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Contour Integrals

Module by: Lawrence Baggett. E-mail the author

Summary: We discuss next what appears to be a simpler notion of integral over a curve. In this one, we really do regard the curve CC as a subset of the complex plane as opposed to two-dimensional real space; we will be integrating complex-valued functions; and we explicitly think of the parameterizations of the curve as complex-valued functions on an interval [a,b].[a,b]. Also, in this definition, a curve CC from z1z1 to z2z2 will be distinguished from its reverse, i.e., the same set CC thought of as a curve from z2z2 to z1.z1.

We discuss next what appears to be a simpler notion of integral over a curve. In this one, we really do regard the curve CC as a subset of the complex plane as opposed to two-dimensional real space; we will be integrating complex-valued functions; and we explicitly think of the parameterizations of the curve as complex-valued functions on an interval [a,b].[a,b]. Also, in this definition, a curve CC from z1z1 to z2z2 will be distinguished from its reverse, i.e., the same set CC thought of as a curve from z2z2 to z1.z1.

Definition 1:

Let CC be a piecewise smooth curve from z1z1 to z2z2 in the plane C,C, parameterized by a (complex-valued) function φ:[a,b]C.φ:[a,b]C. If ff is a continuous, complex-valued function on C,C, The contour integral of f from z1z1 to z2z2 along CC will be denoted by Cf(ζ)dζCf(ζ)dζ or more precisely by Cz1z2f(ζ)dζ,Cz1z2f(ζ)dζ, and is defindd by

C z 1 z 2 f ( ζ ) d ζ = a b f ( φ ( t ) ) φ ' ( t ) d t . C z 1 z 2 f ( ζ ) d ζ = a b f ( φ ( t ) ) φ ' ( t ) d t .
(1)

REMARK There is, as usual, the question about whether this definition depends on the parameterization. Again, it does not. See the next exercise.

The definition of a contour integral looks very like a change of variables formula for integrals. See (Reference) and part (e) of (Reference). This is an example of how mathematicians often use a true formula from one context to make a new definition in another context.

Notice that the only difference between the computation of a contour integral and an integral with respect to arc length on the curve is the absence of the absolute value bars around the factor φ'(t).φ'(t). This will make contour integrals more subtle than integrals with respect to arc length, just as conditionally convergent infinite series are more subtle than absolutely convergent ones.

Note also that there is no question about the integrability of f(φ(t))φ'(t),f(φ(t))φ'(t), because of (Reference). ff is bounded, φ'φ' is improperly-integrable on (a,b),(a,b), and therefore so is their product.

Exercise 1

  1. State and prove the “independence of parameterization” result for contour integrals.
  2. Prove that
    Cz1z2f(ζ)dζ=-Cz2z1f(ζ)dζ.Cz1z2f(ζ)dζ=-Cz2z1f(ζ)dζ.
    (2)
    Just remember how to parameterize the curve in the opposite direction.
  3. Establish the following relation between the absolute value of a contour integral and a corresponding integral with respect to arc length.
    |Cf(ζ)dζ|C|f(s)|ds.|Cf(ζ)dζ|C|f(s)|ds.
    (3)

Not all the usual properties hold for contour integrals, e.g., like those in (Reference) above. The functions here, and the values of their contour integrals, are complex numbers, so all the properties of integrals having to do with positivity and inequalities, except for the one in part (c) of Exercise 1, no longer make any sense. However, we do have the following results for contour integrals, the verification of which is just as it was for (Reference).

Theorem 1

Let CC be a piecewise smooth curve of finite length joining z1z1 to z2.z2. Then the contour integrals of continuous functions on CC have the following properties.

  1. If ff and gg are any two continuous functions on C,C, and aa and bb are any two complex numbers, then
    C(af(ζ)+bg(ζ))dζ=aCf(ζ)dζ+bCg(ζ)dζ.C(af(ζ)+bg(ζ))dζ=aCf(ζ)dζ+bCg(ζ)dζ.
    (4)
  2. If ff is the uniform limit on CC of a sequence {fn}{fn} of continuous functions, then Cf(ζ)dζ=limCfn(ζ)dζ.Cf(ζ)dζ=limCfn(ζ)dζ.
  3. Let {un}{un} be a sequence of continuous functions on C,C, and suppose that for each nn there is a number mn,mn, for which |un(z)|mn|un(z)|mn for all zC,zC, and such that the infinite series mnmn converges. Then the infinite series unun converges uniformly to a continuous function, and Cun(ζ)dζ=Cun(ζ)dζ.Cun(ζ)dζ=Cun(ζ)dζ.

In the next exercise, we give some important contour integrals, which will be referred to several times in the sequel. Make sure you understand them.

Exercise 2

Let cc be a point in the complex plane, and let rr be a positive number. Let CC be the curve parameterized by φ:[-π,π-ϵ]:Cφ:[-π,π-ϵ]:C defined by φ(t)=c+reit=c+rcos(t)+irsin(t).φ(t)=c+reit=c+rcos(t)+irsin(t). For each integer nZ,nZ, define fn(z)=(z-c)n.fn(z)=(z-c)n.

  1. What two points z1z1 and z2z2 does CC join, and what happens to z2z2 as ϵϵ approaches 0?
  2. Compute Cfn(ζ)dζCfn(ζ)dζ for all integers n,n, positive and negative.
  3. What happens to the integrals computed in part (b) when ϵϵ approaches 0?
  4. Set ϵ=π,ϵ=π, and compute Cfn(ζ)dζCfn(ζ)dζ for all integers n.n.
  5. Again, set ϵ=π.ϵ=π. Evaluate
    Ccos(ζ-c)ζ-cdζandCsin(ζ-c)ζ-cdζ.Ccos(ζ-c)ζ-cdζandCsin(ζ-c)ζ-cdζ.
    (5)
    HINT: Make use of the infinite series representations of the trigonometric functions.

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