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.
Integrals with Infinite Integrands
Let f be a continuous function on the interval [ a, b ) which is unbounded as x → b ^{−} (see Fig. 5.2). The integral
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
is an improper integral with infinite integrand at b then the value of the integral is defined to be
provided that this limit exists. See Fig. 5.3.
Examples
Example 1
Evaluate the integral
Solution 1
The integral
is an improper integral with infinite integrand at 8. According to the definition, the value of this integral is
provided the limit exists. Since the integrand is continuous on the interval [2, 8 − ∈], we may calculate this last integral directly. We have
This limit is easy to evaluate: it equals 6 ^{5/3} . We conclude that the integral is convergent and
Example 2
Analyze the integral
Solution 2
This is an improper integral with infinite integrand at 3. We evaluate this integral by considering
This last limit is +∞. We therefore conclude that the improper integral diverges.
You Try It : Evaluate the improper integral .
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
Solution 3
This integral is improper with infinite integrand at 0. The value of the integral is defined to be
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
Clearly the antiderivative is −1/φ( x ). Thus we see that
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 .
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.

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