Improper Integrals - Lecture Notes | MA 114, Study notes of Mathematics

Material Type: Notes; Class: CALCULUS II; Subject: Mathematics; University: University of Kentucky; Term: Spring 2004;

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Pre 2010

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1 Lecture 17: Improper integrals
1. Consider improper integrals at infinity.
2. Find areas as improper integrals.
3. The comparison theorem.
A question.
Can an region have infinite perimeter and finite area? If so, we can color it in, but
we cannot draw its outline.
Example. Find the area bounded by x= 1, y= 0 and y=ex.
Solution. This is a new problem–until now, we have found the area of regions which
were of finite length in every direction.
We attempt to solve this problem, by first finding the area bounded by x= 1,
y= 0, y=exand x=N. This is
ZN
1
exdx =1
eeN.
If we take the limit as Ngoes to infinity, we expect to obtain a reasonable value for
the infinite area.
area = lim
N→∞
1
eeN=1
e.
Definition of improper integral.
Definition. If ais a real number and fis a function which is Riemann integrable on
each interval [a, N ] for N > a, then we define the improper integral as
Z
a
f(t)dt = lim
N→∞ ZN
a
f(t)dt.
If this limit exists and is finite, we say the improper integral is convergent. If the
limit is infinite or does not exist, we say the improper integral is divergent.
Exercise. Write out the corresponding definition for improper integrals at −∞.
Exercise. Re-read the definition of the Riemann integral Rb
af(t)dt and explain what
difficulties would arise if we let b=in this definition.
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1 Lecture 17: Improper integrals

  1. Consider improper integrals at infinity.
  2. Find areas as improper integrals.
  3. The comparison theorem.

A question.

Can an region have infinite perimeter and finite area? If so, we can color it in, but we cannot draw its outline.

Example. Find the area bounded by x = 1, y = 0 and y = e−x.

Solution. This is a new problem–until now, we have found the area of regions which were of finite length in every direction. We attempt to solve this problem, by first finding the area bounded by x = 1, y = 0, y = e−x^ and x = N. This is

∫ (^) N

1

e−x^ dx =

e

− e−N^.

If we take the limit as N goes to infinity, we expect to obtain a reasonable value for the infinite area.

area = lim N →∞

e

− e−N^ =

e

Definition of improper integral.

Definition. If a is a real number and f is a function which is Riemann integrable on each interval [a, N ] for N > a, then we define the improper integral as

∫ (^) ∞

a

f (t) dt = lim N →∞

∫ (^) N

a

f (t) dt.

If this limit exists and is finite, we say the improper integral is convergent. If the limit is infinite or does not exist, we say the improper integral is divergent.

Exercise. Write out the corresponding definition for improper integrals at −∞.

Exercise. Re-read the definition of the Riemann integral

∫ (^) b a f^ (t)^ dt^ and explain what difficulties would arise if we let b = ∞ in this definition.

Example. How should we treat the integral ∫ (^) ∞

−∞

x dx?

Solution. If we break the integral up as follows:

∫ (^) ∞

−∞

x dx =

∫ (^0)

−∞

x dx +

∫ (^) ∞

0

x dx,

then each of the integrals on the right is divergent.

Example. Evaluate the following improper integrals, or state if the integral diverges.

∫ (^0)

−∞

e^2 x^ dx

∫ (^0)

−∞

e−^2 x^ dx

∫ (^) ∞

1

1 + x^2

dx.

Example. Determine for which p, the integral

∫ (^) ∞

1

xp^

dx

converges.

Solution. If p 6 = 1, then

∫ (^) ∞

1

xp^

dx = lim N →∞

∫ (^) N

1

x−p^ dx = lim N →∞

x−p+ −p + 1

∣∣ ∣∣ ∣

x=N

x=

= lim N →∞

N −p+ −p + 1

−p + 1

This limit will exist and be finite if the exponent of N is negative. That is the integral converges if p > 1. The limit will be infinite and the integral diverges if p < 1. The case p = 1 is treated separately because it involves the logarithm. If p = 1, then the integral is divergent.