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Improper integrals of the second kind



 

Def 6.Let a function is continuous on one of these three sets: a)
b) , c) with discontinuity point a, b, c respectively. One introduces the next three improper integrals of the second kind (integrals of discontinuous functions over a finite interval) namely

; ( 14 )

; ( 15 )

. ( 16 )

Notions of convergence or divergence are introduced in the same way that for improper integrals of the first kind.

Def. 7.The principal value of the integral (16) is called the next limit

. ( 16 )

Ex. 7. Improper integrals

( 17 )

are convergent for and divergent for .

■Let’s study the first integral .

a) If we have (divergence);

b) In the case

Ex. 8. Investigate the integral for convergence ( is discontinuity point).

.

The first integral is usual one because of its integrant is continuous on the segment , and the second is divergent improper integral . Therefore the given improper integral diverges.

Ex. 9. Find the principal value of the next divergent integral .

.

Ex. 10. Find the area of an infinite figure bounded by the lines , , , (fig. 7).

.

Note 2(Newton-Leibniz formula). Evaluation of improper integrals of the se-cond kind can be represented in the form of

Fig. 7 Newton-Leibniz formula. Let for example a function is continuous on an interval and for any its primitive we denote

.

Then

Ex. 11. .

Ex. 12. If then ,

.

Note 3(change of a variable and integration by parts in improper integrals).In process of evaluation of improper integrals we can use change of a variable and in-tegration by parts.

Ex. 12. Integrals of the example 7 can be reduced by change of a variable to the integral of the same example. In particular

Ex. 13.

Ex. 14.

Ex. 15. Prove yourselves that .

Ex. 16. Prove that .

Convergence tests

 

We’ll state and prove theorems for an improper integral

with an integrand continuous on the interval , but they are valid for all other improper integrals.

Theorem 1.Let for continuous on functions and sufficient large x one has

.

If the integral converges then the integral also converges.

If the integral diverges then the integral also diverges.

■Let for example the integral converges, It follows that for any b > a

and so there exists the limit

,

that is the integral converges.■

Ex. 17. The integral diverges, because of for any

,

and the integral

diverges.

Ex. 18. Prove convergence of the integral .

■It’s known that for any the inequality holds. Let’s represent the given integral as follows



.

The first and the third improper integrals converge, because of

,

and the integrals

are convergent. Therefore the given integral converges.■

Ex. 19. Prove yourselves divergence of the integral .

Ex. 20. Investigate the integral for convergence.

For any positive x , and the given integral diverges on account of divergence of the integral .

Theorem 2.Let for continuous on functions and sufficient large x one has

.

If the integrals converge then the integral also converges.

■Validity of the theorem follows from the inequality

and preceding theorem.■

Theorem 3(absolute convergence).If for continuous on an interval an integral

converges then the integral

converges and is called absolutely convergent.

■Proving follows from the inequality

and the theorem 2.■

Ex. 21. The integral absolutely converges, for

,

and the integral

converges.

Ex. 22. Prove absolute convergence of the integrals , .

 





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