Math 104: Improper Integrals (With Solutions), Lecture notes of Calculus

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Math 104: Improper Integrals (With Solutions)

Ryan Blair

University of Pennsylvania

Tuesday March 12, 2013

Outline

1 Improper Integrals

Improper integrals

Definite integrals

∫ (^) b

a

f (x)dx were required to have

finite domain of integration [a, b] finite integrand f (x) < ±∞

Improper integrals (^1) Infinite limits of integration (^2) Integrals with vertical asymptotes i.e. with infinite discontinuity

Infinite limits of integration

Definition Improper integrals are said to be convergent if the limit is finite and that limit is the value of the improper integral. divergent if the limit does not exist.

Example 1

Find (^) ∫ (^) ∞

0

e−x^ dx.

(if it even converges)

Solution: ∫ (^) ∞

0

e−x^ dx = lim b→∞

∫ (^) b

0

e−x^ dx = lim b→∞

[

− e−x

]b 0 = lim b→∞

−e−b^ + e^0 = 0 + 1= 1.

So the integral converges and equals 1.

Example 2

Find (^) ∫ (^) ∞

−∞

1 + x^2

dx.

(if it even converges)

Solution: By definition, ∫ (^) ∞

−∞

1 + x^2

dx =

∫ (^) c

−∞

1 + x^2

dx +

c

1 + x^2

dx,

where we get to pick whatever c we want. Let’s pick c = 0. ∫ (^0)

−∞

1 + x^2

dx = lim b→−∞

[

arctan(x)

] 0

b

= lim b→−∞ [arctan(0) − arctan(b)]

= 0 − lim b→−∞ arctan(b) =

π 2

Example 3, the p-test

The integral (^) ∫ ∞

1

xp^

dx

(^1) Converges if p > 1; (^2) Diverges if p ≤ 1. For example: ∫ (^) ∞

1

x^3 /^2

dx = lim b→∞

[ 2

x^1 /^2

]b 1

while (^) ∫ ∞

1

x^1 /^2

dx = lim b→∞

[

x

]b

1

= lim b→∞

b − 2 = ∞,

and (^) ∫ (^) ∞

1

x

dx = lim b→∞

[

ln(x)

]b 1

= lim b→∞

ln(b) − 0 = ∞.

Convergence vs. Divergence

In each case, if the limit exists (or if both limits exist, in case 3!), we say the improper integral converges.

If the limit fails to exist or is infinite, the integral diverges. In case 3, if either limit fails to exist or is infinite, the integral diverges.

Example 5

Find

0

(x − 1)^2 /^3

dx, if it converges.

Solution:

Example 5

Find

0

(x − 1)^2 /^3

dx, if it converges.

Solution: We might think just to do ∫ (^3)

0

(x − 1)^2 /^3

dx =

[

3(x − 1)^1 /^3

] 3

0

Example 5

Find

0

(x − 1)^2 /^3

dx, if it converges.

Solution: We might think just to do ∫ (^3)

0

(x − 1)^2 /^3

dx =

[

3(x − 1)^1 /^3

] 3

0

but this is not okay: The function f (x) = (^) (x−^1 1) 2 / 3 is undefined when x = 1, so we need to split the problem into two integrals. ∫ (^3)

0

(x − 1)^2 /^3

dx =

0

(x − 1)^2 /^3

dx +

1

(x − 1)^2 /^3

dx.

The two integrals on the right hand side both converge and add up to 3[1 + 2^1 /^3 ], so

0

1 (x−1)^2 /^3 dx^ = 3[1 + 2

1 / 3 ].

Tests for convergence and divergence

The gist: (^1) If you’re smaller than something that converges, then you converge. (^2) If you’re bigger than something that diverges, then you diverge.

Theorem Let f and g be continuous on [a, ∞) with 0 ≤ f (x) ≤ g (x) for all x ≥ a. Then 1

a f^ (x)^ dx converges if^

a g^ (x)^ dx converges. 2

a g^ (x)^ dx diverges if^

a f^ (x)^ dx diverges.

Limit Comparison Test

Theorem If positive functions f and g are continuous on [a, ∞) and

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 7: Let f (x) = √x^1 +1 ; consider ∫ (^) ∞

1

x + 1

dx.

Does the integral converge or diverge?

Example 7, continued

Solution: We note that f looks a lot like g (x) = √^1 x , and ∫ (^) ∞ 1 g^ (x)^ dx^ diverges by the^ p-test. Furthermore,

lim x→∞

f (x) g (x)

x √ x + 1

so the LCT says

1 √^1 x+1 dx^ diverges.