6.4  The Fundamental Theorems of Integration

    Let f be analytic in the simply connected domain D.  The theorems in this section show that an antiderivative F can be constructed by contour integration.  A consequence will be the fact that in a simply connected domain, the integral of an analytic function f along any contour joining    is the same, and its value is given by  .  As a result, we can use the antiderivative formulas from calculus to compute the value of definite integrals.  The next two theorems are generalizations of the Fundamental Theorems of Calculus.  

Theorem 6.8 (Indefinite Integrals or Antiderivatives).  Let f(z) be analytic in the simply connected domain D.  If    is a fixed value in D and if C is any contour in D with initial point    and terminal point z, then the function

              

is well-defined and analytic in D, with its derivative given by  .   

Proof.

Proof of Theorem 6.8 is in the book.
Complex Analysis for Mathematics and Engineering

Remark 6.2.  It is important to stress that the line integral of an analytic function is independent of path.  In Example 6.9 we showed that  ,  where and were different contours joining  .  Because the integrand   is an analytic function, Theorem 6.8 lets us know ahead of time that the value of the two integrals is the same; hence one calculation would have sufficed.  If you ever have to compute a line integral of an analytic function over a difficult contour, change the contour to something easier.  You are guaranteed to get the same answer.  Of course, you must be sure that the function you're dealing with is analytic in a simply connected domain containing your original and new contours.

Exploration.

    If we set in Theorem 6.8, then we obtain the following familiar result for evaluating a definite integral of an analytic function.

Theorem 6.9 (Definite Integrals).  Let f(z)  be analytic in a simply connected domain D.  If    and    are two points in D joined by a contour C, then

            ,  

where F(z) is any antiderivative of f(z) in D.

Proof.

Proof of Theorem 6.9 is in the book.
Complex Analysis for Mathematics and Engineering

    Theorem 6.9 gives an important method for evaluating definite integrals when the integrand is an analytic function in a simply connected domain.  In essence, it permits you to use all the rules of integration that you learned in calculus.  When the conditions of Theorem 6.9 are met, applying it is generally much easier than parametrizing a contour.

Example 6.17.  Show that   where is the principal branch of the square root function and C is the line segment joining  .  

Remark. Sometimes we write this as   .  

Solution.  We showed in Example 3.10 (see Section 3.2) that if  ,  then  ,  where the principal branch of the square root function is used in both the formulas for F(z) and F'(z).  We note that C is contained in the simply connected domain , which is the open disk of radius 4 centered at the midpoint of the segment C.  Since    is analytic in the domain and is an anti-derivative of , Theorem 6.9 guarantees that  

              

Explore Solution 6.17.

Example 6.18.  Show that  ,  where C is the line segment between .  

Solution.  An antiderivative of    is  .  Because F(z) is entire, we use Theorem 6.9 to conclude that

              

Explore Solution 6.18.

Example 6.19. We let be the simply connected domain which is the z-plane slit along the negative x-axis, shown in Figure 6.32.  We know that is analytic in D, and has an antiderivative for all z in D.  If C is a contour in D that joins the point to the point , then Theorem 6.9 implies that  

            .  

Figure 6.32  The simply connected domain D shown in Examples 6.19 and 6.20.

Example 6.20. Show that  ,  where C is the unit circle  ,  taken with positive orientation.

Solution.  We let C be that circle with the point -1 omitted, as shown in Figure 6.32(b).  The contour C is contained in the simply connected domain D of Example 6.19.  We know that    is analytic in D, and has an antiderivative  ,  for all  .  Therefore, if we let approach -1 on C through the upper half-plane and approach -1 on C through the lower half-plane,  

              

Explore Solution 6.20.

Extra Example 1.  Show that   .  

Explore Solution for Extra Example 1.

Exercises for Section 6.4.  The Fundamental Theorems of Integration