Lecture. Improper integral
In this lecture we will again speak about integrals. We already know that the integral may be either uncertain or certain. Today there is a new term "improper integral". What is it connected with?
Well, firstly, we consider integrals of the first and second kind. If we remember the concept of the definite integral, it is connected with calculation of a confined function integral, set on the interval [a, b]. And improper integral of the first kind is connected with computation of the integral of a function set on unconfined intervals, along the line [a, + ∞), (-∞, a] or on the whole number line.
Let us consider the first case in detail. Let's start. So, let function y = f (x) defined on the interval [a, + ∞). We assume that at any interval [a, r], where r ≥ a, and there is integral F (r). So, this is the upper limit function r. Thus, each value r from a to + ∞ is assigned number F (r). Then the improper integral of the function f on the interval [a, + ∞) is the limit of this function. This is the definition. At x tending to plus infinity.
Thus, improper integral is the integral limit from to r of function f when r tends to plus infinity. When calculating the limit we can get a number as the answer, and, if not a number, then either infinity, or it does not exist. So, improper integral, we say, converges if the limit is finite and diverges if the limit value is infinity, or it does not exist. So, we say about the improper integrals they converge or diverge.
What is the geometric sense? If function f (x) is non-negative, what is F (r)? This is the area of the curvilinear trapezium predetermined by the function on the interval [a, r]. Then in the limit we get an improper integral on the interval [a, + ∞). This is the area of the unlimited curvilinear trapezoid, which is given by a continuous function on the interval [a, + ∞).
So how do we calculate that? Let's set that Newton-Leibniz formula is also applicable for improper integrals. Let the derivative of F equal to f, then F is a primitive. Applying Newton-Leibniz formula on the interval from a to r, we see that the following equality works.
Let's see what this formula will result in while calculating the improper integral. So, we have to calculate the limit of the difference f (r) - f (a). f (a) is the number, then the whole thing is in calculation of the limit of f on plus infinity. We will denote this limit f from positive infinity, then, using the common symbols, we write down that the improper integral is F (x), vertical line with indices a and plus infinity.
Let's look at another example. Explore the convergence of the improper integral. And, of course, if the integral converges, it is logical to find the value of the improper integral. Thus, dx, divided by x in power α. Well, firstly, we see that at different values of α the formula in the table of integrals will be different. Therefore, the first case. We take α = 1. Calculating by Newton-Leibniz formula, we see, this is the logarithm of x, the vertical line, from one to plus infinity.
Thus, the logarithm of plus infinity minus logarithm of one. How to calculate the limit at infinity plus logarithm? On the graph, of course. So, if the variable tends to plus infinity, then the value of the logarithm also tends to plus infinity. So we get positive infinity - 0. The answer is positive infinity, the integral diverges.
The second case. α is not one. In this case we have to use the tabulated integral for the power function. So. Record and get the next limit. So, to calculate limit x in power 1-α by plus infinity. Let's try to do it. So, make the substitution: 1-α is denoted by β. We get the power function limit. The power function depending on parameter β looks differently. Look, if β values are positive, there are three types of β: β <1, β = 1 and β> 1.
In any case, the limit of this function on plus infinity is equal to plus infinity. But if β is less than one, the limit on plus infinity is 0. Therefore we write 0. Returning to α, where α is more than one, and plus infinity, if α-1. So, we can go back to the computation of the improper integral. What do we see? If α> 1, this limit is equal to 0.
And we see that the integral is equal to 1 / (α-1), the integral converges. But if it is less than 1, then we get the answer - plus infinity. Overall limit is plus infinity, the integral diverges. Thus, we conclude: combining values of α = 1, the integral diverges. The conclusion regards the two cases: α> 1 - integral converges. We are see the answer. If α≤1, then the improper integral diverges.
It is logical to define the improper integral for the second case in the interval (-∞, a] in a similar way. To calculate it, we will use Newton-Leibniz formula. Absolutely similarly.
An d here is the case of (-∞, + ∞). So, a is an arbitrary point of the number line. We say that the integral converges if and only if, by definition, when both integrals converge. If at least one of them diverge, then the integral of the third kind also diverges. Well, we have chosen a randomly. What if with a different value of the value of the integral will change? Let's say we take a different value b. Thus, on the one hand, calculation of the integral is the sum of two, when end a is used, and the second case when one of the integration limits is equal to b.
Well, to be specific a and b are related by the inequality: a <b. So, why are these two values the same? See, in the first integral the interval [a, + ∞) is divided into sections: [a, b] and [b, + ∞). And for the second integral the interval (-∞, b] is divided into two parts: (-∞, b] and [a, b] We see that the records for each of the integrals on the right side are the sums of three integrals are exactly the same. So, the integral does not depend on the choice of a.
Here is a very interesting integral of the third kind. Explore the convergence of the improper integral. Here the integrand y = 1 / x 2. It is also one of interesting curves that has its name, Lokon Agnesi. It is known for a long time. It is used in applications. So, you need to calculate the area of the figure bounded by Lokon Agnesi and Ox axis. It seems strange, that an unlimited figure may have a finite area. But in theory, it is a true fact. It happens. So.
Firstly, divide into two integrals with ends at 0: (-∞, 0], [0, + ∞). This are tabular integrals. We get the arc tangent on the interval [1, + ∞). Remembering the arctangent graph (it has a horizontal asymptote on plus infinity) thus is y = π / 2. This is the limit of the arc tangent function on plus infinity. Arctangent of one is π / 4, and we get the value of the first summand - π / 4. It is not difficult to calculate the second integral. Also π / 4. As a result the given integral equals to π / 2, it converges. Well, it means that the unlimited trapezoid has a finite area, about one and a half square units.