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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 .

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: