6.2  Contours and Contour Integrals

    In Section 6.1 we learned how to evaluate integrals of the form  ,  where f(t) was complex-valued and    was an interval on the real axis (so that t was real, with  ).  In this section, we define and evaluate integrals of the form  ,  where f(t) is complex-valued and C is a contour in the plane (so that z is complex, with  ).  Our main result is Theorem 6.1, which shows how to transform the latter type of integral into the kind we investigated in Section 6.1 .

    We will use concepts first introduced in Section 1.6.  Recall that to represent a curve C we used the parametric notation

(6-10)              for ,

where x(t) and y(t) are continuous functions.  We now place a few more restrictions on the type of curve to be described. The following discussion leads to the concept of a contour, which is a type of curve that is adequate for the study of integration.

    Recall that C is simple if it does not cross itself, which means that    whenever  , except possibly when and .  A curve C with the property that   is a closed curve.  If     is the only point of intersection, then we say that C is a simple closed curve.  As the parameter t increases from the value to the value , the point starts at the initial point , moves along the curve C , and ends up at the terminal point . If C is simple, then moves continuously from to as t increases and the curve is given an orientation, which we indicate by drawing arrows along the curve.  Figure 6.1 illustrates how the terms simple and closed describe a curve.

 

Figure 6.1  The terms simple and closed used to describe curves.

    The complex-valued function    is said to be differentiable on if both and are differentiable for .  Here we require the one-sided derivatives of and to exist at the endpoints of the interval.  As in Section 6.1, the derivative is

               for   .  

    The curve C defined by Equation (6-10) is said to be a smooth curve if is continuous and nonzero on the interval.  If C is a smooth curve, then C has a nonzero tangent vector at each point , which is given by the vector .  If  ,  then the tangent vector    is vertical.  If  ,  then the slope of the tangent line to C at the point is given by .  Hence for a smooth curve the angle of inclination of its tangent vector is defined for all values of and is continuous.  Thus a smooth curve has no corners or cusps.  Figure 6.2 illustrates this concept.

 

Figure 6.2  The term smooth used to describe curves.

    If C is a smooth curve, then , the differential of arc length, is given by  

            .  

    The function    is continuous, as and are continuous functions, so the length of the curve C is  

(6-11)            .  

    Now consider C to be a curve with parameterization  

               for   .  

    The opposite curve   traces out the same set of points in the plane, but in the reverse order, and has the parametrization  

               for   .  

Since  ,    is merely C traversed in the opposite sense, as illustrated in Figure 6.3.  

 

Figure 6.3  The curve    and its opposite curve  .

    A curve C that is constructed by joining finitely many smooth curves end to end is called a contour.  Let    denote n smooth curves such that the terminal point of the curve    coincides with the initial point of    for  .  We express the contour C by the equation

            .

A synonym for contour is path.  

Example 6.5.  Find a parameterization of the polygonal path    from    shown in Figure 6.4.
Here    is the line from  ,    is the line from  ,  and    is the line from  .  

Figure 6.4  The polygonal path from  .  

Solution.  We express C as three smooth curves, or  .  If we set    and  ,  we can use Equation (1-48) to get a formula for the straight-line segment joining two points:  

            ,  

for .  When simplified, this formula becomes  

            ,   for   .  

Similarly, the segments are given by  

            ,   for   ,   and   

            ,   for   .  

Explore Solution 6.5.

    We are now ready to define the integral of a complex function along a contour C in the plane with initial point A and terminal point B.  Our approach is to mimic what is done in calculus.  We create a partition    of points that proceed along C from A to B and form the differences    for  .  Between each pair of partition points    we select a point    on C, as shown in Figure 6.5, and evaluate the function  .  
These values are used to make a Riemann Sum for the partition:  

(6-12)            .

Figure 6.5  Partition points and function evaluation points
                    for a Riemann sum along the contour C from .

    Assume now that there exists a unique complex number L that is the limit of every sequence of Riemann sums given in Equation (6-12), where the maximum of tends toward 0 for the sequence of partitions. We define the number L as the value of the integral of the function f(z) taken along the contour C.

Definition 6.2 (Complex Integral).  Let C be a contour.  Then  

            ,  

provided that the limit exists in the sense previously discussed.

