as something called gamma functions. If you have heard about those things, they are all improper integrals. So, we will not go much into it because we just want to give you an exposure of something called improper integrals as and when it comes to one of some of the courses, we will study more of them. So, this is one way of extending your integral.