**Unformatted text preview: **Lecture 14: Improper Integrals
Math 31B: Integration and infinite series Baskar Balasubramanyam
October 30, 2017 Unbounded regions We have so far looked at definite integrals of the form
Z b f (x) dx
a for numbers a and b. These are integrals of a function f (x) over
the interval [a, b]. This value is for example the signed area under
the curve f (x) over the interval [a, b].
This is a bounded region. What do we do if the region we are
interested in is unbounded? We need to consider different type of
integrals than the one above. 1 1
0.8
0.6
0.4
0.2
0 1 1.5 2 2.5 3 3.5 4 Figure 1: y = 1/x 2 Improper integrals There are two types of improper integrals:
1. This type of integral is where the interval over which we are
integrating is infinite. For example, integrals of the form
Z ∞
Z a
Z ∞
f (x) dx,
f (x) dx,
or
f (x) dx.
a ∞ −∞ 2. This is the type where the function we are integrating could go
to infinity. For example,
Z 2
Z 1
dx
or
ln(x) dx.
2
−1 x
0 3 Area of an unbounded region
We also need to make sense of area for an unbounded region. If we
had a bounded region (like in the figure below), a R RR
we can compute this as a f (x) dx. Now if we let R increase and
go to infinity, the region becomes unbounded. A reasonable
definition of area for this unbounded region is
Z
lim R→∞ a R f (x) dx. 4 R∞
a f (x) dx
We define
Z ∞ Z
f (x) dx = lim R→∞ a a R f (x) dx. Example 1
R∞
Find 0 e −x dx.
Take R > 0 and
Z R R e −x dx = −e −x = −e −R + e 0 = 1 − e −R .
0 0 So
Z ∞ e
0 −x Z
dx = lim R→∞ 0 R e −x dx = lim 1 − e −R = 1.
R→∞ 5 Convergence or divergence of improper integral We also define
Z a Z
f (x) dx = −∞ lim a R→−∞ R f (x) dx. • If the limit in either of the definitions exists, we say that the
improper integral converges.
• If the limit in either of the definitions does not exist, we say
that the improper integral diverges.
Example 2
Does the integral R∞
3 dx
x4 converge or diverge? 6 Divergent integral
Example 3
Does the integral R −2 dx
−∞ x converge or diverge? Solution.
Take any R < −2. Then
Z −2 −2
dx = ln |x| = ln(2) − ln |R|.
x
R
R
So
Z −2 −∞ dx
= lim ln(2) − ln |R| = Does Not Converge.
R→−∞
x 7 Intervals going to infinity in both directions Now we look at integrals of the form
Z ∞
f (x) dx
−∞ where the interval goes to infinity in both directions. Define
Z ∞
Z a
Z ∞
f (x) dx =
f (x) dx +
f (x) dx
−∞ −∞ a for any a.
Example 4
R ∞ dx
Find −∞ 1+x
2. 8 Solution.
We take a = 0. So, we need to find
Z 0
dx
and
2
−∞ 1 + x Z
0 ∞ dx
.
1 + x2 For R1 < 0 and R2 > 0, we see that
Z R2
Z 0 0 R
dx
dx
−1 −1 2
,
and
=
tan
x
=
tan
x .
2
1 + x2
R1
0
0
R1 1 + x
So
Z 0 dx
π
= lim − tan−1 R1 =
2
R
→−∞
1
+
x
2
1
−∞
Z ∞
dx
π
= lim tan−1 R2 = .
2
R2 →∞
1+x
2
0 So R∞ dx
−∞ 1+x 2 = π.
9 p-integrals over [a, ∞) For which p does the integral R∞
a dx
xp converge? We already say that such an integral would converge when p = 4
and diverge when p = 1.
In fact, the integral
Z
a ∞ dx
=
xp ( a1−p
p−1 if p > 1,
diverges if p ≤ 1. 10 Using L’Hopital’s Rule
It is sometimes necessary to use L’Hopital’s rule to compute the
limit which gives the improper integral.
Example 5
R∞
Calculate 0 xe −x dx.
Solution.
First by integration by parts, we check that
Z
xe −x = −(x + 1)e −x + C .
So, RR
0 xe −x = 1 −
Z
0 R+1
.
eR ∞ Taking limits, we get
R +1
= 1.
R→∞ e R xe −x = 1 − lim 11 Unbounded functions
We now move on to the 2nd type of improper integral. For
R3
−1
example, 0 dx
x . This is improper because the function f (x) = x
goes to infinity as x approached 0.
So, we define
Z
0 Since
Z 3 R we have
Z
0 3 3 dx
= lim+
x
R→0 Z 3 R dx
.
x 3
dx = ln(x) = ln(3) − ln(R),
x
R dx
= ln(3) − lim+ ln(R) = Diverges
x
R→0 12 Improper integrals of type 2 These are integrals of the type
following holds Rb
a f (x) dx, where one of the • f is continuous on [a, b) and f goes to ±∞ as x approaches b.
• f is continuous on (a, b] and f goes to ±∞ as x approaches a.
• f is continuous on [a, b] except for a discontinuity at c in
between a and b, and f goes to ±∞ as x approaches c.
In the first two cases define
Z b
Z R
Z
f (x) dx = lim
f (x) dx and
a R→b − a a b Z
f (x) dx = lim+
R→a b f (x) dx.
R 13 Example
R 2 dx
Calculate 0 (x−1)
2/3 . Here the function has a discontinuity in the
middle at x = 1. So, we write
Z 2
Z 1
Z 2
dx
dx
dx
=
+
.
2/3
2/3
2/3
0 (x − 1)
0 (x − 1)
1 (x − 1)
Then
Z
0 1 dx
= lim
(x − 1)2/3 R1 →1− Z R1 (x − 1)−2/3 dx 0 R1 = lim 3(x − 1)1/3 R1 →1− 0 = lim 3(R1 − 1)1/3 − 3(−1)1/3
R1 →1− =3
14 Similarly
Z
1 2 dx
= lim
(x − 1)2/3 R2 →1+ Z 2 (x − 1)−2/3 dx R2 2
1/3 = lim + 3(x − 1)
R2 →1 = 3(1) 1/3 R2 − lim + 3(R2 − 1)1/3
R2 →1 =3
Putting everything together, we get
Z 2
dx
= 3 + 3 = 6.
2/3
0 (x − 1) 15 p-integrals for (0, a] When do the integrals Ra dx
0 xp Infact
Z
0 a dx
=
xp ( converge?
a1−p
1−p if p < 1,
diverges if p ≥ 1. 16 Tests for convergence Sometimes it is difficult to compute the exact value of an improper
integral to see if it converges. The following comparison tests help
us to sometimes decide if an improper integral converges or diverges
R∞
• If 0 ≤ f (x) ≤ g (x) and a g (x) dx converges, then
R∞
a f (x) dx converges.
R∞
R∞
• If 0 ≤ f (x) ≤ g (x) and a f (x) dx diverges, then a f (x) dx
diverges. 17 Example
Example 6
Does the integral R∞
1 2 e −x converge? RR
2
We cannot compute integrals of the form 1 e −x . So, we cannot
directly conclude if the improper integral converges or not.
We want to use the comparison test now. For x ≥ 1, notice that
x 2 ≥ x and we have
2
0 ≤ e −x ≤ e −x .
We can check that
Z ∞
Z
−x
e = lim
1 So R∞
1 R→∞ 1 R R e −x = lim −e −x = e −1 .
R→∞ 1 2 e −x dx converges.
18 ...

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