    Note that in Definition 6.2, the value of the integral depends on the contour.  In Section 6.3 the Cauchy-Goursat theorem will establish the remarkable fact that, if f(z) is analytic, then is independent of the contour.

Example 6.6.  Use a Riemann sum to get an approximation for the integral    where C is a the line segment joining the point  .  

Solution.  Set n=8 in Equation (6-12) and form the partition  .  For this situation, we have a uniform increment  .  
For convenience we select    for  .  Figure 6.6 shows the points and .  

Figure 6.6  Partition and evaluation points for the Riemann sum .

One possible Riemann sum, then, is  

              

By rounding the terms in this Riemann sum to five decimal digits, we obtain an approximation for the integral:  

            

This result compares favorably with the precise value of the integral, which you will soon see equals  

            .  

Explore Solution 6.6.

    In general, obtaining an exact value for an integral given by Definition 6.2 is a daunting task. Fortunately, there is a beautiful theory that allows for an easy computation of many contour integrals.  Suppose that we have a parametrization of the contour C given by the function for .  That is, C is the range of the function over the interval , as Figure 6.7 shows.

Figure 6.7  A parametrization of the contour C by for .

    It follows that  

               

where and are the points contained in the interval with the property that and , as is also shown in Figure 6.7.  If for all k we multiply the kth term in the last sum by , then we get  

               

    The quotient inside the last summation looks suspiciously like a derivative, and the entire quantity looks like a Riemann sum. Assuming no difficulties, this last expression should equal  ,  as defined in Section 6.1.  Of course, if we're to have any hope of this happening, we would have to get the same limit regardless of how we parametrize the contour C.  As Theorem 6.1 states, this is indeed the case.

Theorem 6.1.  Suppose that is a continuous complex-valued function defined on a set containing the contour C.  Let be any parametrization of C for  .  Then  

            .  

Proof.

    Two important facets of Theorem 6.1 are worth mentioning.  First, Theorem 6.1 makes the problem of evaluating complex-valued functions along contours easy, as it reduces the task to the evaluation of complex - valued functions over real intervals - a procedure that you studied in Section 6.1.   Second, according to Theorem 6.1, this transformation yields the same answer regardless of the parametrization we choose for C.

Example 6.7.  Give an exact calculation of the integral in Example 6.6:    where C is a the line segment joining the point  .  

Solution.  We must compute  ,  where C is the line segment joining  .  According to Equation (1-48), we can parametrize C by  ,  for  .  As  ,  Theorem 6.1 guarantees that  

               

Each integral in the last expression can be done using integration by parts. (There is a simpler way-see Remark 6.1.)  We leave as an exercise to show that the final answer simplifies to  ,  as we claimed in Example 6.6.

Explore Solution 6.7.

Example 6.8.  Evaluate the contour integral    where C is a the upper semicircle with radius 1 centered at  .  

Solution.  The function  ,  for    is a parametrization for C.  We apply Theorem 6.1 with  .  (Note: ), and  .)  Hence  

            .

Explore Solution 6.8.

    To help convince yourself that the value of the integral is independent of the parametrization chosen for the given contour, try working through Example 6.8 with  ,  for  .  

    A convenient bookkeeping device can help you remember how to apply Theorem 6.1.  

Because  ,  you can symbolically equate z with z(t) and with  .  

These identities should be easy to remember because z is supposed to be a point on the contour C parametrized by z(t), and , according to the Leibniz notation for the derivative.

    If  ,  then by the preceding paragraph we have

(6-13)              

where are the differentials for , respectively (i.e., is equated with , etc.).  The expression is often called the complex differential of z. Just as are intuitively considered to be small segments along the x and y axes in real variables, we can think of dz as representing a tiny piece of the contour  C.  Moreover, if we write  

            

we can put Equation (6-11) into the form  

(6-14)            .   

so we can think of as representing the length of .

    Suppose    , and    is a parametrization for the contour C.  Then

(6-15)          

where we are equating u(t) with u(z(t)), x' with x'(t), and so on.

    If we use the differentials given in Equation (6-13), then we can write Equation (6-15) in terms of line integrals of the real-valued functions u(t) and v(t) , giving

(6-16)            ,

which is easy to remember if we recall that symbolically

            .  

