6.4 Contour integrals

We discuss next what appears to be a simpler notion of integral over a curve.In this one, we really do regard the curve $C$ 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 theparameterizations of the curve as complex-valued functions on an interval $[a,b].$ Also, in this definition, a curve $C$ from $z_1$ to $z_2$ will be distinguished from its reverse, i.e., the same set $C$ thought of as a curve from $z_2$ to $z_1.$

Let $C$ be a piecewise smooth curve from $z_1$ to $z_2$ in the plane $C,$ parameterized by a (complex-valued) function $\phi :[a,b]\to C.$ If $f$ is a continuous, complex-valued function on $C,$ The contour integral of f from $z_1$ to $z_2$ along $C$ will be denoted by ${\int }_{C}f\left(\zeta \right)d\zeta$ or more precisely by ${{\int }_C}_{z_1}^{z_2}f\left(\zeta \right)d\zeta ,$ and is defindd by

$${{\int }_C}_{z_1}^{z_2}f\left(\zeta \right)d\zeta ={\int }_a^{b}f\left(\phi \left(t\right)\right){\phi }^{\text{'}}\left(t\right)dt.$$

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 [link] and part (e) of [link] . This is an example of how mathematicians often usea 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 ${\phi }^{\text{'}}\left(t\right).$ This will make contour integrals more subtle than integrals with respect to arc length, just as conditionally convergent infinite seriesare more subtle than absolutely convergent ones.

Note also that there is no question about the integrability of $f\left(\phi \left(t\right)\right){\phi }^{\text{'}}\left(t\right),$ because of [link] . $f$ is bounded, ${\phi }^{\text{'}}$ is improperly-integrable on $(a,b),$ and therefore so is their product.

Not all the usual properties hold for contour integrals, e.g., like those in [link] above. The functions here, and the values of their contour integrals, arecomplex numbers, so all the properties of integrals having to do with positivity and inequalities, except for the one in part (c) of [link] , 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 [link] .

Let $C$ be a piecewise smooth curve of finite length joining $z_1$ to $z_2.$ Then the contour integrals of continuous functions on $C$ have the following properties.

1. If $f$ and $g$ are any two continuous functions on $C,$ and $a$ and $b$ are any two complex numbers, then
   $${\int }_C(af\left(\zeta \right)+bg\left(\zeta \right))d\zeta =a{\int }_{C}f\left(\zeta \right)d\zeta +b{\int }_{C}g\left(\zeta \right)d\zeta .$$
2. If $f$ is the uniform limit on $C$ of a sequence $\left\{f_n\right\}$ of continuous functions, then ${\int }_{C}f\left(\zeta \right)d\zeta =lim{\int }_C{f}_n\left(\zeta \right)d\zeta .$
3. Let $\left\{u_n\right\}$ be a sequence of continuous functions on $C,$ and suppose that for each $n$ there is a number $m_n,$ for which $|u_n\left(z\right)|\le m_n$ for all $z\in C,$ and such that the infinite series $\sum m_n$ converges. Then the infinite series $\sum u_n$ converges uniformly to a continuous function, and ${\int }_C\sum u_n\left(\zeta \right)d\zeta =\sum {\int }_C{u}_n\left(\zeta \right)d\zeta .$

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.