§3.2–Riemann Integration, Part I

Tom Lewis

Fall Term 2006

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 1 / 10

Outline

1 Improper integrals: unbounded intervals

2 Absolute convergence

3 Simple comparison test

4 Improper integrals: unbounded functions

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 2 / 10

Improper integrals: unbounded intervals

Definition

Let a ∈ R and let f be integrable on each interval of the form [a, b]for b > a. Define ∫ +∞

af (x)dx = lim

b→+∞

∫ b

af (x)dx ,

whenever this limit exists.

When this limit exists, we say that the improper integral∫ +∞a f (x)dx

converges; otherwise, we say that the improper integral∫ +∞a f (x)dx

diverges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 3 / 10

Improper integrals: unbounded intervals

Definition

Let a ∈ R and let f be integrable on each interval of the form [a, b]for b > a. Define ∫ +∞

af (x)dx = lim

b→+∞

∫ b

af (x)dx ,

whenever this limit exists.

When this limit exists, we say that the improper integral∫ +∞a f (x)dx

converges; otherwise, we say that the improper integral∫ +∞a f (x)dx

diverges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 3 / 10

Improper integrals: unbounded intervals

Definition

Let a ∈ R and let f be integrable on each interval of the form [a, b]for b > a. Define ∫ +∞

af (x)dx = lim

b→+∞

∫ b

af (x)dx ,

whenever this limit exists.

When this limit exists, we say that the improper integral∫ +∞a f (x)dx

converges; otherwise, we say that the improper integral∫ +∞a f (x)dx

diverges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 3 / 10

Improper integrals: unbounded intervals

Example

By properties of the logarithm,∫ ∞

1

1

xdx = lim

b→∞

∫ b

1

1

xdx = lim

b→∞log(b),

which diverges.

On the other hand,∫ ∞

1

1

x2dx = lim

b→∞

∫ b

1

1

x2dx = lim

b→∞

(1− 1

b

)= 1,

which converges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 4 / 10

Improper integrals: unbounded intervals

Example

By properties of the logarithm,∫ ∞

1

1

xdx = lim

b→∞

∫ b

1

1

xdx = lim

b→∞log(b),

which diverges.

On the other hand,∫ ∞

1

1

x2dx = lim

b→∞

∫ b

1

1

x2dx = lim

b→∞

(1− 1

b

)= 1,

which converges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 4 / 10

Improper integrals: unbounded intervals

Example

By properties of the logarithm,∫ ∞

1

1

xdx = lim

b→∞

∫ b

1

1

xdx = lim

b→∞log(b),

which diverges.

On the other hand,∫ ∞

1

1

x2dx = lim

b→∞

∫ b

1

1

x2dx = lim

b→∞

(1− 1

b

)= 1,

which converges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 4 / 10

Absolute convergence

Definition

Let

f +(x) =|f (x)|+ f (x)

2and f −(x) =

|f (x)| − f (x)

2.

f + and f − are called the positive and negative and parts of f respectively.

Theorem

f + and f − are nonnegative functions with f + ≤ |f | and f − ≤ |f |.f = f + − f − and |f | = f + + f −.

Theorem

If∫∞a |f (x)|dx converges, then

∫∞a f (x)dx converges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 5 / 10

Absolute convergence

Definition

Let

f +(x) =|f (x)|+ f (x)

2and f −(x) =

|f (x)| − f (x)

2.

f + and f − are called the positive and negative and parts of f respectively.

Theorem

f + and f − are nonnegative functions with f + ≤ |f | and f − ≤ |f |.f = f + − f − and |f | = f + + f −.

Theorem

If∫∞a |f (x)|dx converges, then

∫∞a f (x)dx converges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 5 / 10

Absolute convergence

Definition

Let

f +(x) =|f (x)|+ f (x)

2and f −(x) =

|f (x)| − f (x)

2.

f + and f − are called the positive and negative and parts of f respectively.

Theorem

f + and f − are nonnegative functions with f + ≤ |f | and f − ≤ |f |.

f = f + − f − and |f | = f + + f −.

