CC Comparison of Improper Integrals

Section 5.11 Comparison of Improper Integrals

Motivating Questions

What are some typical improper integrals that we can classify as convergent or divergent?
When can the convergence behavior of one improper integral be used to determine the convergence behavior of another improper integral?

In the previous section, we learned how to compute improper integrals -- integrals involving certain functions over unbounded integrals, as well as functions that become infinite at a point within or at the endpoint of the interval of integration.

🔗

Recall the important classes of improper integrals we examined in Section 5.10.

🔗

Convergence Behavior of Important Classes of Improper Integrals.

🔗

$\int_{1}^{\infty} \frac{1}{x^{p}} d x$ converges if $p > 1$ and diverges if $p \leq 1$
$\int_{0}^{1} \frac{1}{x^{p}} d x$ converges if $p < 1$ and diverges if $p \geq 1$
$\int_{0}^{\infty} \frac{1}{e^{a x}} d x$ converges if $a > 0$ and diverges if $a \leq 0$

🔗

In this section, we will learn how to use improper integrals that we know converge or diverge (such as those above) to determine the convergence of more complex integrals.

🔗

Subsection 5.11.1 Comparing Improper Integrals

🔗

Sometimes we may encounter an improper integral for which we cannot easily evaluate the limit of the corresponding proper integral. For instance, consider

\int_{1}^{\infty} \frac{1}{1 + x^{3}} d x .

While it is hard (or perhaps impossible) to find an antiderivative for

\frac{1}{1 + x^{3}},

we can still determine whether or not the improper integral converges or diverges by comparison to a simpler one. Observe that for all

x > 0,

1 + x^{3} > x^{3},

and therefore

\frac{1}{1 + x^{3}} < \frac{1}{x^{3}} .

🔗

It therefore follows that

\int_{1}^{b} \frac{1}{1 + x^{3}} d x < \int_{1}^{b} \frac{1}{x^{3}} d x

🔗

for every

b > 1 .

We know that

lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^{3}} d x = \int_{1}^{\infty} \frac{1}{x^{3}} d x

converges, since it is an improper integral of the form

\int_{1}^{\infty} \frac{1}{x^{p}} d x

with

p > 1 .

By the inequality above, if

lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^{3} + 1} d x

exists, it must be less than or equal to

\int_{1}^{\infty} \frac{1}{x^{3}} d x;

in particular, if

lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^{3} + 1} d x

exists, it is finite. By making some technical considerations, it can be proved that in fact

lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^{3} + 1} d x

must exist, and therefore

\int_{1}^{\infty} \frac{1}{x^{3} + 1} d x

is a convergent improper integral.

🔗

The comparison of these improper integrals can be visualized by comparing the area under each function. The area under the curve

\frac{1}{x^{3}}

is shown in green, and the area under the curve

\frac{1}{x^{3} + 1}

is shown in blue.

🔗
Figure 5.106. Comparing the area under the curves $\frac{1}{x^{3}}$ and $\frac{1}{x^{3} + 1}$

🔗

In particular,

\int_{1}^{\infty} \frac{1}{1 + x^{3}} d x

must converge, even though we never explicitly evaluated the corresponding limit of proper integrals. We use this idea and similar ones in the examples that follow.

Example 5.107.

Explain why $x^{2} + x + 1 > x^{2}$ for all $x \geq 1,$ and hence determine if $\int_{1}^{\infty} \frac{1}{x^{2} + x + 1} d x$ converges or diverges by comparison to $\int_{1}^{\infty} \frac{1}{x^{2}} d x .$
Observe that for each $x > 1,$ $\ln (x) < x .$ Explain why

$\int_{2}^{b} \frac{1}{x} d x < \int_{2}^{b} \frac{1}{\ln (x)} d x$

for each $b > 2 .$ Why must it be true that $\int_{2}^{b} \frac{1}{\ln (x)} d x$ diverges?
Explain why $\sqrt{\frac{x^{4} + 1}{x^{4}}} > 1$ for all $x > 1 .$ Then, determine whether or not the improper integral

$\int_{1}^{\infty} \frac{1}{x} \cdot \sqrt{\frac{x^{4} + 1}{x^{4}}} d x$

converges or diverges.

