COMPLEX ANALYSIS: Maple Worksheets, 2009
(c) John H. Mathews Russell W. Howell
mathews@fullerton.edu howell@westmont.edu

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COMPLEX ANALYSIS: for Mathematics & Engineering, 4th Ed, 2001, ISBN: 0-7637-1425-9
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CHAPTER 6 COMPLEX INTEGRATION

Section 6.1 Complex Integrals

In Chapter 3 we defined the derivative of a complex function. We now turn our attention to the problem of integrating complex functions. We will find that integrals of analytic functions are well behaved and that many properties from Calculus carry over to the complex case. To introduce the integral of a complex function, we start by defining what we mean by the integral of a complex-valued function of a real variable.

Definition 6.1: Integral of f(t)

Let , where and are real-valued functions of the real variable for a t b, then

.

We generally evaluate integrals of this type by finding the antiderivatives of and and evaluating the definite integrals on the right side of the equation. That is, if and for a t b, we have

.

Example 6.1, Page 202. Show that .

> f:='f': t:='t': u:='u': v:='v':
f := t -> (t+I)^3:
`f(t) ` = f(t);
`f(t) ` = evalc(f(t));
u := u -> t^3 - 3*t:
v := v -> - 3*t^2 + 1:
`u(t) ` = u(t);
`v(t) ` = v(t);
Int(u(t)+I*v(t),t) = int(u(t),t) + I*int(v(t),t);
defint := int(u(t),t=0..1) + I*int(v(t),t=0..1):
Int(f(t),t=0..1) = defint;

Or we could do the integral directly, with complex function for the integrand.

> f:='f': g:='g': t:='t': T:='T':
f := t -> (t+I)^3:
g := t -> subs(T=t, int(f(T),T)):
`f(t) ` = f(t);
`g(t) = `,Int(f(t),t) = g(t);
`g(1) ` = g(1),` and `,
`g(0) ` = g(0);
`g(1) - g(0) ` = g(1) - g(0);
defint := int(f(t), t=0..1):
Int(f(t),t=0..1) = defint;

Example 6.2, Page 202. Show that .

> f:='f': t:='t': u:='u': v:='v':
f := t -> exp(t+I*t):
`f(t) ` = f(t);
`f(t) ` = evalc(f(t));
u := u -> exp(t)*cos(t):
v := v -> exp(t)*sin(t):
`u(t) ` = u(t);
`v(t) ` = v(t);
Int(u(t)+I*v(t),t) = int(u(t),t) + I*int(v(t),t);
defint := int(u(t),t=0..Pi/2) + I*int(v(t),t=0..Pi/2):
Int(f(t),t=0..pi/2) = defint;

Or we could do the integral directly, with complex function for the integrand.

> f:='f': g:='g': t:='t': T:='T':
f := t -> exp(t+I*t):
g := t -> subs(T=t, int(f(T),T)):
`f(t) ` = f(t);
`g(t) = `,Int(f(t),t) = g(t);
`g(Pi/2) ` = g(Pi/2),` and `,
`g(0) ` = g(0);
`g(Pi/2) - g(0) ` = g(Pi/2) - g(0);
defint := int(f(t), t=0..Pi/2):
Int(f(t),t=0..pi/2) = defint;
Int(f(t),t=0..pi/2) = evalc(defint);

Example 6.3, Page 203. Show that
.

> c:='c': d:='d': S:='S': t:='t': u:='u': v:='v':
s := int((c + I*d)*(u(t) + I* v(t)), t): S = s;
s := expand(s): S = s;
s := evalc(expand(s)): S = s;

Example 6.4, Page 204. Show that .

> f:='f': g:='g': t:='t': T:='T':
f := t -> exp(I*t):
g := t -> subs(T=t, int(f(T),T)):
`f(t) ` = f(t);
`g(t) = `,Int(f(t),t) = g(t);
`g(Pi) ` = g(Pi),` and `,
`g(0) ` = g(0);
`g(Pi) - g(0) ` = g(Pi) - g(0);
defint := int(f(t), t=0..Pi):
Int(f(t),t=0..pi) = defint;

End of Section 6.1.