Improper Integrals

Improper Integrals

Integrals over an infinite interval such as ∫ e-x dx 0

∞

Improper Integrals

Integrals over an infinite interval such as ∫ e-x dx or

0

∞

integrals of f(x) that goes to ∞ within the interval of

integration such as ∫ Ln(x) dx, 0

1

Improper Integrals

Integrals over an infinite interval such as ∫ e-x dx or

0

∞

integrals of f(x) that goes to ∞ within the interval of

integration such as ∫ Ln(x) dx, are called

improper integrals. 0

1

Improper Integrals

Integrals over an infinite interval such as ∫ e-x dx or

0

∞

integrals of f(x) that goes to ∞ within the interval of

integration such as ∫ Ln(x) dx, are called

improper integrals. 0

1

Let F(x) be the antiderivative of f(x).

Improper Integrals

Integrals over an infinite interval such as ∫ e-x dx or

0

∞

integrals of f(x) that goes to ∞ within the interval of

integration such as ∫ Ln(x) dx, are called

improper integrals. 0

1

Let F(x) be the antiderivative of f(x). We define ∫a

∞

f(x) dx as lim f(x) dx

∫a

u

u ∞

Improper Integrals

Integrals over an infinite interval such as ∫ e-x dx or

0

∞

integrals of f(x) that goes to ∞ within the interval of

integration such as ∫ Ln(x) dx, are called

improper integrals. 0

1

Let F(x) be the antiderivative of f(x). We define ∫a

∞

f(x) dx as lim f(x) dx = F(x)| = lim F(x) – F(a)

∫a

u

u ∞ x ∞a

∞

Improper Integrals

Integrals over an infinite interval such as ∫ e-x dx or

0

∞

integrals of f(x) that goes to ∞ within the interval of

integration such as ∫ Ln(x) dx, are called

improper integrals. 0

1

Let F(x) be the antiderivative of f(x). We define

Example: Find

∫0

∞

e-x dx

∫a

∞

f(x) dx as lim f(x) dx = F(x)| = lim F(x) – F(a)

∫a

u

u ∞ x ∞a

∞

Improper Integrals

Integrals over an infinite interval such as ∫ e-x dx or

0

∞

integrals of f(x) that goes to ∞ within the interval of

integration such as ∫ Ln(x) dx, are called

improper integrals. 0

1

Let F(x) be the antiderivative of f(x). We define

Example: Find

∫0

∞

e-x dx

∫0

∞

e-x dx = -e-x|

0

∞

∫a

∞

f(x) dx as lim f(x) dx = F(x)| = lim F(x) – F(a)

∫a

u

u ∞ x ∞a

∞

Improper Integrals

Integrals over an infinite interval such as ∫ e-x dx or

0

∞

integrals of f(x) that goes to ∞ within the interval of

integration such as ∫ Ln(x) dx, are called

improper integrals. 0

1

Let F(x) be the antiderivative of f(x). We define

Example: Find

∫0

∞

e-x dx

∫0

∞

e-x dx = -e-x| = lim (-e-x) – (-e0)

x ∞0

∞

∫a

∞

f(x) dx as lim f(x) dx = F(x)| = lim F(x) – F(a)

∫a

u

u ∞ x ∞a

∞

Improper Integrals

Integrals over an infinite interval such as ∫ e-x dx or

0

∞

integrals of f(x) that goes to ∞ within the interval of

integration such as ∫ Ln(x) dx, are called

improper integrals. 0

1

Let F(x) be the antiderivative of f(x). We define

Example: Find

∫0

∞

e-x dx

∫0

∞

e-x dx = -e-x| = lim (-e-x) – (-e0) = 0 – (-1) = 1

x ∞0

∞

∫a

∞

f(x) dx as lim f(x) dx = F(x)| = lim F(x) – F(a)

∫a

u

u ∞ x ∞a

∞

Improper Integrals

Let F(x) be the antiderivative of f(x) and lim f(x) ∞.

x a

Improper Integrals

Let F(x) be the antiderivative of f(x) and lim f(x) ∞.

We define x a

∫a

b

f(x) dx = F(x)| = F(b) – limF(x) x aa

b

Improper Integrals

Let F(x) be the antiderivative of f(x) and lim f(x) ∞.

We define x a

∫a

b

f(x) dx = F(x)| = F(b) – limF(x) x a

Example: Find

∫0

1

x-1/2 dx

a

b

Improper Integrals

Let F(x) be the antiderivative of f(x) and lim f(x) ∞.

We define x a

∫a

b

f(x) dx = F(x)| = F(b) – limF(x) x a

Example: Find

∫0

1

x-1/2 dx

∫0

1

x-1/2 dx = 2x1/2 |

1

0

a

b

Improper Integrals

Let F(x) be the antiderivative of f(x) and lim f(x) ∞.

We define x a

∫a

b

f(x) dx = F(x)| = F(b) – limF(x) x a

Example: Find

∫0

1

x-1/2 dx

∫0

1

x-1/2 dx = 2x1/2 | = 2 – lim(2x1/2)

1

0 x 0

a

b

Improper Integrals

Let F(x) be the antiderivative of f(x) and lim f(x) ∞.

