Improper integrals of the second kind
Def 6.Let a function is continuous on one of these three sets: a)
; ( 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
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. 15. Prove yourselves that .
Ex. 16. Prove that .
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
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
Ex. 22. Prove absolute convergence of the integrals , .