ASSIGNING VALUE TO THE VALUELESS... THE CAUCHY
PRINCIPLE VALUE METHOD AND RELATED IDEAS IN
DIVERGENT SERIES
Jeffrey Paul Wheeler
Department of Mathematics, The University of Pittsburgh, Pittsburgh,
Pennsylvania, 15260, USA
[email protected]
Abstract
We discuss a question thoughtful Calculus 2 students ask regarding the value of a
divergent integral where the integrand is an odd function. We as well address some
techniques regarding divergent series.
1. Giving Value to Valueless Integrals
1.1. Taking Advantage of Symmetry
Often in the preliminary material to Calculus one is introduced to the notion of an
odd function; namely f (x) is odd if for all x in its domain, f (−x) = −f (x). One
also learns that odd functions have the property that their graphs are symmetric
with respect to the origin.
This is particularly useful when integrating. If we are integrating an odd function
over a symmetric interval, we may take advantage of the symmetry of the graph
and realize
Z a
f (x)dx = 0
−a
2010 Mathematics Subject Classification: Primary 11P99; Secondary 05E15, 20D60.
Key words and phrases: Cauchy Principle Value, Divergent Integrals, Divergent Series.
1
where a is any real number and f (x) an odd function.
For example, one of my favorite questions to ask on my first Calc 2 midterm is
Z
π
3
sin5 xdx.
−π
3
Students paying some level of attention recognize that sin x is an odd function, which
when raised to an odd power will result in an odd function. Hence the integral is
0. Less attentive students spend 10 minutes on the integral with a high percentage
of them making a mistake somewhere.
1.2. Improper Integrals
The first theorem a student encounters after learning the definition of the definite
integral is
Theorem 1.
If f is continuous on [a, b] or if f has only a finite number of jump discontinuities
over [a, b], then f is integrable over [a, b].
Hence there are two types of integrals that behave badly:
• integrals over an infinite interval and
• integrals over an interval that includes a vertical asymptote.
R∞
Ra
R∞
The first kind are integrals of the form a f (x)dx, −∞ f (x)dx, or −∞ f (x)dx.
Rb
The second kind are of the form a f (x)dx where there is a discontinuity at c for
a ≤ c ≤ b. Though the discontinuity is a concern, jump discontinuities are easy to
deal with; the real concern here, as mentioned, is when x = c is a vertical asymptote.
Hence our work-around for the first situation is
Definition 2.
1. If
Rt
a
f (x)dx exists for every t ≥ a, then
Z
∞
Z
f (x)dx := lim
t→∞
a
2
t
f (x)dx.
a
2. If
Rb
s
f (x)dx exists for every s ≤ b, then
Z
b
Z
b
f (x)dx.
f (x)dx := lim
s→−∞
−∞
s
R∞
Rb
If for a f (x)dx and −∞ f (x)dx the corresponding limits exist, then the
integrals are said to be convergent. If the limits do not exist, we say the
integrals are divergent.
R∞
Rb
3. If a f (x)dx and −∞ f (x)dx are convergent, then
Z
∞
c
Z
Z
f (x)dx :=
−∞
∞
f (x)dx +
f (x)dx
−∞
c
where c is any real number.
What to do then with the second type of improper integral mentioned above? We
do what is natural and define
Definition 3.
1. If f (x) is continuous on [a, b) but discontinuous at b, then
b
Z
t
Z
f (x)dx := lim
t→b−
a
f (x)dx.
a
2. If f (x) is continuous on (a, b] but discontinuous at a, then
Z
b
Z
b
f (x)dx := lim
s→a+
a
f (x)dx.
s
Rb
f (x)dx is said to be convergent if the limit exists but is divergent otha
erwise.
3. If f (x) has a discontinuity at c where a ≤ c ≤ b, then
Z
b
c
Z
f (x)dx :=
a
Z
f (x)dx +
a
b
f (x)dx.
c
Note well that the limits in 1 and 2 are directional limits.
3
For example:
2
Z
0
1
dx := lim−
x−2
t→2
t
Z
1
dx
x−2
0
(1)
t
= lim− ln |x − 2|]0
(2)
= lim− (ln |t − 2| − ln 2)
(3)
t→2
t→2
= −∞.
