Improper Integrals Inﬁnite Intervals Area Interpretation Theorem 1 Functions with Inﬁnite Discontinuities Comparison Test Comparison Test Improper Integrals

I That is integrals of the type

Z ∞ 1 Z 1 1 Z ∞ 1 A) 3 dx B) 3 dx C) 2 1 x 0 x −∞ 4 + x

1 I Note that the function f (x) = x3 has a discontinuity at x = 0 and the F.T.C. does not apply to B.

I Note that the limits of integration for integrals A and C describe intervals that are inﬁnite in length and the F.T.C. does not apply.

Improper Integrals Inﬁnite Intervals Area Interpretation Theorem 1 Functions with inﬁnite discontinuities Comparison Test Comparison Test Improper Integrals

R b In this section, we will extend the concept of the definite integral a f (x)dx to functions with an infinite discontinuity and to infinite intervals.

Annette Pilkington Improper Integrals 1 I Note that the function f (x) = x3 has a discontinuity at x = 0 and the F.T.C. does not apply to B.

I Note that the limits of integration for integrals A and C describe intervals that are inﬁnite in length and the F.T.C. does not apply.

Improper Integrals Inﬁnite Intervals Area Interpretation Theorem 1 Functions with inﬁnite discontinuities Comparison Test Comparison Test Improper Integrals

R b In this section, we will extend the concept of the definite integral a f (x)dx to functions with an infinite discontinuity and to infinite intervals.

I That is integrals of the type

Z ∞ 1 Z 1 1 Z ∞ 1 A) 3 dx B) 3 dx C) 2 1 x 0 x −∞ 4 + x

Annette Pilkington Improper Integrals I Note that the limits of integration for integrals A and C describe intervals that are inﬁnite in length and the F.T.C. does not apply.

Improper Integrals Inﬁnite Intervals Area Interpretation Theorem 1 Functions with inﬁnite discontinuities Comparison Test Comparison Test Improper Integrals

R b In this section, we will extend the concept of the definite integral a f (x)dx to functions with an infinite discontinuity and to infinite intervals.

I That is integrals of the type

Z ∞ 1 Z 1 1 Z ∞ 1 A) 3 dx B) 3 dx C) 2 1 x 0 x −∞ 4 + x

1 I Note that the function f (x) = x3 has a discontinuity at x = 0 and the F.T.C. does not apply to B.

Annette Pilkington Improper Integrals Improper Integrals Inﬁnite Intervals Area Interpretation Theorem 1 Functions with inﬁnite discontinuities Comparison Test Comparison Test Improper Integrals

R b In this section, we will extend the concept of the definite integral a f (x)dx to functions with an infinite discontinuity and to infinite intervals.

I That is integrals of the type

Z ∞ 1 Z 1 1 Z ∞ 1 A) 3 dx B) 3 dx C) 2 1 x 0 x −∞ 4 + x

1 I Note that the function f (x) = x3 has a discontinuity at x = 0 and the F.T.C. does not apply to B.

I Note that the limits of integration for integrals A and C describe intervals that are inﬁnite in length and the F.T.C. does not apply.

Annette Pilkington Improper Integrals Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

An Improper Integral of Type 1 R t (a) If a f (x)dx exists for every number t ≥ a, then Z ∞ Z t f (x)dx = lim f (x)dx a t→∞ a provided that limit exists and is finite. R b (c) If t f (x)dx exists for every number t ≤ b, then Z b Z b f (x)dx = lim f (x)dx −∞ t→−∞ t provided that limit exists and is finite. R ∞ R b The improper integrals a f (x)dx and −∞ f (x)dx are called Convergent if the corresponding limit exists and is finite and divergent if the limit does not exists. R ∞ R a (c) If (for any value of a) both a f (x)dx and −∞ f (x)dx are convergent, then we define Z ∞ Z a Z ∞ f (x)dx = f (x)dx + f (x)dx −∞ −∞ a If f (x) ≥ 0, we can give theAnnette definite Pilkington integralImproper above Integrals an area interpretation. R ∞ 1 R t I By definition 1 x dx = limt→∞ 1 1/x dx ˛t I = limt→∞ ln x˛ = limt→∞(ln t − ln 1) ˛1 I = limt→∞ ln t = ∞ R ∞ 1 I The integral 1 x dx diverges.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Z ∞ Z t Z b Z b f (x)dx = lim f (x)dx f (x)dx = lim f (x)dx a t→∞ a −∞ t→−∞ t

If f (x) ≥ 0, we can give the definite integral above an area interpretation; namely that if the improper integral converges, the area under the curve on the infinite interval is finite. Example Determine whether the following integrals converge or diverge: Z ∞ 1 Z ∞ 1 dx, 3 dx, 1 x 1 x