Answer.

converges
diverges
diverges

Solution.

Since $x \geq 1,$ it follows that $x + 1 \geq 1 > 0,$ so adding $x^{2}$ to both sides of this inequality shows that

$x^{2} + x + 1 > x^{2}$

for all $x \geq 1 .$ Taking reciprocals (which reverses the inequality),

$\frac{1}{x^{2} + x + 1} < \frac{1}{x^{2}}$

and thus for any $b > 1,$

$\int_{1}^{b} \frac{1}{x^{2} + x + 1} d x < \int_{1}^{b} \frac{1}{x^{2}} d x .$

Taking the limit as $b \to \infty,$ it follows that

$\int_{1}^{\infty} \frac{1}{x^{2} + x + 1} d x < \int_{1}^{\infty} \frac{1}{x^{2}} d x,$

and since we know that $\int_{1}^{\infty} \frac{1}{x^{2}} d x$ converges, we conclude that $\int_{1}^{\infty} \frac{1}{x^{2} + x + 1} d x$ also converges.
Since for each $x > 1,$ $\ln (x) < x,$ it also holds that $\ln (x) < x$ for all $x > 2 .$ Taking reciprocals,

$\frac{1}{x} < \frac{1}{\ln (x)}$

and thus

$\int_{2}^{b} \frac{1}{x} d x < \int_{2}^{b} \frac{1}{\ln (x)} d x$

for each $b > 2 .$ Taking the limit as $b \to \infty,$ we have

$\int_{2}^{\infty} \frac{1}{x} d x \leq \int_{2}^{\infty} \frac{1}{\ln (x)} d x$

and since we know that $\int_{1}^{\infty} \frac{1}{x} d x$ diverges, it follows that $\int_{2}^{b} \frac{1}{\ln (x)} d x$ also diverges.
Since $x^{4} + 1 > x^{4}$ for all $x,$ it follows that

$\frac{x^{4} + 1}{x^{4}} > 1$

for all $x > 1,$ and thus taking square roots,

$\sqrt{\frac{x^{4} + 1}{x^{4}}} > 1$

for all $x > 1 .$ Integrating from $1$ to $b$ and then letting $b \to \infty,$ we can next say that

$\int_{1}^{\infty} \sqrt{\frac{x^{4} + 1}{x^{4}}} d x \geq \int_{1}^{\infty} 1 d x$

It is obvious that $\int_{1}^{\infty} 1 d x$ diverges since the integrand does not tend to $0 .$ By the last inequality above, we can conclude that

$\int_{1}^{\infty} \frac{1}{x} \cdot \sqrt{\frac{x^{4} + 1}{x^{4}}} d x$

diverges.

🔗

Part (b) demonstrates that if an improper integral can be bounded below by one that diverges, then the improper integral in question also diverges. The same logic can be used to show that if an improper integral can be bounded above by one that converges, it also converges. We formalize these principles as the Comparison Test for Improper Integrals.

🔗

Comparison Test for Improper Integrals.

🔗

f (x) \geq g (x) \geq 0

and

\int_{a}^{\infty} f (x) d x

converges, then

\int_{a}^{\infty} g (x) d x

converges.

🔗

f (x) \geq g (x) \geq 0

and

\int_{a}^{\infty} g (x) d x

diverges, then

\int_{a}^{\infty} f (x) d x

diverges.

🔗

For Part (b), we claimed that

\int_{2}^{\infty} \frac{1}{x} d x

diverges, but we’ve technically only discussed that

\int_{1}^{\infty} \frac{1}{x} d x

diverges. However, since

\int_{1}^{\infty} \frac{1}{x} d x = \int_{2}^{\infty} \frac{1}{x} d x + \int_{1}^{2} \frac{1}{x} d x

and

\int_{1}^{2} \frac{1}{x} d x

is finite,

\int_{2}^{\infty} \frac{1}{x} d x

must diverge; otherwise, the right-hand side of the equation would be finite. In fact,

\int_{a}^{\infty} \frac{1}{x} d x

diverges for any real number

a \geq 1

for the same reason. More generally:

🔗

.

