Calculus 1120, Spring 2012 Dan Barbasch October 11, 2012 Dan - - PowerPoint PPT Presentation

Calculus 1120, Spring 2012

Dan Barbasch October 11, 2012

Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 1 / 6

Improper Integrals

Integration via Riemann sums assumes functions are continuous over a bounded interval (or at least piecewise continuous over finitely many bounded intervals). Deviating from this leads to improper integrals. infinite intervals ∞ e−x dx. unbounded functions, vertical asymptotes 1 1

3

√x dx. The functions in the two examples are > 0, so you can think of the integrals as representing areas below the curves and above the x−axis.

Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 2 / 6

Improper Integrals

Type I: y = f (x) continuous on [a, ∞) ∞

a

f (x) dx = lim

A→∞

A

a

f (x) dx. Similarly for a

−∞

f (x) dx. Example: ∞ e−x dx := lim

R→∞

R e−x dx. Type II: y = f (x) continuous on [a, b) then b

a

f (x) dx = lim

A→b−

A

a

f (x) dx. Similarly if y = f (x) is continuous on (a, b], b

a

f (x) dx = lim

A→a+

b

A

f (x) dx. Example: 1 1

3

√x dx := lim

δ→0+

1

δ

1

3

√x dx.

Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 2 / 6

Improper Integrals

Important: If more than one type occurs in the integral, MUST divide into a sum and treat each term separately. If the interval is infinite in both directions, and there are asymptotes, we need to separate into a sum, and examine each of them. ∞

−∞

1

3

√x dx = −a

−∞

1

3

√x dx+

−a

1

3

√x dx + b 1

3

√x dx+ ∞

b

1

3

√x dx. It does not matter what a and b are, except −a < 0 and b > 0. This integral is improper because there are two type I and two type II issues!

Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 2 / 6

1

−2

1 x dx =

• lim

A→0−

A

−2

1 x dx

• +
• lim

B→0+

1

B

1 x dx

• Contrast this with

lim

R→0+

−R

−2

1 x dx + 1

R

1 x dx

• = − ln 2.

−R

−2

1 x dx+ 1

R

1 x dx = ln |x| |−R

−2 + ln |x| |1 R=

=(ln R − ln 2) + (ln 1 − ln R) = − ln 2. This DOES NOT SAY that 1

−2

1 x dx converges and equals − ln 2. The order in which you take the integrals and the limit makes a difference!

Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 3 / 6

lim

R→0+[ln R − ln 1] + lim R→0+[ln 1 − ln R] = −∞ + ∞.

Since each of the limits are infinite separately, 1

−1 1 x dx diverges.

We need to know limA→0+ ln A = −∞ and other limits of this type.

Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 3 / 6

Comparison Tests

Direct Comparison: If 0 ≤ f (x) ≤ g(x) for x ≥ a , then

1

∞

a

g(x) dx converges ⇒ ∞

a

f (x) dx converges.

2

∞

a

f (x) dx diverges ⇒ ∞

a

g(x) dx diverges. LHS = ∞

a

f (x) dx ≤ RHS = ∞

a

g(x) dx. If the RHS is finite, so is the smaller one. If the LHS is infinite, so is the bigger one.

Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 4 / 6

Comparison Tests

Example: ∞ 1 √ 1 + x4 dx. For example,

1 √ 1+x4 ≤ 1 √ 1+x2 .

• 1

√ 1 + x2 dx = 1 2 ln |x +

• 1 + x2| + C from the tables.

A dx √ 1 + x2 = 1 2

• ln |A +
• 1 + A2| − ln |0 +

√ 1|

• −

→ ∞ as A → ∞. ∞ dx √ 1 + x4 = 1 dx √ 1 + x4 + ∞

1

dx √ 1 + x4 ≤ ≤ 1 dx √ 1 + x4 + ∞

1

dx √ 1 + x2 = ∞ DOES NOT GIVE ANY CONCLUSION.

Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 4 / 6

Comparison Tests

We need a better sense of how

1 √ 1+x4 behaves as x → ∞.

1 √ 1 + x4 = 1

• x4(1 + 1/x4)

= 1 x2 1 + 1/x4 = 1 x2 · 1

• 1 + 1/x4

But limx→∞

1

√

1+1/x4 = 1, so

1 √ 1 + x4 ≈ 1 x2 for x → ∞. In fact

1

√

1+1/x4 ≤ 1, so

∞ dx √ 1 + x4 = 1 dx √ 1 + x4 + ∞

1

dx √ 1 + x4 ≤ ≤ 1 dx √ 1 + x4 + ∞

1

dx x2 < ∞. The integral converges. We had to split the intergral into a sum 1

0 +

∞

1

because 1/x2 has a vertical asymptote at x = 0. A

1

dx x2 = −1 x

• A

1

= − 1 A − 1

• −

→ 1 as A → ∞.

Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 4 / 6

Important Special Cases:

1 1

1 xp dx converges for 0 < p < 1 and diverges for p ≥ 1.

2 ∞

1 1 xp dx diverges for 0 < p ≤ 1 and converges for p > 1.

By Direct Computation, see the text. The calculations we just did, are encoded in what is called the Limit Comparison Test.

Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 5 / 6

Limit Comparison Test: 0 ≤ f (x), g(x) for x ≥ a , and lim

x→∞

f (x) g(x) = L, with 0 ≤ L ≤ ∞.

1 0 < L < ∞, (f (x) ≈ g(x))

∞

a

g(x) dx converges ⇔ ∞

a

f (x) dx converges.

2 L = 0, (eventually f (x) << g(x) )

∞

a

g(x) dx converges ⇒ ∞

a

f (x) dx converges ∞

a

f (x) dx diverges ⇒ ∞

a

g(x) dx diverges

3 If L = ∞, (eventually g(x) << f (x) )

∞

a

f (x) dx converges ⇒ ∞

a

g(x) dx converges ∞

a

g(x) dx diverges ⇒ ∞

a

g(x) dx diverges. Case (1) is the more common one. Remark: These tests also work for integrals over finite intervals for functions with vertical asymptotes.

Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 5 / 6

lim f (x)

g(x) = L, 0 < L < ∞

⇔ L2g(x) ≤ f (x) ≤ L1f (x). You can apply the comparison test both ways, f (x) ≤ L2g(x) AND g(x) ≤ f (x)

L1 .

b

a

f (x) dx converges ⇔ b

a

g(x) dx converges . lim f (x)

g(x) = 0 means f (x) << g(x) eventually.

b

a

g(x) dx diverges ⇒ b

a

g(x) dx diverges lim f (x)

g(x) = ∞ means g(x) << f (x) eventually.

b

a

f (x) dx converges ⇒ b

a

g(x) dx converges

Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 5 / 6

f (x) =

1 √ 1+x4 and g(x) = 1 x2 .

f (x) g(x) = 1

• 1 + 1/x4 −

→ 1 as x → ∞. f (x) = e−x2/2 and g(x) = e−x/2. f (x) g(x) = e(−x2+x)/2 − → 0 as x → ∞ because x − x2 → −∞ as x → ∞. (Other) Important Special Cases:

1 1

1 xp dx converges for 0 < p < 1 and diverges for p ≥ 1.

2 ∞

1 1 xp dx diverges for 0 < p ≤ 1 and converges for p > 1.

3 ∞

a

e−x dx converges

4 a

0 ln x dx diverges

5 ∞

2 1 ln x dx diverges.

We write f (x) ≈ g(x) in case (1) of the limit comparison test.

Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 5 / 6

More Examples: (a) ∞ dt t2 + sin t (b) ∞ ex dx e2x − 1, (c) ∞

2

dx (ln x)p .

Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 5 / 6

∞ dt t2 + sin t = 1 dt t2 + sin t + ∞

1

dt t2 + sin t . 1 t2 + sin t = 1 t2 · 1 1 + sin t/t2 ≈ 1 t2 for t → ∞ 1 t2 + sin t = 1 t · 1 t + sin t/t ≈ 1 t for t → 0+ Necessary Facts: limt→0 sin t

t

= 1. | sin t| ≤ 1.

Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 5 / 6

Remark: The solution is incomplete. We need to check that t2 + sin t = 0 for any t > 0. This is true because on the one hand for t > 1, t2 + sin t > 1 + sin t ≥ 0, so no zeroes. On the other hand, for 0 < t ≤ 1, t2 + sin t is increasing by using the 1st derivative test; it has derivative 2t + cos t which is positive because both 2t > 0 and cos t > 0 for 0 < t ≤ 1. Since t2 + sin t equals 0 for t = 0, being increasing, it is > 0 for 0 < t ≤ 1. This is NOT the case for the analogous problem from last semester, ∞ dt t2 − sin t . t2 − sin t = 0 for t = 0, and its derivative is 2t − cos t equal to −1 at t = 0. Thus the function is decreasing so < 0 for t > 0. But at t = 1, the value is 12 − sin 1 > 0. So by the intermediate value theorem t2 − sin t must be = 0 for at least one value 0 < c < 1. BONUS CREDIT is awarded for a complete careful analysis of all the integrals involved in determining whether ∞ dt t2 − sin t converges or not. You should attempt this only after having done all the homework problems which are easier.

Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 5 / 6

∞ ex dx e2x − 1 = 1 ex dx e2x − 1 + ∞

1

ex dx e2x − 1. ex e2x − 1 = ex e2x · 1 1 − e−2x = 1 ex · 1 1 − e−2x ≈ 1 ex for x → ∞

Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 5 / 6

For x → 0+, ex → 1 but e2x − 1 → 0. So

ex e2x−1 → ∞, but how fast?

The integral is improper. lim

x→0

e2x − 1 x = 2. ex e2x − 1 = 1 x · x e2x − 1 · ex ≈ 1 2x as x → 0+.

Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 5 / 6

L’Hospital’s Rule

Useful Techniques:

1 limx→x0

f (x)−f (x0) x−x0

= f ′(x0).

2 If an expression → 0 or → ±∞, force factor out the term which is

responsible for the behaviour. Generalization: Suppose that limx→a f (x) = ±∞ and limx→a g(x) = ±∞, or limx→a f (x) = 0 and limx→a g(x) = 0, and the functions are differentiable. Then lim

x→a

f (x) g(x) = lim

x→a

f ′(x) g′(x) = . . . .

Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 6 / 6