But for any k > 0,
Z
k
∞
1
dx := lim
t→∞
x2
t
Z
k
1
dx
x2
t
(4)
1
x k
1 1
= lim (− + )
t→∞
t
k
1
= .
k
= lim −
(5)
t→∞
(6)
1.3. The Dilemma
Now I wish for us to consider the following for any real number a:
Z a
1
dx.
x
−a
Certainly x1 is an odd function and we are integrating over a symmetric interval, so
should we not have that
Z a
1
dx = 0?
x
−a
4
2
-2
1
-1
2
-2
-4
Figure 1: The Graph of f (x) =
4
1
x
Of course, there is the problem that x1 is not continuous at 0, so we must proceed
as follows:
Z a
Z s
Z a
1
1
1
dx := lim
dx + lim
dx
(7)
−
+
s→0
t→0
−a x
−a x
t x
s
a
= lim ln |x|]−a + lim ln |x|]t
(8)
= lim− (ln |s| − ln a) + lim+ (ln a − ln t)
(9)
s→0−
t→0+
s→0
t→0
= lim− ln |s| − lim+ ln t
(10)
= −∞ + ∞
(11)
s→0
t→0
=???.
In Calc 2, we teach the students to say the integral is divergent, but a better answer
is that the integral could be anything. More importantly we note that the matter
at hand is that we do not know the rate at which the limits are converging. If they
happen to converge at different rates, say t approaches 0 twice as fast as s, then
the above problem becomes1 :
Z s
Z a
Z a
1
1
1
dx := lim
dx + lim
dx
(12)
−
+
x
x
x
s→0
t→0
−a
t
−a
Z 0−
Z a
1
1
:= lim
dx + lim
dx
(13)
→0 −a
→0 0+2 x
x
−
a
= lim ln |x|]−a + ln |x|]2
(14)
→0
= lim (ln − ln a + ln a − ln 2)
(15)
= lim (ln − ln 2)
(16)
2
(17)
→0
→0
= lim ln
→0
= − ln 2.
If we assume that the individual limits approach 0 at the same rate, however, we
1 This
is essentially the example found in [4].
5
have
Z
a
−a
Z s
Z a
1
1
1
dx := lim−
dx + lim+
dx
x
s→0
t→0
−a x
t x
Z a
Z 0−
1
1
dx + lim
dx
:= lim
→0 0+ x
→0 −a
x
−
a
= lim ln |x|]−a + ln |x|]
→0
= lim (ln − ln a + ln a − ln )
→0
(18)
(19)
(20)
(21)
= 0,
which agrees with our intuition (though that does not make it correct). This is the
idea behind Cauchy’s Principal Value Method, which is a way to assign a value to a
divergent integral by assuming each limit converges at the same rate. The Method
works even if the integrand is not odd and the interval is not symmetric. We remark
that the notation for this is
Z
b
P.V.
f (x)dx.
a
Hence in our example,
Z
a
−a
1
dx is divergent
x
but
Z
a
P.V.
−a
1
dx = 0.
x
Cauchy’s Principal Value Method can only be acceptable if applying it to a convergent integral does not change the value of the integral (which it does not). It
should also be stated that we have presented the idea of the method for divergent
integrals over R. A similar, but more involved, technique can be used for divergent
line integrals over C.
1.4. The Problem with Our Solution
We present an example Dave Rusin of the University of Texas [5] offers to his
10+4x
students, namely finding the area under the curve f (x) = (5x+x
2 )3 over the interval
[−2, 1]. Note there are singularities at x = −5 (which is outside the desired interval)
and x = 0.
6
0.6
0.4
0.2
-6
-4
2
-2
-0.2
-0.4
-0.6
Figure 2: The Graph of f (x) =
Z
1
P.V.
−2
10+4x
(5x+x2 )3 .
Z 1
10 + 4x
10 + 4x
dx
+
dx
(22)
2
3
2 3
−2 (5x + x )
0 (5x + x )
Z −
Z 1
10 + 4x
10 + 4x
:= lim
dx
+
dx
(23)
2 3
2 3
→0
−2 (5x + x )
(5x + x )
−
1 !
1
1
= lim − 2
−
(24)
→0
x (5 + x)2 −2 x2 (5 + x)2 1
1
1
−1
+
−
+ 2
= lim 2
→0 (5 − )2
4(5 − 2)2
1(5 + 1)2
(5 + )2
(25)
2
2
−(25 + 10 + ) + (25 − 10 + )
1
1
= lim
+
−
→0
2 (5 − )2 (5 + )2
36 36
(26)
−20
= lim 2
(27)
→0 (5 − )2 (5 + )2
10 + 4x
dx =
(5x + x2 )3
Z
0
= −∞.