We define x a

∫a

b

f(x) dx = F(x)| = F(b) – limF(x) x a

Example: Find

∫0

1

x-1/2 dx

∫0

1

x-1/2 dx = 2x1/2 | = 2 – lim(2x1/2) = 2

1

0 x 0

a

b

Improper Integrals

Let F(x) be the antiderivative of f(x) and lim f(x) ∞.

We define x a

∫a

b

f(x) dx = F(x)| = F(b) – limF(x) x a

Example: Find

∫0

1

x-1/2 dx

∫0

1

x-1/2 dx = 2x1/2 | = 2 – lim(2x1/2) = 2

1

0 x 0

Example: Find

∫0

1

x-2 dx

a

b

Improper Integrals

Let F(x) be the antiderivative of f(x) and lim f(x) ∞.

We define x a

∫a

b

f(x) dx = F(x)| = F(b) – limF(x) x a

Example: Find

∫0

1

x-1/2 dx

∫0

1

x-1/2 dx = 2x1/2 | = 2 – lim(2x1/2) = 2

1

0 x 0

Example: Find

∫0

1

x-2 dx

∫0

1

x-2 dx = –x–1 | 1

0

a

b

Improper Integrals

Let F(x) be the antiderivative of f(x) and lim f(x) ∞.

We define x a

∫a

b

f(x) dx = F(x)| = F(b) – limF(x) x a

Example: Find

∫0

1

x-1/2 dx

∫0

1

x-1/2 dx = 2x1/2 | = 2 – lim(2x1/2) = 2

1

0 x 0

Example: Find

∫0

1

x-2 dx

∫0

1

x-2 dx = –x–1 | = -1 – lim(-x-1)

1

0 x 0

a

b

Improper Integrals

Let F(x) be the antiderivative of f(x) and lim f(x) ∞.

We define x a

∫a

b

f(x) dx = F(x)| = F(b) – limF(x) x a

Example: Find

∫0

1

x-1/2 dx

∫0

1

x-1/2 dx = 2x1/2 | = 2 – lim(2x1/2) = 2

1

0 x 0

Example: Find

∫0

1

x-2 dx

∫0

1

x-2 dx = –x–1 | = -1 – lim(-x-1) = -1 – (-∞) = ∞

1

0 x 0

a

b

Improper Integrals

Let F(x) be the antiderivative of f(x) and lim f(x) ∞.

We define x a

∫a

b

f(x) dx = F(x)| = F(b) – limF(x) x a

Example: Find

∫0

1

x-1/2 dx

∫0

1

x-1/2 dx = 2x1/2 | = 2 – lim(2x1/2) = 2

1

0 x 0

Example: Find

∫0

1

x-2 dx

∫0

1

x-2 dx = –x–1 | = -1 – lim(-x-1) = -1 – (-∞) = ∞

1

0 x 0

If the improper integral exists, we say it converges.

a

b

Improper Integrals

Let F(x) be the antiderivative of f(x) and lim f(x) ∞.

We define x a

∫a

b

f(x) dx = F(x)| = F(b) – limF(x) x a

Example: Find

∫0

1

x-1/2 dx

∫0

1

x-1/2 dx = 2x1/2 | = 2 – lim(2x1/2) = 2

1

0 x 0

Example: Find

∫0

1

x-2 dx

∫0

1

x-2 dx = –x–1 | = -1 – lim(-x-1) = -1 – (-∞) = ∞

1

0 x 0

If the improper integral exists, we say it converges. If the improper integral fails to exist or its infinite, we say it diverges.

a

b

Improper Integrals(The Floor theorem)

Improper Integrals(The Floor theorem)

y = f(x)

y = g(x)∞

Improper Integrals(The Floor theorem) If f(x) > g(x) > 0 and g(x) dx = ∞, ∫

a

b

y = f(x)

y = g(x)∞

Improper Integrals(The Floor theorem) If f(x) > g(x) > 0 and g(x) dx = ∞, then f(x) = ∞. ∫

a

b

∫a

b

y = f(x)

y = g(x)∞

Improper Integrals(The Floor theorem) If f(x) > g(x) > 0 and g(x) dx = ∞, then f(x) = ∞. ∫

a

b

∫a

b

y = f(x)

y = g(x)∞

(The Ceiling theorem)

Improper Integrals(The Floor theorem) If f(x) > g(x) > 0 and g(x) dx = ∞, then f(x) = ∞. ∫

a

b

∫a

b

y = f(x)

y = g(x)∞

(The Ceiling theorem)

y = f(x)

y = g(x)

N

Improper Integrals(The Floor theorem) If f(x) > g(x) > 0 and g(x) dx = ∞, then f(x) = ∞. ∫

a

b

∫a

b

y = f(x)

y = g(x)∞

(The Ceiling theorem)

If f(x) > g(x) > 0 and f(x) dx = N converges ∫a

b

y = f(x)

y = g(x)

N

Improper Integrals(The Floor theorem) If f(x) > g(x) > 0 and g(x) dx = ∞, then f(x) = ∞. ∫

a

b

∫a

b

y = f(x)

y = g(x)∞

(The Ceiling theorem)