Annette Pilkington Improper Integrals ˛t I = limt→∞ ln x˛ = limt→∞(ln t − ln 1) ˛1 I = limt→∞ ln t = ∞ R ∞ 1 I The integral 1 x dx diverges.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Z ∞ Z t Z b Z b f (x)dx = lim f (x)dx f (x)dx = lim f (x)dx a t→∞ a −∞ t→−∞ t

R ∞ 1 R t I By deﬁnition 1 x dx = limt→∞ 1 1/x dx

Annette Pilkington Improper Integrals I = limt→∞ ln t = ∞ R ∞ 1 I The integral 1 x dx diverges.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Z ∞ Z t Z b Z b f (x)dx = lim f (x)dx f (x)dx = lim f (x)dx a t→∞ a −∞ t→−∞ t

R ∞ 1 R t I By deﬁnition 1 x dx = limt→∞ 1 1/x dx ˛t I = limt→∞ ln x˛ = limt→∞(ln t − ln 1) ˛1

Annette Pilkington Improper Integrals R ∞ 1 I The integral 1 x dx diverges.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Z ∞ Z t Z b Z b f (x)dx = lim f (x)dx f (x)dx = lim f (x)dx a t→∞ a −∞ t→−∞ t

R ∞ 1 R t I By deﬁnition 1 x dx = limt→∞ 1 1/x dx ˛t I = limt→∞ ln x˛ = limt→∞(ln t − ln 1) ˛1 I = limt→∞ ln t = ∞

Annette Pilkington Improper Integrals Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Z ∞ Z t Z b Z b f (x)dx = lim f (x)dx f (x)dx = lim f (x)dx a t→∞ a −∞ t→−∞ t

R ∞ 1 R t I By deﬁnition 1 x dx = limt→∞ 1 1/x dx ˛t I = limt→∞ ln x˛ = limt→∞(ln t − ln 1) ˛1 I = limt→∞ ln t = ∞ R ∞ 1 I The integral 1 x dx diverges.

Annette Pilkington Improper Integrals R ∞ 1 R t 1 I By deﬁnition 1 x3 dx = limt→∞ 1 x3 dx −2 ˛t x ˛ I = limt→∞ −2 ˛1 “ −1 1 ” I = limt→∞ 2t2 + 2 1 1 I = 0 + 2 = 2 . R ∞ 1 1 I The integral 1 x3 dx converges to 2 .

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Z ∞ Z t Z b Z b f (x)dx = lim f (x)dx f (x)dx = lim f (x)dx a t→∞ a −∞ t→−∞ t

Example Determine whether the following integrals converge or diverge: Z ∞ 1 Z ∞ 1 dx, 3 dx, 1 x 1 x

Annette Pilkington Improper Integrals −2 ˛t x ˛ I = limt→∞ −2 ˛1 “ −1 1 ” I = limt→∞ 2t2 + 2 1 1 I = 0 + 2 = 2 . R ∞ 1 1 I The integral 1 x3 dx converges to 2 .

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Z ∞ Z t Z b Z b f (x)dx = lim f (x)dx f (x)dx = lim f (x)dx a t→∞ a −∞ t→−∞ t

Example Determine whether the following integrals converge or diverge: Z ∞ 1 Z ∞ 1 dx, 3 dx, 1 x 1 x

R ∞ 1 R t 1 I By deﬁnition 1 x3 dx = limt→∞ 1 x3 dx

Annette Pilkington Improper Integrals “ −1 1 ” I = limt→∞ 2t2 + 2 1 1 I = 0 + 2 = 2 . R ∞ 1 1 I The integral 1 x3 dx converges to 2 .

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Z ∞ Z t Z b Z b f (x)dx = lim f (x)dx f (x)dx = lim f (x)dx a t→∞ a −∞ t→−∞ t

Example Determine whether the following integrals converge or diverge: Z ∞ 1 Z ∞ 1 dx, 3 dx, 1 x 1 x

R ∞ 1 R t 1 I By deﬁnition 1 x3 dx = limt→∞ 1 x3 dx −2 ˛t x ˛ I = limt→∞ −2 ˛1

Annette Pilkington Improper Integrals 1 1 I = 0 + 2 = 2 . R ∞ 1 1 I The integral 1 x3 dx converges to 2 .

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Z ∞ Z t Z b Z b f (x)dx = lim f (x)dx f (x)dx = lim f (x)dx a t→∞ a −∞ t→−∞ t

Example Determine whether the following integrals converge or diverge: Z ∞ 1 Z ∞ 1 dx, 3 dx, 1 x 1 x

R ∞ 1 R t 1 I By deﬁnition 1 x3 dx = limt→∞ 1 x3 dx −2 ˛t x ˛ I = limt→∞ −2 ˛1 “ −1 1 ” I = limt→∞ 2t2 + 2

Annette Pilkington Improper Integrals R ∞ 1 1 I The integral 1 x3 dx converges to 2 .