🔗

Suppose

f (x)

is continuous on the interval

[a, b]

If $\int_{a}^{\infty} f (x) d x$ converges, then $\int_{b}^{\infty} f (x) d x$ also converges.
If $\int_{a}^{\infty} f (x) d x$ diverges, then $\int_{b}^{\infty} f (x) d x$ also diverges.

🔗

We’ll finish this section with a few guided examples of applications of the Comparison Test. A useful strategy for these types of problems goes as follows: first, make an educated guess about whether the given integral converges or diverges; then, based on this guess, bound the integrand above or below by a simpler function whose integral over the same interval converges or diverges, respectively.

Example 5.108.

Use the Comparison Test for Improper Integrals to determine whether the following improper integrals converge or diverge. Be sure to justify any inequalities used when applying the Test.

$\int_{1}^{\infty} \frac{\sin^{2} (x)}{x^{2}} d x$
$\int_{1}^{\infty} \frac{1}{\sqrt{x^{3} + 6}} d x$
$\int_{10}^{\infty} \frac{3 x}{(x + 4)^{2}} d x$
$\int_{1}^{\infty} \frac{2 + \cos (x)}{\sqrt{x + 12}} d x$

Hint.

Since $- 1 \leq \sin (x) \leq 1$ for any $x \in R,$ we have $0 \leq \sin^{2} (x) \leq 1$ for any $x \in R .$
When $x$ is large, observe that $\sqrt{x^{3} + 6} \approx \sqrt{x^{3}} = x^{3 / 2} .$
When $x$ is large, $\frac{3 x}{(x + 4)^{2}} \approx \frac{3 x}{x^{2}} = \frac{3}{x} .$
Since $\cos (x) \geq - 1$ for all $x \in R,$ the numerator is at least $1 .$ For the denominator, when $x$ is large we have $\sqrt{x + 12} \approx \sqrt{x} .$

Answer.

Converges, via comparison with e.g. $\frac{1}{x^{2}} .$
Converges, via comparison with e.g. $\frac{1}{x^{3 / 2}} .$
Diverges, via comparison with e.g. $\frac{3}{4 x} .$
Diverges, via comparison with e.g. $\frac{1}{2 \sqrt{x}}$ on the interval $[4, \infty) .$

Solution.

From the hint, we know $0 \leq \sin^{2} (x) \leq 1$ for any $x \in R .$ Therefore, $0 \leq \frac{\sin^{2} (x)}{x^{2}} \leq \frac{1}{x^{2}}$ for any $x \in R .$ (In particular, this inequality holds on the interval $[1, \infty) .$ ) Since $\int_{1}^{\infty} \frac{1}{x^{2}} d x$ converges, by the Comparison Test it follows that $\int_{1}^{\infty} \frac{\sin^{2} (x)}{x^{2}} d x$ converges.
From the hint, we should expect our improper integral to behave like $\int_{1}^{\infty} \frac{1}{x^{3 / 2}} d x,$ which converges. Indeed, we can compare our integral directly to this one.

For any $x \in R,$ we have $x^{3} + 6 > x^{3} .$ Therefore, $\sqrt{x^{3} + 6} > \sqrt{x^{3}} = x^{3 / 2}$ whenever both $x^{3} + 6$ and $x^{3}$ are non-negative, i.e., for all $x \geq 0 .$ This, along with the assumption that we use the positive square root, yields the inequality $\frac{1}{x^{3 / 2}} > \frac{1}{\sqrt{x^{3} + 6}} > 0$ for all $x > 0 .$ Specifically, the inequality holds on the interval $[1, \infty) .$ Because $\int_{1}^{\infty} \frac{1}{x^{3 / 2}} d x$ converges, by the Comparison Test we conclude that $\int_{1}^{\infty} \frac{1}{\sqrt{x^{3} + 6}} d x$ converges.
From the hint, we should expect our imroper integral to behave like $\int_{10}^{\infty} \frac{3}{x} d x,$ which diverges. However, since $(x + 4)^{2} > x^{2}$ for all $x \in R$ it’s actually the case that $\frac{3 x}{(x + 4)^{2}} < \frac{3 x}{x^{2}} = \frac{3}{x}$ for $x \in R .$ So we cannot compare our function to $\frac{3}{x}$ to conclude divergence. More care is needed to find an appropriate lower bound for $\frac{3 x}{(x + 4)^{2}} .$

