Improper Integrals: Type I and Type II, Study notes of Calculus

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8.8 Improper integrals

Recall that in the study of definite integral ∫ (^) b

a

f (x) dx,

we assume that

• the interval [a, b] is finite
• the function f is bounded. Such integrals are said to be proper.

Improper integrals

• Type I: the interval is infinite: [a, ∞), (−∞, b] or (−∞, ∞).
• Type II: the function f is unbounded. Idea: improper integrals are calculated as limits. Type I: Integral over infinite intervals.

1. If f (x) is continuous on [a, ∞), then ∫ (^) ∞

a

f (x) dx = lim b→∞

∫ (^) b

a

f (x) dx.

2. If f (x) is continuous on (−∞, b], then ∫ (^) b

−∞

f (x) dx = lim a→−∞

∫ (^) b

a

f (x) dx.

3. If f (x) is continuous on (−∞, ∞), then ∫ (^) ∞

−∞

f (x) dx =

∫ (^) a

−∞

f (x) dx +

a

f (x) dx

where a is any real number. In each case, if the limit is finite, we say the improper integral converges and that the limit is the value of the improper integral. If the limit fails to exist, the improper integral diverges. Note that

−∞ f^ (x)^ dx^ converges if and only if both^

∫ (^) a

∫ (^) ∞^ −∞^ f^ (x)^ dx^ and a f^ (x)^ dx^ converge. Example: Evaluate (^) ∫ ∞

0

dx e^3 x

If p 6 = 1, (^) ∫ b

1

xp

dx =

x^1 −p 1 − p

]xx==1b =

b^1 −p^ − 1 1 − p Thus, ∫ (^) ∞

1

xp

dx = lim b→∞

∫ (^) b

1

xp

dx = lim b→∞

b^1 −p^ − 1 1 − p

=

1 p− 1 , p >^1 +∞, p < 1

If p = 1, then ∫ (^) ∞

1

dx x

= lim b→∞

∫ (^) b

1

dx x

= lim b→∞

(ln b − ln 1) = ∞.

Therefore, (^) ∫ (^) ∞

1

xp

dx

converges for p > 1 and diverges for p ≤ 1.

Direct Comparison Test Let f and g be continuous on [a, +∞) with 0 ≤ f (x) ≤ g (x)

for all x ≥ a. Then,

a f^ (x)^ dx^ converges if^

a g^ (x)^ dx^ converges;

a g^ (x)^ dx^ diverges if^

a f^ (x)^ dx^ diverges;

Remark: Usually, compare with the important example (^) x^1 p for some p. Example: (^) ∫ ∞

2

dx x

1 + x^2

So, compare with (^) x^12. Since for all x ≥ 2 , 0 ≤

x

1 + x^2

x^2

and

2

1 x^2 dx^ converges, we have ∫ ∞

2

dx x

1 + x^2 also converges.

So, compare with (^) x^12. Since for all x ≥ 2 , 0 ≤

x

1 + x^2

x^2

and

2

1 x^2 dx^ converges, we have ∫ ∞

2

dx x

1 + x^2 also converges. How about (^) ∫ (^) ∞

2

dx x

x^2 − 1

Limit Comparison Test Suppose f and g are two nonnegative functions on [a, ∞), and if

lim x→∞

f (x) g (x)

= L, 0 < L < ∞.

Then, (^) ∫ (^) ∞

a

f (x) dx and

a

g (x) dx

both converge or both diverge.

Limit Comparison Test Suppose f and g are two nonnegative functions on [a, ∞), and if

lim x→∞

f (x) g (x)

= L, 0 < L < ∞.

Then, (^) ∫ (^) ∞

a

f (x) dx and

a

g (x) dx

both converge or both diverge. Example: Does the integral ∫ (^) ∞

2

dx x

x^2 − 1 converge? Solution: Let f (x) =

x

x^2 − 1

and g (x) =

x^2

then

x^ lim→∞

f (x) g (x)

= (^) xlim→∞

x

x^2 − 1

· x^2 = (^) xlim→∞

x √ x^2 − 1

So, (^) ∫ (^) ∞

2

dx x

x^2 − 1

converges because

2

1 x^2 dx^ converges. Example:Does the integral ∫ (^) ∞

0

5 + sin

tan^2

x^3

− 4 ex+log

x 5

e^3 x^

dx

converge?

then

x^ lim→∞

f (x) g (x)

= (^) xlim→∞

x

x^2 − 1

· x^2 = (^) xlim→∞

x √ x^2 − 1

So, (^) ∫ (^) ∞

2

dx x

x^2 − 1

converges because

2

1 x^2 dx^ converges. Example:Does the integral ∫ (^) ∞

0

5 + sin

tan^2

x^3

− 4 ex+log

x 5

e^3 x^

dx

converge? Note that

0 ≤

5 + sin

tan^2

x^3

− 4 ex+log

x 5

e^3 x^

e^3 x

Type II Improper integral: integrals of unbounded function.

Important Example: Evaluate ∫ (^1)

0

xp

dx

For p ≤ 0 , the integral

0

1 xpdx^ is a proper integral.^ For^ p >^0 , it is an improper integral. In this case, we define ∫ (^1)

0

xp

dx = lim a→0+

a

xp

dx

For p 6 = 1, ∫ (^1)

a

xp

dx =

x^1 −p 1 − p

|^1 a

1 − a^1 −p 1 − p

→

∞ if p > 1 1 1 −p if^ p <^1