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Z ∞ Z t Z b Z b f (x)dx = lim f (x)dx f (x)dx = lim f (x)dx a t→∞ a −∞ t→−∞ t

Example Determine whether the following integrals converge or diverge: Z ∞ 1 Z ∞ 1 dx, 3 dx, 1 x 1 x

R ∞ 1 R t 1 I By deﬁnition 1 x3 dx = limt→∞ 1 x3 dx −2 ˛t x ˛ I = limt→∞ −2 ˛1 “ −1 1 ” I = limt→∞ 2t2 + 2 1 1 I = 0 + 2 = 2 .

Annette Pilkington Improper Integrals Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Z ∞ Z t Z b Z b f (x)dx = lim f (x)dx f (x)dx = lim f (x)dx a t→∞ a −∞ t→−∞ t

Example Determine whether the following integrals converge or diverge: Z ∞ 1 Z ∞ 1 dx, 3 dx, 1 x 1 x

R ∞ 1 R t 1 I By deﬁnition 1 x3 dx = limt→∞ 1 x3 dx −2 ˛t x ˛ I = limt→∞ −2 ˛1 “ −1 1 ” I = limt→∞ 2t2 + 2 1 1 I = 0 + 2 = 2 . R ∞ 1 1 I The integral 1 x3 dx converges to 2 .

Annette Pilkington Improper Integrals Improper Integrals Inﬁnite Intervals Area Interpretation Theorem 1 Functions with inﬁnite discontinuities Comparison Test Comparison Test Area Interpretation

1.0

0.8 0.8

0.6 0.6

0.4 0.4

0.2 0.2

5 10 15 20 25 30 5 10 15 20 25 30

R ∞ 1 Since 1 x dx diverges, the area under the curve y = 1/x on the interval [1, ∞) (shown on the left above) is not ﬁnite.

R ∞ 1 3 Since 1 x3 dx converges, the area under the curve y = 1/x on the interval [1, ∞) (shown on the right above) is ﬁnite.

Annette Pilkington Improper Integrals R 0 x R 0 x I By deﬁnition −∞ e dx = limt→−∞ t e dx 0 x ˛ I = limt→−∞ e ˛ ˛t 0 t I = limt→−∞(e − e )

I = 1 − 0 = 1. R 0 x I The integral −∞ e dx converges to 1.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Z ∞ Z t Z b Z b f (x)dx = lim f (x)dx f (x)dx = lim f (x)dx a t→∞ a −∞ t→−∞ t

Example Determine whether the following integral converges or diverges: R 0 x −∞ e dx

Annette Pilkington Improper Integrals 0 x ˛ I = limt→−∞ e ˛ ˛t 0 t I = limt→−∞(e − e )

I = 1 − 0 = 1. R 0 x I The integral −∞ e dx converges to 1.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Z ∞ Z t Z b Z b f (x)dx = lim f (x)dx f (x)dx = lim f (x)dx a t→∞ a −∞ t→−∞ t

Example Determine whether the following integral converges or diverges: R 0 x −∞ e dx R 0 x R 0 x I By deﬁnition −∞ e dx = limt→−∞ t e dx

Annette Pilkington Improper Integrals 0 t I = limt→−∞(e − e )

I = 1 − 0 = 1. R 0 x I The integral −∞ e dx converges to 1.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Z ∞ Z t Z b Z b f (x)dx = lim f (x)dx f (x)dx = lim f (x)dx a t→∞ a −∞ t→−∞ t

Example Determine whether the following integral converges or diverges: R 0 x −∞ e dx R 0 x R 0 x I By deﬁnition −∞ e dx = limt→−∞ t e dx 0 x ˛ I = limt→−∞ e ˛ ˛t

Annette Pilkington Improper Integrals I = 1 − 0 = 1. R 0 x I The integral −∞ e dx converges to 1.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Z ∞ Z t Z b Z b f (x)dx = lim f (x)dx f (x)dx = lim f (x)dx a t→∞ a −∞ t→−∞ t

Example Determine whether the following integral converges or diverges: R 0 x −∞ e dx R 0 x R 0 x I By deﬁnition −∞ e dx = limt→−∞ t e dx 0 x ˛ I = limt→−∞ e ˛ ˛t 0 t I = limt→−∞(e − e )

Annette Pilkington Improper Integrals R 0 x I The integral −∞ e dx converges to 1.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Z ∞ Z t Z b Z b f (x)dx = lim f (x)dx f (x)dx = lim f (x)dx a t→∞ a −∞ t→−∞ t

I = 1 − 0 = 1.

Annette Pilkington Improper Integrals Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Z ∞ Z t Z b Z b f (x)dx = lim f (x)dx f (x)dx = lim f (x)dx a t→∞ a −∞ t→−∞ t

I = 1 − 0 = 1. R 0 x I The integral −∞ e dx converges to 1.

