Chapter 8.4, Problem 58E

To determine

To show: Theorem 5(Comparison test) is true using the properties of integrals.

Explanation of Solution

Given information: Theorem 5 related to comparison test is given.

Concept used :

Comparison property of integral:

For integral functions $f\left(x\right)$ and $g\left(x\right)$ on the interval $\left[a,b\right]$ if $0\le f\left(x\right)\le g\left(x\right)$ for all $x\ge a$ , then for any $b>a$ ,

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx\le {\displaystyle \underset{a}{\overset{b}{\int }}g\left(x\right)}dx$

Let the infinite integral of function $g\left(x\right)$ converges, then taking the limit $b\to \infty$ in the inequality ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx\le {\displaystyle \underset{a}{\overset{b}{\int }}g\left(x\right)}dx$ implies that the infinite integral of the function $f\left(x\right)$ is bounded above by the infinite integral of the function $g\left(x\right)$ .As a result, the infinite integral of the function $f\left(x\right)$ must be finite and hence converges.

In other words, ${\displaystyle \underset{a}{\overset{\infty }{\int }}f\left(x\right)}dx$ converges if ${\displaystyle \underset{a}{\overset{\infty }{\int }}g\left(x\right)}dx$ converges.

Similarly, Let the infinite integral of function $f\left(x\right)$ diverges. The divergence of any function implies that the corresponding function must grows to infinity. Therefore, function $f\left(x\right)$ must grow to infinity. Now, taking the limit $b\to \infty$ in the inequality ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx\le {\displaystyle \underset{a}{\overset{b}{\int }}g\left(x\right)}dx$ implies that the infinite integral of the function $g\left(x\right)$ must tends to infinity and hence diverges.

Thus, ${\displaystyle \underset{a}{\overset{\infty }{\int }}g\left(x\right)}dx$ diverges if ${\displaystyle \underset{a}{\overset{\infty }{\int }}f\left(x\right)}dx$ diverges.

As a result, theorem 5 is true by using the property of integrals.