Calculus Of One Real Variable – By Pheng Kim Ving Chapter 11: Techniques Of Integration – Section 11.2: Improper Integrals 11.2 Improper Integrals

Go To Problems & Solutions

 1. Proper And Improper Integrals

Let the function f be continuous on the closed bounded interval [a, b] ([a, b] is bounded if both a and b are (finite) numbers).
See Fig. 1.1. Under these conditions, f attains both a maximum and a minimum values (see Section 1.2.2 Theorem 2.1), # Fig. 1.1

f is continuous on [a, b], # is a (finite) number. In this section we're going to extend our study of definite integrals to include those of functions that are continuous on
unbounded intervals and of functions that are discontinuous at a finite number of points. Such definite integrals will be called
improper integrals. Definite integrals of functions continuous on closed bounded intervals are called proper integrals.

When we say that f is continuous on [a, b), we mean that f is: (1) continuous on (a, b), (2) right-continuous at a, and
(3) not left-continuous (and thus is discontinuous) at b. See Figs. 3.1, 3.2, and 3.3. Similarly for when we say that f is
continuous on (a, b] or (a, b).

 2. Improper Integrals Over Half-Open Unbounded Intervals  Here,

#  #  # ## Convergence And Divergence This observation motivates the definitions of convergent and divergent  improper integrals, as seen in the following
definition.

### Definition 2.1 ### Example 2.1

Determine if each of the following improper integrals converges. Sketch a graph in each case. #### Solution

a.  b.  EOS

 3. Improper Integrals Over Half-Open Bounded Intervals  #  #  #  # ## Definition 3.1 ### Example 3.1

Determine whether each of the following improper integrals converges. Sketch a graph in each case. #### Solution

a.  b.  EOS

 4. Improper Points And The 4 Basic Types Of Improper Integrals Each of the 4 improper integrals defined in Definitions 2.1 and 3.1 has only 1 improper point. There's only 1 limit to
handle for each of them. So each of them is said to be of a basic type. An improper integral is said to be of a basic type
if it has only 1 improper point. As we'll see later on in this section, all other types are based on these basic ones. Note
that an improper integral is a finite number if it exists.

 5. Improper Integrals Over Open Intervals {5.1} Part 3.

Now let f be continuous on the open bounded interval (a, b). Refer to Fig. 5.4. The definite integral of f over (a, b) is also an
improper integral; it has 2 improper points, a and b, not just 1. Again there are 3 situations for the behavior of f at or
near each of a and b. So there are 6 situations. We show the situation where f is undefined at a and b and unbounded
near them in Fig. 5.4. The improper integral of f we're going to define now is the same for all 6 situations. Let c be an
arbitrary point such that a < c < b. The improper integral of f over (a, b) is defined to be the sum of the basic-type
improper integrals of f over the half-open finite intervals (a, c] and [c, b), and converges iff both the basic types
converge. #  #  #  # ## Definition 5.1 In each case, the improper integral on the left-hand side converges iff both the basic-type improper integrals on the right-hand side do.

Remark that the point c should be chosen so that it makes the calculations of the improper integrals on the right-hand
side as simple as possible.

### Example 5.1

Determine whether or not the integral: converges. See Fig. 5.5. # Improper Integral For Example 5.1.

Solution EOS

 6. Non-Basic Types Of Improper Integrals

Each of the improper integrals defined in Definition 5.1 has 2 improper points. Each is of a non-basic type. We say that an
improper integral is of a non-basic type if it has more than 1 improper points. There are more than 1 limits to handle
for a non-basic-type improper integral.

 7. Breaking Non-Basic-Type Improper Integrals

Part 5 shows the necessity that non-basic-type improper integrals must be broken into (ie, expressed as a sum of ) separate
basic-type improper integrals, and the way to break them. There we break the given improper integrals into 2 basic types.

A non-basic-type improper integral will be broken into basic types. There are non-basic types that must be broken into
more than 2 basic types.

### Example 7.1

Determine whether the following improper integral converges or diverges. Sketch a graph. Solution  EOS 8. Examining Definite Integrals For Improper Points ### Example 8.1

Find: See Fig. 8.1. # Fig. 8.1

Integral For Example 8.1.

#### Incorrect Solution EOS

Correct Solution

The integrand 1/x is undefined & so is discontinuous at x = 0. Thus: EOS

Recall that the fundamental theorem of calculus applies only if the interval of integration is finite and closed of the form
[a, b] and the integrand is continuous on [a, b]. In the incorrect solution, the theorem is incorrectly applied to a function
that is not continuous at 0 and hence not continuous on [–1, 1].

 9. Areas Of Unbounded Regions

The region under the graph of y = 1/x, above the x-axis, and to the right of the vertical line x = 1, colored in Fig. 9.1,
extends to infinity on the right-hand side. It's an unbounded region. Its area is found in Example 2.1.a and is infinite. This
unbounded region has an infinite area.

The region under the graph of y = 1/x2, above the x-axis, and to the right of the vertical line x = 1, colored in Fig. 9.2,
extends to infinity on the right-hand side. It's an unbounded region. Its area is found in Example 2.1.b and is 1. This
unbounded region has a finite area. {9.1} Example 5.1. # This unbounded region has an infinite area. # This unbounded region has a finite area. # This unbounded region has a finite area.

In general, some unbounded regions have infinite areas while others have finite areas. This is true whether a region
extends to infinity along the x-axis or along the y-axis or both, and whether it extends to infinity on one or both sides of an
axis.

###### Problems & Solutions

1. Determine whether each of the following improper integrals converges, and find its value if it does. Solution d. Using the method of integration by parts let u = ln x and dv = dx, so that du = dx/x and v = x. Thus: convergent to –1. 2. Solution

a.   3.  Evaluate each of the following improper integrals or show that it diverges. Solution  d. Utilizing the method of integration by parts let u = x and dv = ex dx, so that du = dx and v = – ex. Consequently: e. We have:  4.
a.  Sketch the curve y = ex. Shade the region that lies above the x-axis, below the curve y = ex, and to the left of the
y-axis.
b.  Find the area of the shaded region.

Solution

a.   5. Prove that, for a > 0, the improper integral: Solution

If p = 1 then we have: 