Theorem

If∫∞a |f (x)|dx converges, then

∫∞a f (x)dx converges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 5 / 10

Absolute convergence

Definition

Let

f +(x) =|f (x)|+ f (x)

2and f −(x) =

|f (x)| − f (x)

2.

f + and f − are called the positive and negative and parts of f respectively.

Theorem

f + and f − are nonnegative functions with f + ≤ |f | and f − ≤ |f |.f = f + − f − and |f | = f + + f −.

Theorem

If∫∞a |f (x)|dx converges, then

∫∞a f (x)dx converges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 5 / 10

Absolute convergence

Definition

Let

f +(x) =|f (x)|+ f (x)

2and f −(x) =

|f (x)| − f (x)

2.

f + and f − are called the positive and negative and parts of f respectively.

Theorem

f + and f − are nonnegative functions with f + ≤ |f | and f − ≤ |f |.f = f + − f − and |f | = f + + f −.

Theorem

If∫∞a |f (x)|dx converges, then

∫∞a f (x)dx converges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 5 / 10

Absolute convergence

Proof.

Since f + and f − are nonnegative and bounded by |f |, the integrals∫ b

af +(x)dx and

∫ b

af −(x)dx

are monotone increasing as b → +∞ and bounded above by∫∞a |f (x)|dx . In particular, both integrals converge as b → +∞.

It follows that

limb→∞

∫ b

af (x)dx = lim

b→∞

(∫ b

af +(x)dx +

∫ b

af −(x)dx

)exists; hence, the improper integral

∫∞a f (x)dx converges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 6 / 10

Absolute convergence

Proof.

Since f + and f − are nonnegative and bounded by |f |, the integrals∫ b

af +(x)dx and

∫ b

af −(x)dx

are monotone increasing as b → +∞ and bounded above by∫∞a |f (x)|dx . In particular, both integrals converge as b → +∞.

It follows that

limb→∞

∫ b

af (x)dx = lim

b→∞

(∫ b

af +(x)dx +

∫ b

af −(x)dx

)exists; hence, the improper integral

∫∞a f (x)dx converges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 6 / 10

Absolute convergence

Proof.

Since f + and f − are nonnegative and bounded by |f |, the integrals∫ b

af +(x)dx and

∫ b

af −(x)dx

are monotone increasing as b → +∞ and bounded above by∫∞a |f (x)|dx . In particular, both integrals converge as b → +∞.

It follows that

limb→∞

∫ b

af (x)dx = lim

b→∞

(∫ b

af +(x)dx +

∫ b

af −(x)dx

)exists; hence, the improper integral

∫∞a f (x)dx converges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 6 / 10

Simple comparison test

Theorem

If f and g are integrable on [a, b] for all b > a, g(x) ≥ |f (x)| on [a,∞),and

∫∞a g(x)dx converges, then

∫∞a f (x)dx converges.

Proof.

The integral∫ ba |f (x)|dx is monotone increasing as b → +∞ and

bounded above by∫∞a g(x)dx .

Thus the improper integral∫∞a |f (x)|dx converges.

By a previous result,∫∞a f (x)dx converges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 7 / 10

Simple comparison test

Theorem

If f and g are integrable on [a, b] for all b > a, g(x) ≥ |f (x)| on [a,∞),and

∫∞a g(x)dx converges, then

∫∞a f (x)dx converges.

Proof.

The integral∫ ba |f (x)|dx is monotone increasing as b → +∞ and

bounded above by∫∞a g(x)dx .

Thus the improper integral∫∞a |f (x)|dx converges.

By a previous result,∫∞a f (x)dx converges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 7 / 10

Simple comparison test

Theorem

If f and g are integrable on [a, b] for all b > a, g(x) ≥ |f (x)| on [a,∞),and

∫∞a g(x)dx converges, then

∫∞a f (x)dx converges.

Proof.

The integral∫ ba |f (x)|dx is monotone increasing as b → +∞ and

bounded above by∫∞a g(x)dx .