But if we use the u-substitution u = 5x + x2 we have du = (5 + 2x)dx and when
x = −2, u = 5(−2) + (−2)2 = −6 and when x = 1, u = 5(1) + 12 = 6. Thus
Z
1
P.V.
−2
10 + 4x
dx = P.V.
(5x + x2 )3
Z
6
−6
2
du
u3
Z 6
2
2
du
+
du
3
→0
u3
−6
0+ u
−
6 !
−1
−1
= lim
− 2
2
→0
u −6
u −1
1
1
1
+
−
+
= lim
→0
2
36 36 2
Z
:= lim
= 0.
7
(28)
0−
(29)
(30)
(31)
If we try to “fix” the situation in the original integral by letting the limit on the
left approach 0 twice as fast as the limit on the right (since the interval on the left
of the singularity is twice as long as the interval on the right), we get:
Z 1
Z 0
Z 1
10 + 4x
10 + 4x
10 + 4x
dx =
dx +
dx
(32)
2 )3
2 )3
(5x
+
x
(5x
+
x
(5x
+ x2 )3
−2
−2
0
Z −2
Z 1
10 + 4x
10 + 4x
:= lim
dx
+
dx
(33)
2 3
→0
(5x + x2 )3
−2
(5x + x )
−2
1 !
1
1
= lim − 2
− 2
(34)
→0
x (5 + x)2 −2
x (5 + x)2 −1
1
1
1
= lim
+
−
+
(35)
→0 42 (5 − 2)2
36 36 2 (5 + )2
−(5 + )2 + 4(5 − 2)2
= lim
(36)
→0
42 (5 − 2)2 (5 + )2
−25 − 10 − 2 + 100 − 80 + 162
= lim
(37)
→0
42 (5 − 2)2 (5 + )2
75 − 90 + 152
= lim 2
(38)
→0 4 (5 − 2)2 (5 + )2
= ∞.
Dr. Rusin recommends not using substitution when using Cauchy’s Principal Value
Method.
2. Giving Value to Valueless Series
Now we consider ways to assign value to another valueless object.
2.1. Valueless because of Futility
At face value, an infinite sum has no meaning. Consider
1 1 1
1
1 + + + + ··· + n + ··· .
2 4 8
2
What could it mean to perform this task? Even if I were to live forever and have
whatever calculating tool available, this is a job that never finishes (hence the “· · · ”
at the end). Even God can not preform this task2 .
2 I quote C.S. Lewis: “His Omnipotence means power to do all that is intrinsically possible, not
to do the intrinsically impossible. You may attribute miracles to Him, but not nonsense. There is
8
A reasonable thing to do would be to consider partial sums. We put S0 = 1, S1 =
1 + 21 , . . . , Sk = 1 + 12 + 14 + · · · + 21k . This leads to the values in table 2.1.
Certainly it seems that the sequence of partial sums is converging to 2. This is, in
fact, true and can be shown. Hence it seems that a perfectly natural way to give
meaning to a series is to:
Definition 4.
We define
∞
X
an = a0 + a1 + a2 + · · · := lim Sk
k→∞
n=0
where Sk := a0 + a1 + · · · + ak is the k-th partial sum. If the limit exists, the series
is said to be convergent, else the series is said to be divergent.
This is precisely what is done in standard Calc 2 courses.
2.2. Assigning Value to Divergent Series
A fun series to consider is one commonly known as Grandi’s Series:
∞
X
(−1)n = 1 − 1 + 1 − 1 + 1 − 1 + · · · .
n=0
The partial sums form the sequence
{1, 0, 1, 0, 1, 0, . . . },
which (in the classic sense) diverges.
The series is due to Guido Ubaldus, a.k.a Guido Grandi (1671-1742) who did the
following:
0 = 0 + 0 + 0 + 0 + ···
(39)
= (1 − 1) + (1 − 1) + (1 − 1) + · · ·
(40)
= 1 + (−1 + 1) + (−1 + 1) + (−1 + 1) + · · ·
(41)
= 1 + 0 + 0 + 0 + 0 + ···
(42)
= 1.
no limit to His power. If you choose to say, ’God can give a creature free will and at the same time
withhold free will from it,’ you have not succeeded in saying anything about God: meaningless
combinations of words do not suddenly acquire meaning simply because we prefix to them the
two other words, ’God can.’ It remains true that all things are possible with God: the intrinsic
impossibilities are not things but nonentities. It is no more possible for God than for the weakest
of His creatures to carry out both of two mutually exclusive alternatives; not because His power
meets an obstacle, but because nonsense remains nonsense even when we talk it about God.” [3]
9
Grandi thought he had proved the existence of God because “something had been
created out of nothing”.
