Education.com
Try
Brainzy
Try
Plus

Improper Integrals Help

By — McGraw-Hill Professional
Updated on Apr 25, 2014

Introduction to Improper Integrals

The theory of the integral that we learned earlier enables us to integrate a continuous function f ( x ) on a closed, bounded interval [ a, b ]. See Fig. 5.1. However, it is frequently convenient to be able to integrate an unbounded function, or a function defined on an unbounded interval. In this section and the next we learn to do so, and we see some applications of this new technique. The basic idea is that the integral of an unbounded function is the limit of integrals of bounded functions; likewise, the integral of a function on an unbounded interval is the limit of the integral on bounded intervals.

Indeterminate Forms 5.3 Improper Integrals: A First Look

Fig. 5.1

Integrals with Infinite Integrands

Let f be a continuous function on the interval [ a, b ) which is unbounded as xb (see Fig. 5.2). The integral

Indeterminate Forms 5.3 Improper Integrals: A First Look

Indeterminate Forms 5.3 Improper Integrals: A First Look

Fig. 5.2

is then called an improper integral with infinite integrand at b . We often just say “improper integral” because the source of the improperness will usually be clear from context. The next definition tells us how such an integral is evaluated.

If

Indeterminate Forms 5.3 Improper Integrals: A First Look

is an improper integral with infinite integrand at b then the value of the integral is defined to be

Indeterminate Forms 5.3 Improper Integrals: A First Look

provided that this limit exists. See Fig. 5.3.

Indeterminate Forms 5.3 Improper Integrals: A First Look

Fig. 5.3

Examples

Example 1

Evaluate the integral

Indeterminate Forms 5.3 Improper Integrals: A First Look

Solution 1

The integral

Indeterminate Forms 5.3 Improper Integrals: A First Look

is an improper integral with infinite integrand at 8. According to the definition, the value of this integral is

Indeterminate Forms 5.3 Improper Integrals: A First Look

provided the limit exists. Since the integrand is continuous on the interval [2, 8 − ∈], we may calculate this last integral directly. We have

Indeterminate Forms 5.3 Improper Integrals: A First Look

This limit is easy to evaluate: it equals 6 5/3 . We conclude that the integral is convergent and

Indeterminate Forms 5.3 Improper Integrals: A First Look

Example 2

Analyze the integral

Indeterminate Forms 5.3 Improper Integrals: A First Look

Solution 2

This is an improper integral with infinite integrand at 3. We evaluate this integral by considering

Indeterminate Forms 5.3 Improper Integrals: A First Look

This last limit is +∞. We therefore conclude that the improper integral diverges.

You Try It : Evaluate the improper integral Indeterminate Forms 5.3 Improper Integrals: A First Look .

Improper integrals with integrand which is infinite at the left endpoint of integration are handled in a manner similar to the right endpoint case:

Example 3

Evaluate the integral

Indeterminate Forms 5.3 Improper Integrals: A First Look

Solution 3

This integral is improper with infinite integrand at 0. The value of the integral is defined to be

Indeterminate Forms 5.3 Improper Integrals: A First Look

provided that this limit exists.

Since 1/( x ln 2 x ) is continuous on the interval [∈, 1/2] for ∈ > 0, this last integral can be evaluated directly and will have a finite real value. For clarity, write φ( x ) = ln x , φ′( x ) = 1/ x . Then the (indefinite) integral becomes

Indeterminate Forms 5.3 Improper Integrals: A First Look

Clearly the antiderivative is −1/φ( x ). Thus we see that

Indeterminate Forms 5.3 Improper Integrals: A First Look

Now as ∈ → 0 + we have ln ∈ → −∞ hence 1/ln ∈ → 0. We conclude that the improper integral converges to 1/ln 2.

You Try It : Evaluate the improper integral Indeterminate Forms 5.3 Improper Integrals: A First Look .

Many times the integrand has a singularity in the middle of the interval of integration. In these circumstances we divide the integral into two pieces for each of which the integrand is infinite at one endpoint, and evaluate each piece separately.

View Full Article
Add your own comment

Ask a Question

Have questions about this article or topic? Ask
Ask
150 Characters allowed