Thus the improper integral∫∞a |f (x)|dx converges.

By a previous result,∫∞a f (x)dx converges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 7 / 10

Simple comparison test

Theorem

If f and g are integrable on [a, b] for all b > a, g(x) ≥ |f (x)| on [a,∞),and

∫∞a g(x)dx converges, then

∫∞a f (x)dx converges.

Proof.

The integral∫ ba |f (x)|dx is monotone increasing as b → +∞ and

bounded above by∫∞a g(x)dx .

Thus the improper integral∫∞a |f (x)|dx converges.

By a previous result,∫∞a f (x)dx converges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 7 / 10

Simple comparison test

Theorem

If f and g are integrable on [a, b] for all b > a, g(x) ≥ |f (x)| on [a,∞),and

∫∞a g(x)dx converges, then

∫∞a f (x)dx converges.

Proof.

The integral∫ ba |f (x)|dx is monotone increasing as b → +∞ and

bounded above by∫∞a g(x)dx .

Thus the improper integral∫∞a |f (x)|dx converges.

By a previous result,∫∞a f (x)dx converges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 7 / 10

Simple comparison test

Example

Since| cos(x)|

x2≤ 1

x2for all x ≥ 1

and since∫∞1

1x2 dx converges, it follows that

∫∞1

cos(x)x2 dx converges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 8 / 10

Improper integrals: unbounded functions

Definition

Suppose that f is defined but unbounded on (a, b]. If f is boundedand integrable on [c , b] for all a < c < b, then let∫ b

af (x)dx = lim

c→a+

∫ b

cf (x)dx ,

whenever this limit exists.

When this limit exists, we say that the improper integral∫ ba f (x)dx

converges; otherwise, we say that the improper integral∫ ba f (x)dx

diverges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 9 / 10

Improper integrals: unbounded functions

Definition

Suppose that f is defined but unbounded on (a, b]. If f is boundedand integrable on [c , b] for all a < c < b, then let∫ b

af (x)dx = lim

c→a+

∫ b

cf (x)dx ,

whenever this limit exists.

When this limit exists, we say that the improper integral∫ ba f (x)dx

converges; otherwise, we say that the improper integral∫ ba f (x)dx

diverges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 9 / 10

Improper integrals: unbounded functions

Definition

Suppose that f is defined but unbounded on (a, b]. If f is boundedand integrable on [c , b] for all a < c < b, then let∫ b

af (x)dx = lim

c→a+

∫ b

cf (x)dx ,

whenever this limit exists.

When this limit exists, we say that the improper integral∫ ba f (x)dx

converges; otherwise, we say that the improper integral∫ ba f (x)dx

diverges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 9 / 10

Improper integrals: unbounded functions

Example

By properties of the natural logarithm, the integral∫ 1

0

1

x= lim

c→0+

∫ 1

c

1

x= lim

c→0+log(c),

diverges.

However, the integral∫ 1

0

1

x1/2= lim

c→0+

∫ 1

c

1

x1/2= lim

c→0+(2− 2c1/2) = 2,

converges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 10 / 10

Improper integrals: unbounded functions

Example

By properties of the natural logarithm, the integral∫ 1

0

1

x= lim

c→0+

∫ 1

c

1

x= lim

c→0+log(c),

diverges.

However, the integral∫ 1

0

1

x1/2= lim

c→0+

∫ 1

c

1

x1/2= lim

c→0+(2− 2c1/2) = 2,

converges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 10 / 10

Improper integrals: unbounded functions

Example

By properties of the natural logarithm, the integral∫ 1

0

1

x= lim

c→0+

∫ 1

c

1

x= lim

c→0+log(c),

diverges.

However, the integral∫ 1

0

1

x1/2= lim

c→0+

∫ 1

c

1

x1/2= lim

c→0+(2− 2c1/2) = 2,

converges.

Tom Lewis () §3.2–Riemann Integration, Part I Fall Term 2006 10 / 10