Annette Pilkington Improper Integrals R ∞ 1 I Since this is a continuous function, we can calculate 0 4+x2 dx and R 0 1 −∞ 4+x2 dx ˛t R ∞ 1 R t 1 1 −1 x ˛ I 0 4+x2 dx = limt→∞ 0 4+x2 dx = limt→∞ 2 tan 2 ˛0 1 −1 t −1 1 π π I = limt→∞ 2 (tan 2 − tan 0) = 2 ( 2 − 0) = 4 . ˛0 R 0 1 R 0 1 1 −1 x ˛ I −∞ 4+x2 dx = limt→−∞ t 4+x2 dx = limt→−∞ 2 (tan 2 ) ˛t 1 −1 −1 t 1 (−π) π I = limt→−∞ 2 (tan 0 − tan 2 ) = 2 (0 − 2 ) = 4 . R ∞ 1 I The integral −∞ 4+x2 dx converges and is equal to Z 0 1 Z ∞ 1 π π π dx + dx = + = . −∞ 4 + x2 0 4 + x2 4 4 2

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals R ∞ R a If (for any value of a) both a f (x)dx and −∞ f (x)dx are convergent, then we define Z ∞ Z a Z ∞ f (x)dx = f (x)dx + f (x)dx −∞ −∞ a Example Determine whether the following integral converges or diverges:

Z ∞ 1 dx −∞ 4 + x2

Annette Pilkington Improper Integrals ˛t R ∞ 1 R t 1 1 −1 x ˛ I 0 4+x2 dx = limt→∞ 0 4+x2 dx = limt→∞ 2 tan 2 ˛0 1 −1 t −1 1 π π I = limt→∞ 2 (tan 2 − tan 0) = 2 ( 2 − 0) = 4 . ˛0 R 0 1 R 0 1 1 −1 x ˛ I −∞ 4+x2 dx = limt→−∞ t 4+x2 dx = limt→−∞ 2 (tan 2 ) ˛t 1 −1 −1 t 1 (−π) π I = limt→−∞ 2 (tan 0 − tan 2 ) = 2 (0 − 2 ) = 4 . R ∞ 1 I The integral −∞ 4+x2 dx converges and is equal to Z 0 1 Z ∞ 1 π π π dx + dx = + = . −∞ 4 + x2 0 4 + x2 4 4 2

Z ∞ 1 dx −∞ 4 + x2

R ∞ 1 I Since this is a continuous function, we can calculate 0 4+x2 dx and R 0 1 −∞ 4+x2 dx

Annette Pilkington Improper Integrals 1 −1 t −1 1 π π I = limt→∞ 2 (tan 2 − tan 0) = 2 ( 2 − 0) = 4 . ˛0 R 0 1 R 0 1 1 −1 x ˛ I −∞ 4+x2 dx = limt→−∞ t 4+x2 dx = limt→−∞ 2 (tan 2 ) ˛t 1 −1 −1 t 1 (−π) π I = limt→−∞ 2 (tan 0 − tan 2 ) = 2 (0 − 2 ) = 4 . R ∞ 1 I The integral −∞ 4+x2 dx converges and is equal to Z 0 1 Z ∞ 1 π π π dx + dx = + = . −∞ 4 + x2 0 4 + x2 4 4 2

Z ∞ 1 dx −∞ 4 + x2

R ∞ 1 I Since this is a continuous function, we can calculate 0 4+x2 dx and R 0 1 −∞ 4+x2 dx ˛t R ∞ 1 R t 1 1 −1 x ˛ I 0 4+x2 dx = limt→∞ 0 4+x2 dx = limt→∞ 2 tan 2 ˛0

Annette Pilkington Improper Integrals ˛0 R 0 1 R 0 1 1 −1 x ˛ I −∞ 4+x2 dx = limt→−∞ t 4+x2 dx = limt→−∞ 2 (tan 2 ) ˛t 1 −1 −1 t 1 (−π) π I = limt→−∞ 2 (tan 0 − tan 2 ) = 2 (0 − 2 ) = 4 . R ∞ 1 I The integral −∞ 4+x2 dx converges and is equal to Z 0 1 Z ∞ 1 π π π dx + dx = + = . −∞ 4 + x2 0 4 + x2 4 4 2

Z ∞ 1 dx −∞ 4 + x2

Annette Pilkington Improper Integrals 1 −1 −1 t 1 (−π) π I = limt→−∞ 2 (tan 0 − tan 2 ) = 2 (0 − 2 ) = 4 . R ∞ 1 I The integral −∞ 4+x2 dx converges and is equal to Z 0 1 Z ∞ 1 π π π dx + dx = + = . −∞ 4 + x2 0 4 + x2 4 4 2

Z ∞ 1 dx −∞ 4 + x2

Annette Pilkington Improper Integrals R ∞ 1 I The integral −∞ 4+x2 dx converges and is equal to Z 0 1 Z ∞ 1 π π π dx + dx = + = . −∞ 4 + x2 0 4 + x2 4 4 2

Z ∞ 1 dx −∞ 4 + x2

R ∞ 1 I Since this is a continuous function, we can calculate 0 4+x2 dx and R 0 1 −∞ 4+x2 dx ˛t R ∞ 1 R t 1 1 −1 x ˛ I 0 4+x2 dx = limt→∞ 0 4+x2 dx = limt→∞ 2 tan 2 ˛0 1 −1 t −1 1 π π I = limt→∞ 2 (tan 2 − tan 0) = 2 ( 2 − 0) = 4 . ˛0 R 0 1 R 0 1 1 −1 x ˛ I −∞ 4+x2 dx = limt→−∞ t 4+x2 dx = limt→−∞ 2 (tan 2 ) ˛t 1 −1 −1 t 1 (−π) π I = limt→−∞ 2 (tan 0 − tan 2 ) = 2 (0 − 2 ) = 4 .