Observe that if $x \geq 4,$ we have $2 x \geq x + 4 .$ (In fact, these inequalities are equivalent.) So $(2 x)^{2} \geq (x + 4)^{2}$ for $x \geq 4$ and thus we may write $\frac{3 x}{(x + 4)^{2}} \geq \frac{3 x}{(2 x)^{2}} = \frac{3}{4 x} \geq 0 .$ Now, $\int_{10}^{\infty} \frac{3}{4 x} d x = \frac{3}{4} \int_{10}^{\infty} \frac{1}{x} d x$ diverges, so it follows from the Comparison Test that $\int_{10}^{\infty} \frac{3 x}{(x + 4)^{2}} d x$ diverges.
From the hint, for large enough $x$ the function $\frac{2 + \cos (x)}{\sqrt{x + 12}}$ is approximately bounded below by $\frac{1}{\sqrt{x}} .$ Since $\int_{1}^{\infty} \frac{1}{\sqrt{x}} d x$ diverges, we should expect our integral to diverge as well. However, a direct comparison to $\frac{1}{\sqrt{x}}$ won’t show divergence, since $\frac{1}{\sqrt{x + 12}} < \frac{1}{\sqrt{x}} .$

Note, however, that for $x \geq 4$ we have $x + 12 \leq 4 x .$ (These inequalities are indeed equivalent.) So when $x \geq 4,$ we know $\frac{2 + \cos (x)}{\sqrt{x + 12}} \geq \frac{1}{\sqrt{4 x}} = \frac{1}{2 \sqrt{x}} > 0 .$ Since $\int_{4}^{\infty} \frac{1}{2 \sqrt{x}} d x = \frac{1}{2} \int_{4}^{\infty} \frac{1}{\sqrt{x}} d x$ diverges, by the Comparison Test $\int_{4}^{\infty} \frac{2 + \cos (x)}{\sqrt{x + 12}} d x$ diverges. It follows that $\int_{1}^{\infty} \frac{2 + \cos (x)}{\sqrt{x + 12}} d x$ diverges as well.

🔗

Subsection 5.11.2 Summary

🔗

There are three important classes of improper integrals discussed in this section.
- One important class of improper integrals is given by
  
  $\int_{1}^{\infty} \frac{1}{x^{p}} d x$
  
  where $p$ is a positive real number. We can show that this improper integral converges whenever $p > 1$ and diverges whenever $p \leq 1 .$
- A related class of improper integrals is
  
  $\int_{0}^{1} \frac{1}{x^{p}} d x,$
  
  which converges for $p < 1$ and diverges for $p \geq 1 .$
- Another class of improper integrals is
  
  $\int_{0}^{\infty} e^{- a x} d x,$
  
  which converges when $a > 0$ and diverges when $a \leq 0 .$
These important classes of improper integrals are used for comparisons in the Comparison Test for Improper Integrals. The Comparison Test for Improper Integrals allows us to determine if an improper integral converges or diverges without having to calculate the antiderivative. The actual test states the following:

If $f (x) \geq g (x) \geq 0$ and $\int_{a}^{\infty} f (x) d x$ converges, then $\int_{a}^{\infty} g (x) d x$ converges.

If $f (x) \geq g (x) \geq 0$ and $\int_{a}^{\infty} g (x) d x$ diverges, then $\int_{a}^{\infty} f (x) d x$ diverges.

🔗

Exercises 5.11.3 Exercises

🔗

1. Determining convergence or divergence of various improper integrals.

🔗

Determine, with justification, whether each of the following improper integrals converges or diverges.

$\int_{e}^{\infty} \frac{\ln (x)}{x} d x$
$\int_{e}^{\infty} \frac{1}{x \ln (x)} d x$
$\int_{e}^{\infty} \frac{1}{x (\ln (x))^{2}} d x$
$\int_{e}^{\infty} \frac{1}{x (\ln (x))^{p}} d x,$ where $p$ is a positive real number
$\int_{0}^{1} \frac{\ln (x)}{x} d x$
$\int_{0}^{1} \ln (x) d x$

Prev Top Next