    We emphasize that Equation (6-16) is merely a notational device for applying Theorem 6.1. You should carefully apply Theorem 6.1as illustrated in Examples 6.7 and 6.8 before using any shortcuts suggested by the latter.

Example  6.9.  Show that  ,   
where   is the line segment from  ,  and is the portion of the parabola    joining  ,  as indicated in Figure 6.8.  

Figure 6.8  The two contours and joining  .

Solution.  The line segment joining is given by the slope intercept formula  ,  which can be written as  .  
If we choose the parametrization    and  ,  we can write segment as

               and   

for  .

Along we have  .  Applying Theorem 6.1 gives  

            .  

We now multiply out the integrand and put it into its real and imaginary parts:  

               

Similarly, we can parametrize the portion of the parabola    joining by    and and   so that  

               and   

for  .

Along we have  .  Theorem 6.1 now gives  

              

Explore Solution 6.9 (a)

Explore Solution 6.9 (b)

Extra Example 1.  Evaluate the contour integrals  of    starting at the points   .  
(a)  Use the line segment joining the points.  
(b)  Use a portion of a parabola joining the points.
Remark. The example in the text used the function   which is difficult for hand computations but it is not a challenge for Mathematica, hence we choose to use the function .  

Explore Solution for Extra Example 1 (a)

Explore Solution for Extra Example 1 (b)

Extra Example 2.  Evaluate the contour integrals  of    starting at the points   .  
(a)  Use the line segment joining the points.  
(b)  Use a portion of a parabola joining the points.
Remark.  This example illustrates the situation when f(z) is not analytic.

Explore Solution for Extra Example 2 (a)

Explore Solution for Extra Example 2 (b)

    In Example 6.9, the value of the two integrals is the same.  This outcome doesn't hold in general, as Example 6.10 shows.

Example 6.10.  (a)   Show that      but that   ,  

where is the semicircular path from , in the upper half plane, and is the polygonal path from , respectively, shown in Figure 6.9.  

Figure 6.9  The two contours and joining  .  

Solution.  We parametrize the semicircle as

               and   ,  

for  .  

Applying Theorem 6.1, we have  ,  so  

                
            
            and
            
              

We parametrize in three parts, one for each line segment:

            
where    in each case.  Integrating over these three line segments we obtain  

            ,

            ,

            .

We get our answer by adding the three integrals along the three segments:  

              

Note that the value of the contour integral along isn't the same as the value of the contour integral along , although both integrals have the same initial and terminal points.

Explore Solution 6.10 (a)

Explore Solution 6.10 (b)

     Contour integrals have properties that are similar to those of integrals of a complex function of a real variable, which you studied in Section 6.1.  If C is given by Equation (6-10), then the integral for the opposite contour -C is  

            

    Using the change of variable    in this last equation and the property that  ,  we obtain  
    
(6-17)            .  

    If two functions f and g can be integrated over the same path of integration C, then their sum can be integrated over C, and we have the familiar result  

            .  

    Constant multiples also behave as we would expect:

            .  

    If two contours and are placed end to end so that the terminal point of coincides with the initial point of , then the contour is a continuation of , and  

(6-18)            .

    If the contour C has two parametrizations  

               for   ,   and

               for   ,   

and there exists a differentiable function    such that  

(6-19)            ,    ,    and        for   ,

then we say that is a reparametrization of the contour C.  If f is continuous on C, then we have  

(6-20)            .  

    Equation (6-20) shows that the value of a contour integral is invariant under a change in the parametric representation of its contour if the reparametrization satisfies Equations (6-19).

    We now give two important inequalities relating to complex integrals.

Theorem 6.2 (Absolute Value Inequality).  If   is a continuous function of the real parameter t, then  

(6-21)            .  

Proof.

Theorem 6.3 (ML - Inequality).  If   is continuous on the contour C, then  

(6-23)            .  

Proof.

Example 6.11.  Use Inequality (6-23) to show that  ,
where C is the straight-line segment from  .  

Solution.  Here  ,  and the terms    represent the distance from the point z to the points , respectively.  Referring to Figure 6.10 and using a geometric argument, we get

            ,   for z on C.

Figure 6.10  The distances   for z on C.

Thus we have  

            .

Because L, the length of C, equals 1, Inequality (6-23) implies that  

            .

Exercises for Section 6.2.  Contours and Contour Integrals