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Chapter 8 Improper Integrals

Definition 0.1. Improper Integrals

Integrals with infinite limits of integration and integrals of functions that become infinite at a pointwithin the interval of integration are improper integrals.

Example 1. Examples of Improper Integrals

(a)

0

ex dx is an improper integral because

.

(b)

21

1

2x 3 dx is an improper integral because

.

Definition 0.2. Type 1 : Improper Integrals with Infinite Integration Limits

(a) If

ta

f(x) dx exists for every t a, then

a

f(x) dx = limt

ta

f(x) dx

provided that this limit exists (as a finite number).

(b) If

bt

f(x) dx exists for every t b, thenb

f(x) dx = limt

bt

f(x) dx

provided that this limit exists (as a finite number).

Remark. In the above two cases, if the limit exists and is finite, then we say that the improperintegral is convergent. Otherwise, we say that the improper integral is divergent.

(c) If both

a

f(x) dx and

a

f(x) dx are convergent, then we define

f(x) dx =

a

f(x) dx +

a

f(x) dx

where a is any real number.Remark. This improper integral diverges if either of the improper integrals on the right diverges.

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Example 2. Evaluate I =

0

ex dx.

Remark. Since ex > 0, I can be interpreted as the area of the infinite region in the first quadrantthat lies under the curve y = ex and above the x-axis.

Example 3. Find, if possible, I = 0

cos x dx.

Example 4. Find I =

1

1 + x2dx.

Example 5. For which numbers p does the improper integral I =

1

1

xpdx converges?

Example 6. Evaluate I =

x

1 + x2dx or show that it diverges.

Exercise 1.

f(x)x dx may not equal limtt

tf(x) dx

(a) Show that

x dx is divergent.

(b) Show that limt

tt

x dx = 0.

Remark. This shows that we cant define

f(x)x dx = limt

tt

f(x) dx

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Type 2 : Improper Integrals with Discontinuous Integrands

(a) Iff is continuous on [a, b] except for an infinite discontinuity at a, thenba

f(x) dx = limta+

bt

f(x) dx

provided that this limit exists (as a finite number).(b) Iff is continuous on [a, b] except for an infinite discontinuity at b, thenb

a

f(x) dx = limtb

ta

f(x) dx

provided that this limit exists (as a finite number).

(c) Iff is continuous on [a, b] except at c (a, b), where f has an infinite discontinuity, then

b

a

f(x) dx = c

a

f(x) dx + b

c

f(x) dx

provided that both the improper integrals on the right converge. If both converge, then the value

of the improper integral

ba

f(x) dx is the sum of the two values.

Remark. For each of the case (a), (b) and (c), the improper integral is convergent if the appropriatelimit or limits exist and are finite. Otherwise, the improper integral is divergent.

Example 7. Evaluate I =

10

1

xdx.

Example 8. Show that I = 1

2

1

x2dx diverges.

Example 9. Evaluate I =

42

dx

(3 x)2/3 if possible.

Reading Assignment 1. Investigate the convergence of I =

41

dx

(x 2)2 dx

Answer. Note that I is an improper integral because f(x) = 1(x2)2

has an infinite discontinuity atx = 2.

I = 2

1

dx

(x 2)2

dx I1

+4

2

dx

(x 2)2

dx I2

where I1 = limc2

c1

(x 2)2 dx = limc2

(x 2)1 dxc1

= limc2

1 1

c 2

= .

Thus I1 is divergent. This implies that I is also divergent. [We do not have to evaluate I2.]

Remark.

IF we ignore the fact that the integral is improper, then we are led to the incorrect conclusion thatI = 3/2!!!

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Tests for ConvergenceIt is often difficult to determine the convergence of a given improper integral by direct integration.But, we can gain some information by comparison with integrals of known behavior. The principaltests for convergence are the direct comparison (or domination) and the limit comparison tests.

Theorem 0.1. Direct Comparison Test (for Type 1 Improper Integrals)

Suppose that f and g are continuous functions with

0 f(x) g(x) x a.

(i) If

a

g(x) dx converges, then

a

f(x) dx converges.

(ii) If

a

f(x) dx diverges, then

a

g(x) dx diverges.

Example 10. Use the Direct Comparison Test to test the integrals for convergence.

(a)

1

sin2 x

x2 dx

(b)

1

1 + ex

xdx

Reading Assignment 2. Use the Direct Comparison Test to test the integral I =

2

1x2 1 dx

for convergence.

Answer.

(a) For x 2, we have 0 < x2 1 x2 so that 0