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Improper Integrals: Convergence and Divergence, Cheat Sheet of Differential and Integral Calculus

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Typology: Cheat Sheet

2022/2023

Uploaded on 03/16/2023

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Improper Integrals
Let f: (a, b]โ†’Runbounded at x=a(x=acritical point) (f: [a, b)โ†’
Runbounded at x=b,x=bcritical point) is a function integrable on
[a+๎˜, b] ([a, b โˆ’๎˜]) for all ๎˜ > o. We say that integral Rb
af(x)dx converges
if lim๎˜โ†’0Rb
a+๎˜f(x)dx (lim๎˜โ†’0Rbโˆ’๎˜
af(x)dx) exist.
Examples: Determine if the following integral is convergent or divergent.
1. R1
0
1
โˆšxdx.
Critical point is x= 0 ( 1
โˆšxis not bounded at x= 0.
lim
๎˜โ†’0Z1
0+๎˜
1
โˆšxdx = lim
๎˜โ†’0[2(โˆšx|1
๎˜)] = 2 lim
๎˜โ†’0(โˆš1โˆ’โˆš๎˜)=2.
Thus integral R1
0
1
โˆšxdx converges.
2. R1
0
1
x2dx.
Critical point is x= 0 ( 1
x2is not bounded at x= 0.
lim
๎˜โ†’0Z1
๎˜
1
x2dx =โˆ’lim
๎˜โ†’0(1
x|1
๎˜) = โˆ’lim
๎˜โ†’0(1
1โˆ’1
๎˜) = +โˆž.
Thus integral R1
0
1
x2dx diverges.
3. R2
1
1
xโˆ’2dx.
x= 2 is critical point.
lim๎˜โ†’0R2โˆ’๎˜
1
1
xโˆ’2dx = lim๎˜โ†’0(ln |xโˆ’2||2โˆ’๎˜
1) = lim๎˜โ†’0(ln(2 โˆ’x)|2โˆ’๎˜
1)
= lim๎˜โ†’0(ln(2 โˆ’(2 โˆ’๎˜)) โˆ’ln(2 โˆ’1)) = lim๎˜โ†’0ln ๎˜=โˆ’โˆž.
Thus integral R2
1
1
xโˆ’2dx diverges.
1
pf3

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Improper Integrals

Let f : (a, b] โ†’ R unbounded at x = a (x = a critical point) (f : [a, b) โ†’

R unbounded at x = b , x = b critical point) is a function integrable on

[a + , b] ([a, b โˆ’ ]) for all  > o. We say that integral

โˆซ (^) b a f^ (x)dx^ converges

if limโ†’ 0

โˆซ (^) b a+ f^ (x)dx^ (limโ†’^0

โˆซ (^) bโˆ’ a f^ (x)dx) exist.

Examples: Determine if the following integral is convergent or divergent.

0

โˆš^1 x

dx.

Critical point is x = 0 ( โˆš^1 x is not bounded at x = 0.

lim โ†’ 0

0+

x

dx = lim โ†’ 0

[2(

x|

1  )] = 2 lim โ†’ 0 (

Thus integral

0

โˆš^1 x dx^ converges.

0

1 x^2 dx. Critical point is x = 0 ( (^) x^12 is not bounded at x = 0.

lim โ†’ 0



x^2

dx = โˆ’ lim โ†’ 0

x

|^1  ) = โˆ’ lim โ†’ 0

Thus integral

0

1 x^2 dx^ diverges.

1

1 xโˆ’ 2 dx. x = 2 is critical point.

limโ†’ 0

1

1 xโˆ’ 2 dx^ = limโ†’^0 (ln^ |x^ โˆ’^2 ||

2 โˆ’ 1 ) = limโ†’^0 (ln(2^ โˆ’^ x)|

2 โˆ’ 1 )

= limโ†’ 0 (ln(2 โˆ’ (2 โˆ’ )) โˆ’ ln(2 โˆ’ 1)) = limโ†’ 0 ln  = โˆ’โˆž.

Thus integral

1

1 xโˆ’ 2 dx^ diverges.

1

2

Theorem 1Let f and g be continuous on (a, b) with 0 < f (x) โ‰ค g(x) for

all x โ‰ฅ a. Then

  1. If

โˆซ (^) b a g(x)dx^ converges, then^

โˆซ (^) b a f^ (x)dx^ converges

  1. If

โˆซ (^) b a f^ (x)dx^ diverges, then^

โˆซ (^) b a g(x)dx^ diverges

Theorem 2 Let positive functions f and g are continuous on (a, b] and

lim xโ†’a

f (x)

g(x)

= L, 0 < L < +โˆž.

Then

โˆซ (^) b a f^ (x)dx^ and^

โˆซ (^) b a g(x)dx^ both converges or both diverges. Theorem 2โ€™ Let positive functions f and g are continuous on [a, b) and

lim xโ†’b

f (x)

g(x)

= L, 0 < L < +โˆž.

Then

โˆซ (^) b a f^ (x)dx^ and^

โˆซ (^) b a g(x)dx^ both converges or both diverges.

Integral โˆซ (^) b

a

(x โˆ’ a)p^

dx,

converges for p < 1 and diverges for p โ‰ฅ 1.

Integral โˆซ (^) b

a

(b โˆ’ x)p^

dx,

converges for p < 1 and diverges for p โ‰ฅ 1.

Examples: Determine if the following integral is convergent or divergent.

0

x+ x

โˆš x+

dx. Critical point is x = 0.

0 < f (x) =

x + 1

x

x + 2

x

x + 1 โˆš x + 2