Annette Pilkington Improper Integrals Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals R ∞ R a If (for any value of a) both a f (x)dx and −∞ f (x)dx are convergent, then we define Z ∞ Z a Z ∞ f (x)dx = f (x)dx + f (x)dx −∞ −∞ a Example Determine whether the following integral converges or diverges:

Z ∞ 1 dx −∞ 4 + x2

Annette Pilkington Improper Integrals I We’ve veriﬁed this for p = 1 above. If p 6= 1 1−p ˛t “ 1−p ” R ∞ 1 R t 1 x ˛ t 1 I 1 xp dx = limt→∞ 1 xp dx = limt→∞ 1−p = limt→∞ 1−p − 1−p ˛1 “ t1−p 1 ” 1 I If p > 1, limt→∞ 1−p − 1−p = − 1−p and the integral converges.

“ t1−p 1 ” t1−p I If p < 1, limt→∞ 1−p − 1−p does not exist since 1−p → ∞ as t → ∞ and the integral diverges.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Theorem

Z ∞ 1 p dx is convergent if p > 1 and divergent if p ≤ 1 1 x

Proof

Annette Pilkington Improper Integrals 1−p ˛t “ 1−p ” R ∞ 1 R t 1 x ˛ t 1 I 1 xp dx = limt→∞ 1 xp dx = limt→∞ 1−p = limt→∞ 1−p − 1−p ˛1 “ t1−p 1 ” 1 I If p > 1, limt→∞ 1−p − 1−p = − 1−p and the integral converges.

“ t1−p 1 ” t1−p I If p < 1, limt→∞ 1−p − 1−p does not exist since 1−p → ∞ as t → ∞ and the integral diverges.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Theorem

Z ∞ 1 p dx is convergent if p > 1 and divergent if p ≤ 1 1 x

Proof

I We’ve veriﬁed this for p = 1 above. If p 6= 1

Annette Pilkington Improper Integrals “ t1−p 1 ” 1 I If p > 1, limt→∞ 1−p − 1−p = − 1−p and the integral converges.

“ t1−p 1 ” t1−p I If p < 1, limt→∞ 1−p − 1−p does not exist since 1−p → ∞ as t → ∞ and the integral diverges.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Theorem

Z ∞ 1 p dx is convergent if p > 1 and divergent if p ≤ 1 1 x

Proof

I We’ve veriﬁed this for p = 1 above. If p 6= 1 1−p ˛t “ 1−p ” R ∞ 1 R t 1 x ˛ t 1 I 1 xp dx = limt→∞ 1 xp dx = limt→∞ 1−p = limt→∞ 1−p − 1−p ˛1

Annette Pilkington Improper Integrals “ t1−p 1 ” t1−p I If p < 1, limt→∞ 1−p − 1−p does not exist since 1−p → ∞ as t → ∞ and the integral diverges.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Theorem

Z ∞ 1 p dx is convergent if p > 1 and divergent if p ≤ 1 1 x

Proof

I We’ve veriﬁed this for p = 1 above. If p 6= 1 1−p ˛t “ 1−p ” R ∞ 1 R t 1 x ˛ t 1 I 1 xp dx = limt→∞ 1 xp dx = limt→∞ 1−p = limt→∞ 1−p − 1−p ˛1 “ t1−p 1 ” 1 I If p > 1, limt→∞ 1−p − 1−p = − 1−p and the integral converges.

Annette Pilkington Improper Integrals Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Infinite Intervals

Theorem

Z ∞ 1 p dx is convergent if p > 1 and divergent if p ≤ 1 1 x

Proof

“ t1−p 1 ” t1−p I If p < 1, limt→∞ 1−p − 1−p does not exist since 1−p → ∞ as t → ∞ and the integral diverges.