Grandi, Euler, and Leibnitz each believed for various reasons that the series summed
to 12 (see [2]). Euler used the geometric series
1 + x + x2 + x3 + · · · =
1
1−x
and put x = −1, getting
1 − 1 + 1 − 1 + 1 − 1 + ··· =
1
1
= .
1 − (−1)
2
This is, of course, no good since this series only converges for |x| < 1. If we want
to play this game we could use
1 − x2
1+x
=
= 1 − x2 + x3 − x5 + x6 − x8 + · · ·
1 + x + x2
1 − x3
and put x = 1, getting
1 − 1 + 1 − 1 + ··· =
2
.
3
Hence some ground rules must be stated for assigning values to divergent series.
Konrad Knopp stated some is his classic text on Series [1]:
1. Every series convergent in the classic sense must be convergent in the new
sense and must have the same limit.
2. Any new process must produce a value for at least one divergent series.
3. Any two processes applied to the same series must produce the same result. (Knopp says “simultaneously applied” here and I am not sure what that
means.)
Knopp’s text list nine processes for assigning value to divergent series. Other techniques are referenced in [2]. We mention but two, both from [1]:
1). The H1 -process - also known as the C1 or M -process.
1 +···+Sn
We define h0n := S0 +S
, i.e. the arithmetic mean of the sequence of partial
Pn+1
sums. We then put
ai = lim h0n .
We take for an example the series 1 + 0 + 0 + 0 + · · · . This converges in the classic
sense and has a limit of 1. Using the H1 process we also get 1 + 0 + 0 + 0 + · · · = 1
(see table 2.2).
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Regarding Grandi’s series, we have that Sn = 12 [1 + (−1)n ]. The h0n values are
states in Table 2.2 These values can be shown to converge to 21 (details are in [1]).
2). The Hölder process - also known as the Hp -process.
h0 +h0 +···+h0
1
n
We define h00n := 0 n+1
, i.e. the arithmetic
P mean of00 the arithmetic means of
the sequence of partial sums. We then put
ai = lim hn . It is the case that the
Hp -process also yields a sum of 12 for Grandi’s series. Again, details are in [1].
References
[1] Theory and Application of Infinite Series, Konrad Knopp, Dover.
[2] Euler and divergent series, Victor Kowalenko, European Journal of Pure and
Applied Mathematics, Vol 4, Number 4, (2011) pages 370-423.
[3] The Problem of Pain, C.S. Lewis, HarperCollins, pg. 18.
[4] Basic Complex Analysis, third edition by Jerrold E. Marsden and Michael J.
Hoffman, W.H. Freeman and Company.
[5] Dave Rusin’s webpage, http://www.math.utexas.edu/users/rusin/408D-12b/CPV.pdf,
the Department of Mathematics, the University of Texas.
11
n
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
Sn
1
1.5
1.75
1.875
1.9375
1.96875
1.984375
1.9921875
1.99609375
1.998046875
1.999023438
1.999511719
1.999755859
1.99987793
1.999938965
1.999969482
1.999984741
1.999992371
1.999996185
1.999998093
1.999999046
1.999999523
1.999999762
1.999999881
1.99999994
1.99999997
1.999999985
1.999999993
Table 1: Partial Sums for
n
0
1
2
Sn
1
1
1
P∞
1
n=0 2n .
h0n
S0
1
1 = 1 =1
S0 +S1
1+1
1+1 = 2 = 1
S0 +S1 +S2
= 1+1+1
=
1+1+1
3
1
Table 2: H1 Process for 1 + 0 + 0 + 0 + · · · .
12
n
0
1
2
3
4
5
h0n
S0
1
1 = 1 =1
1
1+0
1+1 = 2
1+0+1
2
1+1+1 = 3
1+0+1+0
2
1
1+1+1+1 = 4 = 2
1+0+1+0+1
3
1+1+1+1+1 = 5
1+0+1+0+1+0
3
1
1+1+1+1+1+1 = 6 = 2
Table 3: H1 Process for Grandi’s Series.
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