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IMPROPER INTEGRALS
Def. Improper integral. The definite integral
is called an improper integral if
1] at least one of the limits of integration is infinite, or
2] the integrand f(x) has one or more points of discontinuity on the interval a
x
b.
Infinite limits of integration Integrals with one or more infinite limits are given meanings by the following definitions:
1] If f(x) is continuous on the interval a
x
h, we define
2] If f(x) is continuous on the interval h
x
b, we define
3] If f(x) is continuous on the interval h'
x
h, we define
If the limit (or limits) exist, an improper integral is said to be convergent. If a limit doesn’t exist the integral has no value and is said to be divergent.
Discontinuous integrand.
1] If f(x) is continuous on the interval a
x < b, but is discontinuous at x = b, we define
2] If f(x) is continuous on the interval a < x
b, but is discontinuous at x = a, we define
3] If f(x) is continuous for all values on the interval a
x
b except x = c, where a < c < b,
we define
Example 1. Evaluate the integral
Solution. First we integrate from 0 to h thus obtaining a function of h. Then we examine the behavior of this function when h → ∞.
The graphical interpretation is shown in Fig. 1. The area under the curve
from x = 0 to x = h is
As the point h moves to the right, the area continually increases and approaches 1.
Example 2. Evaluate the integral
Solution. We will integrate from 2 to h and then examine the behavior of the resulting function of h as h → ∞.
The integral has no limit and is said to be divergent.
Theorem. Let
be an improper integral in which the function f(x) is discontinuous somewhere in the interval a
x
b . Let Φ(x) be the primitive of f(x) i.e. Sf(x)dx = Φ(x). If Φ(x) is continuous over the
interval a
x
b then integral 1) can be evaluated in the usual way of a regular, proper
integral i.e. without using ε. In other words, if Φ(x) is continuous on interval a
x
b
The proof for the case in which f(x) is continuous over the interval except for a vertical asymptote at x = k where a < k < b is as follows:
Because of the assumed continuity of Φ(x),
We then have
Proofs for the other cases are similar.
Example 3. Evaluate
Solution. The function
has a vertical asymptote at x = 2. However, the primitive
is a continuous function over the interval 1
x
10. We may then evaluate the integral in the
usual way:
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