Improper integrals of Type 2 (a) If f is continuous on [a, b) and is discontinuous at b, then

Z b Z t f (x)dx = lim f (x)dx a t→b− a if that limit exists and is ﬁnite. (b) If f is continuous on (a, b] and is discontinuous at a, then

Z b Z b f (x)dx = lim f (x)dx a t→a+ t if that limit exists and is ﬁnite. R b The improper integral a f (x)dx is called convergent if the corresponding limit exists and Divergent if the limit does not exist. R c (c) If f has a discontinuity at c, where a < c < b, and both a f (x)dx and R b c f (x)dx are convergent, then we deﬁne

Z b Z c Z b f (x)dx = f (x)dx + f (x)dx a a c

Annette Pilkington Improper Integrals 1 I The function f (x) = x−2 is continuous on [0, 2) and is discontinuous at 2. Therefore, we can calculate the integral. ˛t R 2 1 R t 1 ˛ I 0 x−2 dx = limt→2− 0 x−2 dx = limt→2− ln |x − 2| ˛0

I = limt→2− (ln |t − 2| − ln | − 2|) which does not exist since ln |t − 2| → −∞ as t → 2−. R 2 1 I Therefore the improper integral 0 x−2 dx diverges.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Functions with infinite discontinuities

If f is continuous on [a, b) and is discontinuous at b, then

Z b Z t f (x)dx = lim f (x)dx a t→b− a if that limit exists and is ﬁnite.

Example Determine whether the following integral converges or diverges

Z 2 1 dx 0 x − 2

Annette Pilkington Improper Integrals ˛t R 2 1 R t 1 ˛ I 0 x−2 dx = limt→2− 0 x−2 dx = limt→2− ln |x − 2| ˛0

I = limt→2− (ln |t − 2| − ln | − 2|) which does not exist since ln |t − 2| → −∞ as t → 2−. R 2 1 I Therefore the improper integral 0 x−2 dx diverges.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Functions with infinite discontinuities

If f is continuous on [a, b) and is discontinuous at b, then

Z b Z t f (x)dx = lim f (x)dx a t→b− a if that limit exists and is ﬁnite.

Example Determine whether the following integral converges or diverges

Z 2 1 dx 0 x − 2

1 I The function f (x) = x−2 is continuous on [0, 2) and is discontinuous at 2. Therefore, we can calculate the integral.

Annette Pilkington Improper Integrals I = limt→2− (ln |t − 2| − ln | − 2|) which does not exist since ln |t − 2| → −∞ as t → 2−. R 2 1 I Therefore the improper integral 0 x−2 dx diverges.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Functions with infinite discontinuities

If f is continuous on [a, b) and is discontinuous at b, then

Z b Z t f (x)dx = lim f (x)dx a t→b− a if that limit exists and is ﬁnite.

Example Determine whether the following integral converges or diverges

Z 2 1 dx 0 x − 2

1 I The function f (x) = x−2 is continuous on [0, 2) and is discontinuous at 2. Therefore, we can calculate the integral. ˛t R 2 1 R t 1 ˛ I 0 x−2 dx = limt→2− 0 x−2 dx = limt→2− ln |x − 2| ˛0

Annette Pilkington Improper Integrals R 2 1 I Therefore the improper integral 0 x−2 dx diverges.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Functions with infinite discontinuities

If f is continuous on [a, b) and is discontinuous at b, then

Z b Z t f (x)dx = lim f (x)dx a t→b− a if that limit exists and is ﬁnite.

Example Determine whether the following integral converges or diverges

Z 2 1 dx 0 x − 2

I = limt→2− (ln |t − 2| − ln | − 2|) which does not exist since ln |t − 2| → −∞ as t → 2−.

If f is continuous on [a, b) and is discontinuous at b, then

Z b Z t f (x)dx = lim f (x)dx a t→b− a if that limit exists and is ﬁnite.

Example Determine whether the following integral converges or diverges

Z 2 1 dx 0 x − 2

I = limt→2− (ln |t − 2| − ln | − 2|) which does not exist since ln |t − 2| → −∞ as t → 2−. R 2 1 I Therefore the improper integral 0 x−2 dx diverges.

Annette Pilkington Improper Integrals 1 I The function f (x) = x2 is continuous on (0, 1] and is discontinuous at 0. Therefore, we can calculate the integral. 1 R 1 1 R 1 1 −1 ˛ + + ˛ I 0 x2 dx = limt→0 t x2 dx = limt→0 x ˛t (−1) (1) I = limt→0+ (−1 − t ) == limt→0+ ( t − 1) which does not exist since 1 + t → ∞ as t → 0 . R 1 1 I Therefore the improper integral 0 x2 dx diverges.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Functions with infinite discontinuities

If f is continuous on (a, b] and is discontinuous at a, then

Z b Z b f (x)dx = lim f (x)dx a t→a+ t if that limit exists and is ﬁnite. Example Determine whether the following integral converges or diverges

Z 1 1 dx 0 x2

Annette Pilkington Improper Integrals 1 R 1 1 R 1 1 −1 ˛ + + ˛ I 0 x2 dx = limt→0 t x2 dx = limt→0 x ˛t (−1) (1) I = limt→0+ (−1 − t ) == limt→0+ ( t − 1) which does not exist since 1 + t → ∞ as t → 0 . R 1 1 I Therefore the improper integral 0 x2 dx diverges.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Functions with infinite discontinuities