If f(x) > g(x) > 0 and f(x) dx = N converges then

g(x) dx converges also.∫a

b

∫a

b

y = f(x)

y = g(x)

N

Improper Integrals

Note that no conclusion may be drawn if

f(x) > g(x) > 0 and g(x) dx converges, ∫a

b

Improper Integrals

Note that no conclusion may be drawn if

f(x) > g(x) > 0 and g(x) dx converges, (or f(x) dx = ∞).∫a

b

∫a

b

Improper Integrals

The function y = serves as an important

"boundary" for divergence and convergence.

x1

Note that no conclusion may be drawn if

f(x) > g(x) > 0 and g(x) dx converges, (or f(x) dx = ∞).∫a

b

∫a

b

Improper Integrals

The function y = serves as an important

"boundary" for divergence and convergence. The following are the graphs of y = and

x ,1

x1

x2 ,1

x3 1

for x > 1.

Note that no conclusion may be drawn if

f(x) > g(x) > 0 and g(x) dx converges, (or f(x) dx = ∞).∫a

b

∫a

b

Improper Integrals

The function y = serves as an important

"boundary" for divergence and convergence. The following are the graphs of y = and

x ,1

x1

x2 ,1

x3 1

for x > 1.

(1, 1)

y = 1/x

y = 1/x2

y = 1/x3

Note that no conclusion may be drawn if

f(x) > g(x) > 0 and g(x) dx converges, (or f(x) dx = ∞).∫a

b

∫a

b

1 ∞

Improper Integrals

∫1

∞

One checks that

(1, 1)

y = 1/x

y = 1/x2

y = 1/x3

x1 dx = Ln(x)| = ∞,

1

∞

1 ∞

Improper Integrals

∫1

∞

One checks that

(1, 1)

y = 1/x

y = 1/x2

y = 1/x3

x1 dx = Ln(x)| = ∞, but ∫1

∞

x21 dx and ∫1

∞

x31 dx are finite.

1

∞

1 ∞

Improper Integrals

∫1

∞

One checks that

(1, 1)

y = 1/x

y = 1/x2

y = 1/x3

x1 dx = Ln(x)| = ∞, but ∫1

∞

x21 dx and ∫1

∞

x31 dx are finite.

The p-Theorem:

∫1 xp1

∞

dx converges for p > 1, A.

1

∞

1 ∞

Improper Integrals

∫1

∞

One checks that

(1, 1)

y = 1/x

y = 1/x2

y = 1/x3

x1 dx = Ln(x)| = ∞, but ∫1

∞

x21 dx and ∫1

∞

x31 dx are finite.

The p-Theorem:

∫1 xp1

∞

dx converges for p > 1, diverges for p < 1.A.

1

∞

1 ∞

Improper Integrals

y = 1/x

y = 1/x1/2

y = 1/x1/3

0 1

Improper Integrals

∫0

1

x1

On the other hand, over (0, 1]

dx = Ln(x)| = ∞0

1

y = 1/x

y = 1/x1/2

y = 1/x1/3

0 1

Improper Integrals

∫0

1

x1

On the other hand, over (0, 1]

dx = Ln(x)| = ∞0

1

and that x2

1 > x1 over (0, 1]

y = 1/x

y = 1/x1/2

y = 1/x1/3

0 1

Improper Integrals

∫0

1

x1

So ∫0

1

x2

1 dx = ∞

On the other hand, over (0, 1]

dx = Ln(x)| = ∞0

1

and that x2

1 > x1 over (0, 1]

y = 1/x

y = 1/x1/2

y = 1/x1/3

0 1

Improper Integrals

∫0

1

x1

So ∫0

1

x2

1 dx = ∞

On the other hand, over (0, 1]

dx = Ln(x)| = ∞0

1

and that x2

1 > x1 over (0, 1]

y = 1/x

y = 1/x1/2

y = 1/x1/3

∫0

1

x1/2

1 dx and ∫ x1/3

1 dx are finite.

One checks easily that

0

1

0 1

Improper Integrals

∫0

1

x1

So ∫0

1

x2

1 dx = ∞

The p-Theorem:

∫ xp1 dx converges for p < 1, B.

On the other hand, over (0, 1]

dx = Ln(x)| = ∞0

1

and that x2

1 > x1 over (0, 1]

y = 1/x

y = 1/x1/2

y = 1/x1/3

∫0

1

x1/2

1 dx and ∫ x1/3

1 dx are finite.

One checks easily that

0

1

0

1

0 1

Improper Integrals

∫0

1

x1

So ∫0

1

x2

1 dx = ∞

The p-Theorem:

∫ xp1 dx converges for p < 1, diverges for p > 1.B.

On the other hand, over (0, 1]

dx = Ln(x)| = ∞0

1

and that x2

1 > x1 over (0, 1]

y = 1/x

y = 1/x1/2

y = 1/x1/3

∫0

1

x1/2

1 dx and ∫ x1/3

1 dx are finite.

One checks easily that

0

1

0

1

0 1