§3.2--Riemann Integration, Part I - Furman...
Transcript of §3.2--Riemann Integration, Part I - Furman...
§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