If f is continuous on (a, b] and is discontinuous at a, then

Z b Z b f (x)dx = lim f (x)dx a t→a+ t if that limit exists and is ﬁnite. Example Determine whether the following integral converges or diverges

Z 1 1 dx 0 x2

1 I The function f (x) = x2 is continuous on (0, 1] and is discontinuous at 0. Therefore, we can calculate the integral.

Annette Pilkington Improper Integrals (−1) (1) I = limt→0+ (−1 − t ) == limt→0+ ( t − 1) which does not exist since 1 + t → ∞ as t → 0 . R 1 1 I Therefore the improper integral 0 x2 dx diverges.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Functions with infinite discontinuities

If f is continuous on (a, b] and is discontinuous at a, then

Z b Z b f (x)dx = lim f (x)dx a t→a+ t if that limit exists and is ﬁnite. Example Determine whether the following integral converges or diverges

Z 1 1 dx 0 x2

1 I The function f (x) = x2 is continuous on (0, 1] and is discontinuous at 0. Therefore, we can calculate the integral. 1 R 1 1 R 1 1 −1 ˛ + + ˛ I 0 x2 dx = limt→0 t x2 dx = limt→0 x ˛t

Annette Pilkington Improper Integrals R 1 1 I Therefore the improper integral 0 x2 dx diverges.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Functions with infinite discontinuities

If f is continuous on (a, b] and is discontinuous at a, then

Z b Z b f (x)dx = lim f (x)dx a t→a+ t if that limit exists and is ﬁnite. Example Determine whether the following integral converges or diverges

Z 1 1 dx 0 x2

1 I The function f (x) = x2 is continuous on (0, 1] and is discontinuous at 0. Therefore, we can calculate the integral. 1 R 1 1 R 1 1 −1 ˛ + + ˛ I 0 x2 dx = limt→0 t x2 dx = limt→0 x ˛t (−1) (1) I = limt→0+ (−1 − t ) == limt→0+ ( t − 1) which does not exist since 1 + t → ∞ as t → 0 .

If f is continuous on (a, b] and is discontinuous at a, then

Z b Z b f (x)dx = lim f (x)dx a t→a+ t if that limit exists and is ﬁnite. Example Determine whether the following integral converges or diverges

Z 1 1 dx 0 x2

Theorem It is not diﬃcult to show that

Z 1 1 p dx is divergent if p ≥ 1 and convergent if p < 1 0 x

Annette Pilkington Improper Integrals 1 I The function (x−2)2 has a discontinuity at x = 2. Therefore we must R 2 1 R 4 1 check if both improper integrals 0 (x−2)2 dx and 2 (x−2)2 dx converge or diverge. ˛t R 2 1 R t 1 (−1) ˛ I 0 (x−2)2 dx = limt→2− 0 (x−2)2 dx = limt→2− x−2 ˛0 (−1) 1 I = limt→2− ( t−2 − 2 ), which does not exist. R 4 1 R 2 1 R 4 1 I Therefore we can conclude that 0 (x−2)2 dx = 0 (x−2)2 dx + 2 (x−2)2 dx diverges, since this integral converges only if both improper integrals R 2 1 R 4 1 0 (x−2)2 dx and 2 (x−2)2 dx converge.

Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Functions with infinite discontinuities R c If f has a discontinuity at c, where a < c < b, and both a f (x)dx and R b c f (x)dx are convergent, then we define

Z b Z c Z b f (x)dx = f (x)dx + f (x)dx a a c

Example determine if the following integral converges or diverges and if it converges ﬁnd its value. Z 4 1 dx 0 (x − 2)2

Annette Pilkington Improper Integrals ˛t R 2 1 R t 1 (−1) ˛ I 0 (x−2)2 dx = limt→2− 0 (x−2)2 dx = limt→2− x−2 ˛0 (−1) 1 I = limt→2− ( t−2 − 2 ), which does not exist. R 4 1 R 2 1 R 4 1 I Therefore we can conclude that 0 (x−2)2 dx = 0 (x−2)2 dx + 2 (x−2)2 dx diverges, since this integral converges only if both improper integrals R 2 1 R 4 1 0 (x−2)2 dx and 2 (x−2)2 dx converge.

Z b Z c Z b f (x)dx = f (x)dx + f (x)dx a a c

Example determine if the following integral converges or diverges and if it converges ﬁnd its value. Z 4 1 dx 0 (x − 2)2

1 I The function (x−2)2 has a discontinuity at x = 2. Therefore we must R 2 1 R 4 1 check if both improper integrals 0 (x−2)2 dx and 2 (x−2)2 dx converge or diverge.

Annette Pilkington Improper Integrals (−1) 1 I = limt→2− ( t−2 − 2 ), which does not exist. R 4 1 R 2 1 R 4 1 I Therefore we can conclude that 0 (x−2)2 dx = 0 (x−2)2 dx + 2 (x−2)2 dx diverges, since this integral converges only if both improper integrals R 2 1 R 4 1 0 (x−2)2 dx and 2 (x−2)2 dx converge.

