Proof of Comparison Test for Improper Integrals

Comparison Test for Improper Integrals
Comparison Test for Improper Integrals

Proof of Comparison Test for Improper Integrals

University: 香港中文大學

Course: University Mathematics (MATH 1010)

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COMPARISON TEST FOR

IMPROPER INTEGRALS

This supplement provides a proof of Theorem 3 in Section 8.6 (ET Section 7.7).

THEOREM 1 Theorem Comparison Test for Improper Integrals Let f(x)and g(x)

be continuous functions. Assume that f(x)≥g(x) ≥0 for x≥a.

•If ∞

a

f(x)dx converges, then ∞

a

g(x) dx converges.

•If ∞

a

g(x) dx diverges, then ∞

a

f(x)dx diverges.

Informally, the Comparison Test says:

•If the integral of the bigger function converges, then the integral the smaller function

also converges.

•If the integral of the smaller function diverges, then the integral the bigger function

also diverges.

Stated this way, the Comparison Theorem may seem “obvious”. However, a rigorous proof

uses Lemma 2 stated below, which in turn is based on the Upper Bound Property of real

numbers discussed in Appendix B.

Before turning to the proof, we observe that the two parts of Theorem 1 are con-

trapositives of the other (the term “contrapositive” is explained in Appendix A) and for

this reason, the two parts are logically equivalent. To explain this further, assume that the

first part is known to be true and that ∞

a

g(x) dx diverges. Then ∞

a

f(x)dx must

also diverge, for if it converged, the first part would imply that ∞

a

g(x) dx converges.

Similarly, the second part implies the first. It will suffice, therefore, to prove the first part.

Lemma 2 is similar to Theorem 4 in

Appendix B (Theorem 6 in Section 11.1 or

ET Section 10.1) which states that

bounded monotonic sequences converge.

Lemma 2 remains true (with a similar

proof) if the interval (a, ∞)is replaced by a

finite open interval (a, b).

THEOREM 2 Lemma Let G(t) be an increasing function on an interval (a, ∞). As-

sume there exists M>0 such that G(t) ≤Mfor all t∈(a, ∞). Then the following

limit exists:

L=lim

t→∞ G(t)

and L≤M.

Proof Let Sbe the set of values of G(t) on (a, ∞):

S={y:y=G(t) for t>a}

By assumption, Sis bounded by M, that is, y≤Mfor all y∈S. Theorem 1 in Appendix

B states that every bounded set of real numbers has a least upper bound. Let Lbe the least

1

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