Z b Z c Z b f (x)dx = f (x)dx + f (x)dx a a c

Example determine if the following integral converges or diverges and if it converges ﬁnd its value. Z 4 1 dx 0 (x − 2)2

Annette Pilkington Improper Integrals R 4 1 R 2 1 R 4 1 I Therefore we can conclude that 0 (x−2)2 dx = 0 (x−2)2 dx + 2 (x−2)2 dx diverges, since this integral converges only if both improper integrals R 2 1 R 4 1 0 (x−2)2 dx and 2 (x−2)2 dx converge.

Z b Z c Z b f (x)dx = f (x)dx + f (x)dx a a c

Example determine if the following integral converges or diverges and if it converges ﬁnd its value. Z 4 1 dx 0 (x − 2)2

Annette Pilkington Improper Integrals Improper Integrals Infinite Intervals Area Interpretation Theorem 1 Functions with infinite discontinuities Comparison Test Comparison Test Functions with infinite discontinuities R c If f has a discontinuity at c, where a < c < b, and both a f (x)dx and R b c f (x)dx are convergent, then we define

Z b Z c Z b f (x)dx = f (x)dx + f (x)dx a a c

Example determine if the following integral converges or diverges and if it converges ﬁnd its value. Z 4 1 dx 0 (x − 2)2

1 I The function (x−2)2 has a discontinuity at x = 2. Therefore we must R 2 1 R 4 1 check if both improper integrals 0 (x−2)2 dx and 2 (x−2)2 dx converge or diverge. ˛t R 2 1 R t 1 (−1) ˛ I 0 (x−2)2 dx = limt→2− 0 (x−2)2 dx = limt→2− x−2 ˛0 (−1) 1 I = limt→2− ( t−2 − 2 ), which does not exist. R 4 1 R 2 1 R 4 1 I Therefore we can conclude that 0 (x−2)2 dx = 0 (x−2)2 dx + 2 (x−2)2 dx diverges, since this integral converges only if both improper integrals R 2 1 R 4 1 0 (x−2)2 dx and 2 (x−2)2 dx converge.

Annette Pilkington Improper Integrals We have I 1 1 ≤ if x > 1. x2 + x + 1 x2 1 1 I Therefore using f (x) = x2 and g(x) = x2+x+1 in the comparison test above, we can conclude that Z ∞ 1 dx converges 1 x2 + x + 1 since Z ∞ 1 dx converges 1 x2

Improper Integrals Inﬁnite Intervals Area Interpretation Theorem 1 Functions with inﬁnite discontinuities Comparison Test Comparison Test Comparison Test for Integrals

Comparison Test for Integrals Theorem If f and g are continuous functions with f (x) ≥ g(x) ≥ 0 for x ≥ a, then R ∞ R ∞ (a) If a f (x)dx is convergent, then a g(x)dx is convergent. R ∞ R ∞ (b) If a g(x)dx is divergent, then a f (x)dx is divergent. Example Use the comparison test to determine if the following integral is convergent or divergent (using your knowledge of integrals previously calculated). Z ∞ 1 dx 1 x2 + x + 1

Annette Pilkington Improper Integrals 1 1 I Therefore using f (x) = x2 and g(x) = x2+x+1 in the comparison test above, we can conclude that Z ∞ 1 dx converges 1 x2 + x + 1 since Z ∞ 1 dx converges 1 x2

Improper Integrals Inﬁnite Intervals Area Interpretation Theorem 1 Functions with inﬁnite discontinuities Comparison Test Comparison Test Comparison Test for Integrals

We have I 1 1 ≤ if x > 1. x2 + x + 1 x2

We have I 1 1 ≤ if x > 1. x2 + x + 1 x2 1 1 I Therefore using f (x) = x2 and g(x) = x2+x+1 in the comparison test above, we can conclude that Z ∞ 1 dx converges 1 x2 + x + 1 since Z ∞ 1 dx converges 1 x2

Annette Pilkington Improper Integrals We have I 1 1 ≥ if x > 1 x − 1 x 2 1 1 I therfore using f (x) = 1 and g(x) = x in the comparison test, we have x− 2 Z ∞ 1 dx diverges 1 x − 1 2 since Z ∞ 1 dx diverges. 1 x

Improper Integrals Inﬁnite Intervals Area Interpretation Theorem 1 Functions with inﬁnite discontinuities Comparison Test Comparison Test Comparison Test for Integrals

Annette Pilkington Improper Integrals 1 1 I therfore using f (x) = 1 and g(x) = x in the comparison test, we have x− 2 Z ∞ 1 dx diverges 1 x − 1 2 since Z ∞ 1 dx diverges. 1 x

Improper Integrals Inﬁnite Intervals Area Interpretation Theorem 1 Functions with inﬁnite discontinuities Comparison Test Comparison Test Comparison Test for Integrals

Annette Pilkington Improper Integrals